B.

Tech , Sem- II, Mathematics -II (BT -202 )

CLASS NOTES

Module 1: Ordinary Differential Equations-I

Contents
Differential Equations of First Order and First Degree (Leibnitz linear, Bernoulli’s, Exact),
Differential Equations of First Order and Higher Degree, Higher order differential equations with
constants coefficients, Homogeneous Linear Differential equations, Simultaneous Differential
Equations.

Introduction: Differential equations arise from many problems in oscillations of mechanical and
electrical systems, bending of beams, conduction of heat, velocity of chemical reactions
etc.,and as such play a very important role in all modern scientific and engineering studies.

Definitions:
1. Ordinary Differential Equation: An ordinary differential equation is that in which all the
differential coefficients have reference to a single independent variable.

dy
For example: =x+ 1
dx

2. Order and Degree of Ordinary Differential Equation:

The order of a differential equation is the order of highest derivative appearing in it.
The degree of a differential equation is power of highest derivative, while the highest derivative
free from radicals and fractions.

dy 2
For example: 1. =x +1 it is first order and first degree.
dx

2
d y dy
2. 2
− =0It is second order and first degree.
d x dx

In general an ordinary differential equation of nth order is denoted by

Part-I: Solution of First order and First degree Ordinary Differential Equation

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Prerequisite Knowledge
[ I] Variable Separable Form :-

dy f 1 ( x )
=
dx f 2 ( y )
An differential Equation is of the type --------- (1)

can be written as f 2 ( y ) dy =f 1 ( x ) dx
Integrate both sides

∫ f 2 ( y ) dy = ∫ f 1 ( x ) dx +C
F 2 ( y ) =F1 ( x ) dx + C
i.e.
C
Where is the constant of integration .
Which is the required general solution of given equation (1).

dy
x3 −3 y 2=xy 2
Example-1.Solve the differential Equation
dx
Solution: Given equation can be written as

3 dy 2 2
x =xy +3 y
dx

dy
x3 = y 2 ( x+ 3 )
dx

Or
dy
y 2
=
1
x 2( 3
+ 3
x ) dx
It is in variable separable form
Therefore integrate both sides

1 1 3
− + =C
x y 2 x2

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

dy
−x tan ( y−x ) = 1
Example-2. Solve the differential Equation dx

Solution: Put y−x =t , therefore differentiate w.r.t. x

dy dt dy dt
−1= ∴ =1+
dx dx dx dx
Hence the given equation becomes

dt
1+ −x tan t =1
dx
or cot t dt = xdx
Integrate both sides
x2
logsin t = + log c
2
sin t x2 / 2
or =e
c
2
x /2
or sin ( y −x )=c e
[II ] Homogeneous Form :-

A function, in which each term is of same degree is known as a homogenous function.

dy f 1 ( x , y )
=
differential Equation is of the type
dx f 2( x , y )
A --------- (1)
is called a homogeneous differential Equation of order 1 and degree 1, if
f 1 ( x , y ) and f 2 ( x , y ) be two different homogeneous functions of x and y .

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

To find the general solution ,put

y=vx where v =f ( x )
dy dv
∴ =v +x
dx dx

Hence the given equation is reduces to variable separable form and we solve them accordingly
y
v=
and we put x in the last step.

dy
x = y +√ x 2+ y 2
Example . Solve the differential equation dx
Solution: Given equation can be written as

dy y+ √ x 2 + y 2
=
dx x
It is a homogeneous differential Equation of order 1 and degree 1.

To solve put,

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

y=vx where v=f ( x )

dy dv
∴ =v+x
dx dx
Hence the , given equation becomes
dv vx+x √1+v 2
v+x =
dx x
dv dx
=
√ 1+v 2 x

It is in var iable separable form

Integrate both sides
Log [ v+ √1+v 2 ]=log x+log c
or v+ √ 1+v 2 =cx

√
2
y y
+ 1+ 2 = c x
x x
y+ √ x + y =cx
2 2 2

Type-I : Linear differential equation of order 1 and degree 1 :-

A differential equation is said to be Linear if the dependent variable and it’s derivatives appear
only in the first degree and are not multiplied together.

dy
The equation of the type + Py=Q ------- (1) is called a linear differential equation in y
dx

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Where P∧Q be some functions of x .

Solution: Let the solution be y=uv --------- (2) where u∧v be some functions of x.
dy dv du
∴ =u +v
dx dx dx
∴ Equation (1) becomes

dv du
u +v + P uv=Q
dx dx

du dv
[ + Pu¿ v+ u =Q ----------- (3)
dx dx
In (2), one of the two functions u∧v is arbitrary,
du
∴ Let we choose u in such a way that, + Pu=0
dx

du
∴ =−Pdx
u
It is in variable separable form
Integrate both sides
logu=−∫ Pdx
∴ u=e ∫
− Pdx

Hence the equation (3) becomes

dv dv Q
u =Q∴ =
dx dx u
dv
=Q e∫
Pdx
dx
dv =Q e∫ dx
Pdx

It is in variable separable form

Integrate both sides
v=∫ Q e∫ dx +c
Pdx

Hence the required solution is

y=e ∫ [∫ Q e∫ dx +c ]
− Pdx Pdx

Or y e∫ =∫ Q e∫ dx + c
Pdx Pdx

The expression e∫ Pdx is called the integrating factor [I.F.].

2 dy
Example-1: Solvecos x + y =tanx
dx

dy 2 2
Solution: Given equation can be written as + Sec x y=Sec x tanx
dx
dy
It is of the type + Py=Q i.e. it is linear in y
dx

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

∴ P=sec x∧Q=Sec x tanx , I.F. =e∫ =e∫

2
2 2 Pdx Sec x dx=¿ etanx ¿

Hence the general solution is

y e∫ =∫ Q e∫ dx + c
Pdx Pdx

y e =∫ Sec x tanx e dx+ c

tanx 2 tanx

Put tanx=t
2
∴ Sec x dx=dt
Hence the above equation becomes
y e =∫ t e dt +c
tanx t

tanx t t
ye =t e −e +c
tanx tanx
Or y e =e ( tanx−1 ) +c .
3
Example-2: Solve ydx −xdy +3 x2 y 2 e x dx=0

Solution: Given equation can be written as

( y+3 x 2 y 2 e x ) dx=xdy
3

⇒
dy 3

⇒ x = y+ 3 x 2 y 2 e x
dx
dy y 3

⇒ − =3 x y 2 e x
dx x
1 dy 1 3

⇒ 2
− =3 x e x
y dx yx

1 1 dy dz
Put =z ⇒− 2 = , then
y y dx dx

dz z x 3

⇒ + =−3 x e
dx x

This is a linear differential equation¿ z

∫ 1x dx
I.F.=e =e log x =x . Hence solution is

z ⋅ IF =∫ (−3 x e ) ⋅ IF dx +C
3
x

⇒ zx =−∫ 3 x 2 e x dx+C=−e x + C
3 3

x x 3

⇒ =−e +C
y

Example-3: Solve y ( log y ) dx + ( x−log y ) dy =0

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Solution: Given equation can be written as

dx
y ( log y ) + x−log y=0
dy
dx x 1
+ =
dy y ( log y ) y

This is a linear differential equation in x

1 1
∫ y(log dy ∫ t dt
=t=log y . Hence solution is
y) logt
I.F.=e , Put log y=t , then e =e

1 log y
x ⋅ IF =∫ ⋅IFdy + c ⇒ x ⋅log y=∫ dy + c Putlog y=t
y y
2
t2 ( log y )
⇒ x ⋅log y=∫ tdt +c = + c= +c
2 2

Type-II : Non-Linear differential equation of order 1 and degree 1 :-

dy n
The equation of the type + Py=Q y ------- (1) is called a Non- linear differential equation in
dx
y or Bernoulli’s Equation or Leibnitz’s linear equation. Where P∧Q be some functions of x and
n>1 .

Solution: To solve, we divide equation (1) by y n on both sides.

−n dy 1−n
y + P y =Q -------------- (2)
dx

1−n
Put y =t
dy dt−n
∴ ( 1−n ) y =
dx dx
−n dy 1 dt
⇒y =
dx 1−n dx

Hence, the equation (2) becomes.

1 dt
+ Pt=Q
1−n dx

dt
⟹ + ( 1−n ) Pt =( 1−n ) Q
dx

It is linear in t

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

∴I.F. =e∫ ( 1−n ) Pdx

Hence the general solution is

t . I . F .=∫ I . F . ( 1−n ) Q dx +C
Or
y . I . F .=∫ I . F . ( 1−n ) Q dx +C
1−n

dy 3 6
Example-1: Solve x + y= x y ---- (1)
dx

Solution: To solve, we divide equation (1) by x y 6 on both sides.

−5
−6 dy 1 2
∴y + y =x ------- (2)
dx x
−5
y =t

−6 dy dt
∴−5 y =
dx dx

Hence, the equation (2) becomes

−1 dt t 2
+ =x
5 dx x

dt t 2
¿ −5 =−5 x
dx x

It is linear in t
∴I.F. =e ∫ −5 dx
x
=x−5

Hence the general solution is

t . I . F .=∫ I . F . ( 1−n ) Q dx +C
t . x−5=∫ x −5 (−5 x 2 ) . dx +C
0 r t . x−5=−5 ∫ ( x−3 ) . dx +C
−5 −5 5 −2
Or y x = x +C
2
dy
Example-2: Solve xy ( 1+ x y )
2
=1
dx
Solution: Given equation can be written as
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 B.Tech , Sem- II, Mathematics -II (BT -202 )

dx 2 3
⇒ −xy=x y
dy
1 dy y 3
⇒ 2
− =y
x dx x

−1 1 dx dz
Put =z ⇒ 2 =
x x dy dy

dz 3
⇒ + yz= y
dy

This is a linear differential equation¿ z

2
y
IF=e ∫ ydy
=e . 2

Hence the general solution is

z ⋅ IF =∫ y ⋅ IF dy +c
3

2 2
y y y
2
y
2

⇒ z ⋅e =∫ y e dy +c Put e 2 =t ⇒ y e 2 dy=dy
2 3 2

2
y
⇒ z ⋅e =∫ log t dt +c=t ( log t−1 )+ c .
2

( )
2 2
y y 2
−1 2 y
Put the values of z and t , then e =e 2 −1 +c
x 2

Type-III : Exact differential equation of order 1 and degree 1:-

The equation of the type Mdx + Ndy=0 ------- (1) is called an exact differential equation of
∂M ∂N
order 1 and degree 1 iff = Where M ∧N be some functions of x and y .i.e. It is an
∂ y ∂x
exact differential of some functions of x and y .

Verification: If the equation of the type Mdx + Ndy=0 ------- (1) is called an exact differential
of order 1 and degree 1.
Then Mdx + Ndy=du , Where ube some functions of x and y .
∂u ∂u
But du= dx+ dy
∂x ∂y

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
∂u ∂u
Hence, Mdx + Ndy= dx+ dy
∂x ∂y
∂u ∂u
i.e. M = , N=
∂x ∂y
2 2
∂M ∂u ∂N ∂ u
∴ = , =
∂ y ∂ y ∂x ∂ x ∂ x ∂ y
2 2
∂u ∂ u
But, we know that , =
∂ y∂ x ∂x ∂ y

∂M ∂N
Hence from above = .
∂ y ∂x
Hence proved

The Method of Solution of Exact differential equation of order 1 and degree 1 is

∫ Mdx Keeping y constant +∫ ( terms of N not containing x ) dy =C .

Example-1: S h ow t h at t h e equation ( 2 x 3 +3 y ) dx + ( 3 x+ y−1 ) dy=0 is exact. Find it’ s general
solution.

Solution: On comparing the given equation with Mdx + Ndy=0

∴ M =2 x +3 y , N=3 x + y−1
3

∂M ∂N
=3 , =3
∂y ∂x

∂M ∂N
Hence, = Therefore ,it is an exact differential equation.
∂ y ∂x
The general solution is

∫ Mdx Keeping y constant +∫ ( terms of N not containing x ) dy =C

⟹∫ ( 2 x3 +3 y ) dx Keeping y constant +∫ ( y−1 ) dy=C

4 2
x y
⟹2 +3 xy+ − y=C .
4 2
Example-2: S h ow t h at t h e equation ( y 2 e x y + 4 x 3 ) dx + ( 2 xy e x y −3 y 2) dy=0is exact. Find it’ s
2 2

general solution .

Solution: On comparing the given equation with Mdx + Ndy=0

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

∴ M =( y e +4 x )And N=2 xy e x y −3 y 2 ,
2 2
2 xy 3

∂M xy 2
3 xy ∂N
2
xy 2
3 xy 2

Hence , =2 y e + 2 x y e and =2 y e +2 x y e .
∂y ∂x

∂M ∂N
Here, we have = , i.e. Equation is exact differential equation. Hence the general
∂ y ∂x
solution is

∫ Mdx +∫ ( terms of N not containing x ) dy=c ⇒∫ ( y 2 e x y + 4 x 3) dx−∫ 3 y 2 dy =c

2

⇒ e x y + x 4 − y 3=c .

Type-IV : Solution of Non-Exact differential equation of order 1 and degree 1 :-

When the equation is of the type Mdx + Ndy=0 is not exact. Then, we multiply this equation
by some function of x and y or by a suitable factor so that the equation becomes exact. Then
this factor is called I.F. i.e. Integrating factor.

Rule-1 of I.F.:--When the equation is of the type Mdx + Ndy=0 is not exact and if M ∧N be
1
homogeneous functions of x and y and M x + Ny ≠ 0 , Then I . F .=
Mx+ Ny

Example: Solvet h e equation ( x 2 y−2 x y 2 ) dx−( x 3−3 x 2 y ) dy=0

Solution: On comparing the given equation with Mdx + Ndy=0

∴ M =x y −2 x y , N=−( x −3 x y )
2 2 3 2

∂M 2 ∂N 2
=x −4 xy , =−3 x + 6 xy
∂y ∂x

Hence the given equation is not exact.

Since M ∧N be homogeneous functions of x and y .

1 1
Now, Mx + Ny=x 2 y 2 ≠ 0 hence I . F .= =
Mx + Ny x 2 y 2
Therefore, multiply given equation by this I.F

We get, ( 1y − 2x ) dx−( yx −3/ y) dy =0is an exact equation.

2

The general solution is

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

∫ Mdx Keeping y constant +∫ ( terms of N not containing x ) dy =C

⟹∫ ( 1y − 2x )dx Keeping y constant

+∫ ( 3/ y ) dy=C

x
⟹ −2 logx+3 logy=C .
y

Rule-2 of I.F.:-- When the equation is of the type Mdx + Ndy=0 is not Exact and if it is of the
1
type f 1 ( x , y ) ydx + f 2 ( x , y ) xdy=0and Mx−Ny ≠0 , Then I . F .=
Mx−Ny

Example: Solvet h e equation ( 2 x y 2 + y ) dx + ( x +2 x 2 y−x 4 y 3 ) dy =0

Solution: On comparing the given equation with Mdx + Ndy=0

∴ M =2 x y + y , N=x +2 x y −x y
2 2 4 3

∂M ∂N 3 3
=4 xy +1 , =1+ 4 xy−4 x y
∂y ∂x

Hence the given equation is not exact.

Since, It is of the type f 1 ( x , y ) ydx + f 2 ( x , y ) xdy=0 .

1 1
∴ Mx−Ny=x y ≠ 0hence , I . F .= Mx −Ny = 4 4
4 4
x y
Therefore, multiply given equation by this I.F

We get,
2
(
2
x y x y
3
1 1
) (
2
)
+ 4 3 dx + 3 4 + 2 3 −1/ y dy=0 is an exact equation.
x y x y
Hence , the general solution is

∫ Mdx Keeping y constant +∫ ( terms of N not containing x ) dy =C

⟹∫
( x 2y + x 1y ) dx
3 2 4 3
Keeping y constant
+∫ −( 1/ y ) dy=C

1 1
⟹− 22
− 3 3 −logy=C .
x y 3x y

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Rule-3 of I.F.:-- When the equation is of the type Mdx + Ndy=0 is not exact

∂ M ∂N
−
If ∂ y ∂x then, I.F. =e∫ f ( x ) dx.
=f ( x )
N

Rule-4 of I.F.:-- When the equation is of the type Mdx + Ndy=0 is not exact

∂N ∂M
−
If ∂x ∂ y then, I.F. =e∫ g ( y )dy
=g ( y )
M

Example-1 : Solvet h e equation ( x 2 + y 2 + x ) dx+ ( xy ) dy =0

Solution: On comparing the given equation with Mdx + Ndy=0

∴ M =x + y + x , N=xy
2 2

∂M ∂N
=2 y , =y
∂y ∂x

Hence the given equation is not exact.

∂ M ∂N
− 1
Now, ∂ y ∂ x 2 y− y 1 Hence, I.F. =e∫ f ( x ) dx =e∫ x dx =x
= = =f ( x )
N xy x

Therefore, multiply given equation by this I.F

We get, ( x 3 + xy 2+ x 2) dx + ( x 2 y ) dy=0 is an exact equation.

Hence, the general solution is

∫ Mdx Keeping y constant +∫ ( terms of N not containing x ) dy =C

⟹∫ ( x 3+ xy 2+ x 2 ) dx Keeping y constant +∫ ( 0 ) dy=C

4 2 2 3
x x y x
⟹ + + =C .
4 2 3

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Example-2 : Solvet h e equation ( x y 3+ y ) dx+ 2 ( x 2 y 2+ x + y 4 ) dy=0

Solution: On comparing the given equation with Mdx + Ndy=0

∴ M =x y + y , N=2 ( x y + x + y )
3 2 2 4

∂M 2 ∂N 2
=3 x y +1 , =4 x y + 2
∂y ∂x

Hence the given equation is not exact.

∂N ∂M
− 1
Now, ∂x ∂ y 1 then, I.F. =e∫ g ( y )dy =e∫ y dy = y
= =g ( y )
M y
Therefore, multiply given equation by this I.F

We get, ( x y 4 + y 2 ) dx +2 ( x2 y 3 + xy + y 5 ) dy=0 is an exact equation.

Hence , the general solution is

∫ Mdx Keeping y constant +∫ ( terms of N not containing x ) dy =C

⟹∫ ( x y 4 + y 2 ) dx Keeping y constant +∫ ( 2 y 5 ) dy=C

2 4 2 6
x y xy y
⟹ + + =C .
2 1 3
Part-II: Differential Equations of First Order and Higher Degree
The most general form of a differential equation of the first order and of higher degree say of
nth degree can be written as

( )
dy n
( ) ( )
n−1 n−2
dy dy
+a1 ( x , y ) +a2 ( x , y ) + .. . . ..
dx dx dx
dy
+ an−1 ( x , y ) + a ( x , y )=0
… … … dx n
or pn+a1pn-1+a2pn-2+ …….+an-1 p+an= 0 (2.1)

dy
p=
Where dx and a1, a2, . . , an area functions of x and y.
(2.1) can be written as
F(x, y, p) = 0 (2.2)
[I] First-Order differential Equation of Higher Degree solvable for p

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Let (2.2) can be solved for p and can be written as

(p-q1(x,y) ) (p-q2(x,y) ) ………. (p-qn(x,y )) = 0
Equating each factor to zero ,we get equations of the first order and first degree.
One can find solutions of these equations by the methods discussed in the previous part. Let
their solution be given as:
i(x, y,ci )=0, i=1,2,3 ………n (2.3)
Therefore the general solution of (2.1) can be expressed in the form
1(x, y, c ) 2(x, y, c )………n(x, y, c) = 0 (2.4)
where c in any arbitrary constant.
It can be checked that the sets of solutions represented by (2.3) and (2.4) are identical because
the validity of (2.4) in equivalent to the validity of (2.3) for at least one i with a suitable value of
c, namely c=ci

Example1. Solve
xy ( )
dy 2
dx
dy
+( x 2 + y 2 ) +xy=0
dx (2.5)

Solution: This is first-order differential equation of degree 2.

dy
p=
Let dx Equation (2.5) can be written as
xy p2+(x2+y2) p+xy=0 (2.6)
or (xp+y)(yp+x)=0
This implies that
xp+y=0, yp+x=0 (2.7)
By solving equations in (2.7) we get
xy=c1 and x2+y2=c2 respectively
dy dy 1
x + y=0 or + y=0 ,
[ dx dx x Integrating factor
∫ 1x dx
Ι ( x )=e = e log x . This gives
y.x =  o.x dx +c1 or xy=c1 ] ,
dy
y + x=0 , or ydy + xdx=0
[ dx
1 2 1 2
y + 2 x =c
By integration we get 2
or x2+y2 = c2, ]
The general solution can be written in the form
(x2+y2-c2) (xy-c1)=0 (2.8)
2 2
It can be seen that none of the nontrivial solutions belonging to xy=c 1 or x +y =c2 is valid on the
whole real line.
[ II ] Equations Solvable for y

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Let the differential equation given by (2.2) be solvable for y. Then y can be expressed as a
function x and p, that is,
y= f ( x , p ) (2.9)
Differentiating (2.9) with respect to x we get
dy ∂ f ∂ f dp
= + .
dx ∂ x ∂ p dx (2.10)

(2.10) is a first order differential equation of first degree in x and p. It may be solved. Let
solution be expressed in the form
ϕ( x , p , c )=0 (2.11)
The solution of equation (2.9) is obtained by eliminating p between (2.9) and (2.11). If
elimination of p is not possible then (2.9) and (2.11) together may be considered parametric
equations of the solutions of (2.9) with p as a parameter.
Example 2: Solve y2-1-p2=o

Solution: It is clear that the equation is solvable for y, that is

y= √1+ p2 (2.12)

By differentiating (2.12) with respect to x we get

dy 1 1 dp
= .2 p
dx 2 √ 1+ p2 dx
p dp
p=
or √ 1+p 2 dx

or
[
p 1−
1
√ 1+ p ]
dp
2 dx
=0
(2.13)

(2.13) gives p=o or

By solving p=0 in (2.12) we get
y=1
1 dp
1− =0
By √ 1+ p 2 dx

we get a separable equation in variables p and x.

dp
=√ 1+ p2
dx
By solving this we get

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

p= sinh (x+c) (2.14)

By eliminating p from (2.12) and (2.14) we obtain
y=cos h (x+c) (2.15)
(2.15) is a general solution.
Solution y=1 of the given equation is a singular solution as it cannot be obtained by giving a
particular value to c in (2.15).
[ III ] Equations Solvable for x

Let equation (2.2) be solvable for x,

that is x=f(y,p) .. (2.16)
Then as argued in the previous section for y we get a function  such that
(y, p, c) = 0 (2.17)
By eliminating p from (2.16) and (2.17) we get a general solution of (2.2). If elimination
of p with the help of (2.16) and (2.17) is cumbersome then these equations may be considered
parametric equations of the solutions of (2.16) with p as a parameter.

Example 3. Solve
x ( )
dy 3
dx
dy
−12 −8=0
dx
dy
p= , then
Solution: Let dx
xp3-12p-8=0
It is solvable for x, that is,
12 p +8 12 8
x= = 2+ 3
p3 p p … (2.18)
Differentiating (2.18) with respect to y, we get
dx 12 dp 8 dp
=−2 3 −3 4
dy p dy p dy
1 24 dp 24 dp
or =− 3 − 4
p p dy p dy

or dy= −
( 24 24
2
p p
− 3 dp
)
… (2.19)
(2.18) and (2.19) constitute parametric equations of solution of the given differential equation.
[ IV] Differential Equation of First-Order and Higher Degree in x and y – Lagrange’s i.e.
Clairaut’s Form.
Let Equation (2.2) be of the first degree in x and y, then
y = x1(p) + 2 (p) … (2.20)
Equation (2.20) is known as Lagrange’s equation.
If 1(p) = p then the equation
18
 B.Tech , Sem- II, Mathematics -II (BT -202 )

y = xp + 2 (p) .. (2.21)
is known as Clairaut’s equation
By differentiating (2.20) with respect to x, we get
dy dp dp
=ϕ ( p)+ xϕ ' ( p ) +ϕ ' ( p )
dx 1 1 dx 2 dx
dp
p−ϕ 1 ( p)=( xϕ 1' ( p )+ ϕ '2 ( p))
or dx … (2.22)
From (2.22) we get
dp
( x+ ϕ '2 ( p )) =0
dx for 1(p)=p
This gives
dp
=0 '
dx or x+2 (p) =0
dp
=0
dx gives p = c and
by putting this value in (2.21) we get
y=cx+2(c)
This is a general solution of Clairaut’s equation.
'
The elimination of p between x+2 (p) = 0 and (2.21) gives a singular solution.
If 1(p)  p for any p, then we observe from (2.22) that
dp
≠0
dx every where. Division by
dp
[ p−ϕ 1 ( p )]
dx in (2.22) gives

dx ϕ '1 ϕ '2 ( p )
− x
dp p−ϕ1 ( p ) p−ϕ 1 ( p )
=
which is a linear equation of first order in x and thus can be solved for x as a function of
p, which together with (2.20) will form a parametric representation of the general solution of
(2.20)

( dy
Example 4 . Solve dx
−1 )( y−x )=
dy dy
dx dx

dy
p=
Solution: Let dx then,
(p-1) (y-xp)= p
This equation can be written as
p
y=xp +
p−1
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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Differentiating both sides with respect to x we get

dp
dx [
x−
1
( p−1 )2
=0
]
dp
=0
Thus either dx or
dp
=0
dx gives p=c Since
Putting p=c in the equation we get
c
y=cx+
c−1
(y-cx)(c-1)=c
Which is the required solution.

Part-III: Higher order Linear differential Equation with constant coefficients

A differential equation is said to be Linear if the dependent variable and it’s derivatives occur
only in the first degree and are not multiplied together.
Thus, the general linear differential equation of the nth order with constant coefficients is of
the form

n n−1 n−2
d y d y d y dy
n
+a 1 n−1 +a2 n−2
+−−−−−a n−1 + an y=X … … … … (1)
dx dx dx dx

d
Or (Dn +a 1 Dn−1 +a 2 Dn−2 +−−−−−an−1 D+ an ) y =X Since D ≡
dx
Or f ( D )y = X ………………………………………. (1)
Where f ( D )=Dn +a1 Dn−1 +a2 Dn−2+−−−−−a n−1 D+a n

and a 1,a 2 ,…………,a n−1 , a n be the constants , X =f ( x ).

If X =0 , then equation (1) is called a homogeneous linear differential equation of the n th order
with constant coefficients.
If X ≠ 0 , then equation (1) is called a Non-homogeneous linear differential equation of the
n th order with constant coefficients.
Now, the general solution of equation (1) consists of two parts.
i.e. y=C . F .+ P . I .… … … … … … … … .(2)

Where C.F. = Complementary function and P.I. = Particular Integral.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Method of finding C.F.(Complementary function ):-- The Complementary function can written
on the basis of nature of roots of auxiliary equation[ A.E.] of (1) as

f ( D=m )=0
n n−1 n−2
i.e. m +a1 m + a2 m +−−−−−an −1 m+ an =0 ……………………… (3)

On solving the above equation, we get then roots.

Sr. Roots of A.E. Form of C.F.

No
.
1 m1 ' m2 ' m3 ' … … … … … mn ' c1 e
m1 x
+ c2e
m2 x
+… … .. c n e
mn x

(Real and distinct roots )

2 m1=¿m m …… …… …m ¿
2' 3' n' ( c 1 +c 2 x ) e m x +c 3 e m x + … … .. c n e m x
1 3 n

(two equal real roots )

3 m1=¿m ¿
2=¿m3 ' … … …… … mn' ¿ ( c1 + c2 x+ c 3 x 2 ) e m x + c 4 e m x +… … .. c n em x
1 4 n

(three equal real roots )

4 m=α ±i β , m3 ' … … … … … mn ' e
αx
[ c1 cosβx +c 2 sinβx ]+ c 3 e m x + … ….. c n e m x
3 n

(two roots are imaginary )

5 α ± i β , α ± i β , m5 ' … … … … … mn ' e
αx
[ ( c 1+ c 2 x ) cosβx +( c 3 +c 4 x ) sinβx ] +c 5 em x + … … ..c n e m x
5 n

( equal pair of imaginary roots )

6 m=α ± √ β , m3 ' … … … … … mn ' e
αx
[ c1 cosh √ β x+ c 2 sinh √ β x ] +c 3 e m x + … … .. c n e m x
3 n

(two roots a in surds form or irrational )

7 α ± √ β ,α ±i √ β m 5 ' … … … … … m n ' (equal e
αx
[ ( c 1+ c 2 x ) cos h √ β x +( c 3 +c 4 x ) sin h √ β x ]+ c 5 e m x +… …..
5

pair of surds form or irrational )

Method of finding P.I. [ Particular Integral]:--

The Particular Integral of equation f ( D )y = X ………………………………………. (1)
Where f ( D )=Dn +a1 Dn−1 +a2 Dn−2+−−−−−a n−1 D+a n
and a 1,a 2 ,…………,a n−1 , a n be the constants , X =f ( x )

X
is P.I. ¿
f ( D)

Particular Integral for different cases of X :-

Case-1: When X =e ax
In this case

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
ax ax
e e
[i] P.I. ¿ = , providedf ( a ) ≠ 0
f D f (a )
( )
ax ax
e e '
[ii] If f ( a ) =0 then P.I. ¿ =x , provided f ( a ) ≠ 0
f ( D) '
f (a )
ax ax
e 2 e ''
'
Again If f ( a )=0then P.I. ¿ =x , provided f ( a ) ≠ 0
(
f D ) ''
f ( a)
Case-2 : When X =cosax
In this case
cosax cosax
, providedf ( −a ) ≠ 0
2
[i] P.I. ¿ =
f ( D ) f (−a )
2 2

cosax cosax
, provided f (−a ) ≠ 0
' 2
[ii] If f (−a2 )=0then P.I. ¿ =x '
f (D )2
f (−a ) 2

cosax 2 cosax
, provided f ( −a ) ≠ 0
'' 2
Again if f ' (−a2 )=0 then P.I. ¿ =x ' '
f (D ) 2
f ( −a ) 2

Case-3 : When X =sinax

In this case
Sinax Sinax
, provided f (−a ) ≠ 0
2
[i] P.I. ¿ =
f ( D ) f (−a )
2 2

Sinax Sinax
, provided f (−a ) ≠ 0
' 2
[ii] If f (−a2 )=0then P.I. ¿ =x '
f (D )2
f ( −a ) 2

Sinax 2 Sinax
, provided f (−a ) ≠ 0
'' 2
Again if f ' (−a2 )=0 then P.I. ¿ =x ' '
f (D ) 2
f (−a ) 2

m
Case-4 : When X =x , w h ere mis a positive integer .
In this case ,
m
x
P.I. ¿
f ( D)

Take out the lowest degree term common from f ( D )to make the first term unity.The
remaining factor will be of the form ( 1−∅ ( D ) ).Now ,take this factor in the numerator i.e.
( 1−∅ ( D ) )−1. Expand ( 1−∅ ( D ) )−1 in ascending powers of D as far as the term containing Dm and
then operate on x m term by term.
Note:-
1.( 1−x )−1 =1+ x+ x 2 + x 3+ … … … .
2. ( 1−x )−2 =1+ 2 x +3 x 2+ 4 x 3 +… … … .

Case-5 : When X =e ax V where V =f ( x )

In this case ,
ax
e V ax V
P.I. ¿ =e
f ( D) f ( D+a )
Case-6 : When X =xV where V =f ( x )

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

In this case ,
'
xV V f ( D)
P.I. ¿ =x − V
f ( D) f ( D ) ( f (D) )2
X
Case-7 : =∫ Xdx
( D)

X
=e ∫ X e dx
ax −ax
Case-8 :
( D−a )

Example-1 : Solvet h e equation ( D2−4 D+3 ) y =0

Solution: On comparing the given equation with f ( D )y = X

∴ f ( D )= ( D2−4 D+3 ) , X=0

Auxiliary equation is m2−4 m+3=0

2
m −m−3 m+3=0
( m−1 ) ( m−3 )=0
m=1 , 3
Hence C.F. = c 1 e x + c 2 e 3 x
Since X =0 ∴P.I.=0
Hence the complete general solution is

x 3x x 3x
y=C . F .+ P . I .=c 1 e + c2 e +0=c 1 e +c 2 e

Example-2 : Solvet h e equation ( D−2 )2 y=8 ( e2 x + sin 2 x + x 2 )

Solution: On comparing the given equation with f ( D )y = X

∴ f ( D )= ( D−2 ) , X=8 ( e2 x +sin 2 x+ x2 )

2

Auxiliary equation is ( m−2 )2 =0

m=2 , 2

2x
Hence C.F. = ( c 1 +c 2 x ) e

[
8 ( e 2 x + sin 2 x + x 2 )
]
2x 2
X e sin 2 x x
∴P.I. = = =8 + +
f ( D) ( D−2 )2 ( D−2 )2 ( D−2 )2 ( D−2 )2
¿ 8 [P . I 1 + P . I 2 + P . I 3 ]…..(1)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

2x 2x 2x
e e e
Now, P . I 1= 2
= 2
= Hence, this method fails
( D−2 ) ( 2−2 ) 0
2x 2x 2x 2x
e e 2 e 2e
∴ P . I 1= =x =x =x
( D−2 )2 2 ( D−2 ) 2.1 2
sin 2 x sin 2 x sin 2 x sin 2 x 1
P . I 2= = 2 = 2
( D−2 ) D −4 D+4 −2 −4. D+4 −4. D −4
2
= = ∫ sin 2 x dx
1
∴ P . I 2= cos 2 x
8

( )
2 −2 2
x 1 D 2 1 D D 2
P . I 3= 2
= 1− x = [1+ 2 + 3 + … … ..]x
( D−2 ) 4 2 4 2 4

2
1 D D 2
= [1 x2 +2 x 2+3 x + … … ..]
4 2 4
1 2 2x 2
= [ x +2 +3 +0 … … ..+0]
4 2 4
1 2 3
= [ x +2x+ ]
4 2

∴P.I.¿ 8 [P . I 1 + P . I 2 + P . I 3 ]
2x
e 1 1 2 3
=8 [ x 2 + Co 2 x+ ( x + 2 x + ) ¿
2 8 4 2

Hence the complete general solution is

2x
2x e 1
2 1 2 3
y=C . F .+ P . I .=( c 1 +c 2 x ) e +8[ x + cos 2 x + ( x +2 x+ )]
2 8 4 2

Example-3 : Solvet h e equation ( D2−2 D+1 ) y=x e x Sinx

Solution: On comparing the given equation with f ( D )y = X

2 x
∴ f ( D )= D −2 D+ 1, X =x e Sinx

Auxiliary equation is ( m−1 )2=0

m=1 , 1

x
Hence C.F. = ( c 1 +c 2 x ) e

x
X e x Sinx x x Sinx x x Sinx x x Sinx
∴P.I. ¿ = =e =e =e 2 =
f ( D) f ( D) f ( D+1 ) 2
(D+1) −2(D+1)+ 1 (D)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
1 x 1
=e
x
∫ x Sinx dx=e [−xcosx+ sinx ] =e x [ −∫ x cosx dx+∫ Sinx dx ]=e x [−xsinx−2 cosx ]
D D

Hence the complete general solution is

2x x
y=C . F .+ P . I =( c1 +c 2 x ) e +e [−xsinx−2 cosx]

Example-4 : Solvethe equation ( D2 + 4 ) y=x Sinx

Solution: On comparing the given equation with f ( D )y = X

∴ f ( D )= ( D2+ 4 ) , X=x Sinx

Auxiliary equation is m2 +4=0

m=± 2 i=∝± iβ

Hence C.F. = ( c 1 cos 2 x+ c2 sin 2 x )

P.I. ¿
X x Sinx 1
[ Sinx 1
]
= 2 = x− 2 2 D 2 =¿ x− 2 2 D
f ( D) D +4 D +4 D +4 D +4
Sinx
−1+ 4 [ ]
[
¿ x−
1
D +4
2D
2
3
=
]
Sinx xSinx
3
1
− 2 2D
D +4
Sinx xSinx
3
=
3
1
− 2 2
D +4
cosx xSinx 2
3
=
3
−
1 cosx
3 (−1+4 ) 1
xSinx 2
P.I. = − cosx
3 9
Hence the complete general solution is

xSinx 2
y=C . F .+ P . I =( c1 cos 2 x+ c 2 sin 2 x )+ − cosx .
3 9

Example-5 : Solvet h e equation ( D2−4 ) y=2x

Solution: On comparing the given equation with f ( D )y = X

2 x
∴ f ( D )= D −4 , X=2

Auxiliary equation is m2−4=0

m=2 ,−2

Hence C.F. = c 1 e−2 x + c 2 e 2 x

X 2x e (log 2) x e( log2 ) x 2x
∴P.I. ¿ = = = = Since a x =e (loga ) x
f ( D ) D 2−4 D 2−4 ( lo 2 )2−4 ( lo2 )2−4

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Hence the complete general solution is

x
−2 x 2x 2
y=C . F .+ P . I =c 1 e +c 2 e +
( lo 2 )2−4
Example-6: Solvethe equation ( D2−3 D+2 ) y =x+ e x

Solution: On comparing the given equation with f ( D )y = X

∴ f ( D )= ( D2−3 D+2 ) , X=x +e x

Its auxiliary equation is m2−3 m+2=0 ⇒ m=1 ,2

x 2x
⇒ CF=c 1 e + c 2 e and

( )
−1
1
PI = 2
( x +e x )= 1 1− 3 D+ 1 D2 x+ 2
1
e
x

D −3 D+ 2 2 2 2 D −3 D+2

1 3 1 x 1 3 x
¿ x− + x e = x− −x e
2 4 2 D−3 2 4

Hence the complete general solution is

x 2x 1 3 x
y=CF + PI =c 1 e +c 2 e + x− −x e .
2 4

Part-IV: Homogeneous linear differential equations

There are two types of homogeneous linear differential equations, which can be reduced to the
linear differential equations with constant coefficient by convenient substitutions.

Type-1: Cauchy’s Homogeneous linear differential equation:

An equation is of the type

n n−1 n −2
n d y n−1 d y n−2 d y
x n
+ a1 x n −1
+ a2 x n −2
+−−−−+a n y= X … … … …(1)
dx dx dx

is called a Cauchy’s Homogeneous linear differential equation.

W h ere a 1,a 2 ,…………,a n−1 , a n be the constants , X =f ( x )

26
 B.Tech , Sem- II, Mathematics -II (BT -202 )

Steps to solve: Equation (1) can be reduced to a linear differential equation with constant
Coefficients by using the following steps:
z 1 dz
Step 1: Put x=e so t h at logx=z =≫ =
x dx

dy dy dz 1 dy
Now = =
dx dz dx x dz
dy dy d
∴x = =Dy w h ere D ≡
dx dz dz

2
2 d y
Similarly x 2
=D ( D−1 ) y ,
dx
3 n
d y
3 nd y
x 3
=D ( D−1 ) ( D−2 ) y∧so on x n
=D ( D−1 ) ( D−2 ) … . ( D−n+1 ) y
dx dx
Step 2 : Putting all the values in equation (1) then it will reduces into linear differential equation
with constant Coefficients in terms of z .
Step 3: Find the complete general solution is y=C . F .+ P . I
Step 4: Finally, we replace z by logx .

Type-2: Legendre’s Homogeneous linear differential equation:--

An equation is of the type

n n−1 n−2
n d y n−1 d y n−2 d y
(ax +b) n
+a1 (ax+ b) n−1
+a2 (ax+ b) n−2
+−−−−+ an y=X … … … … (1)
dx dx dx

is called a Legendre’s Homogeneous linear differential equation.

W h ere a 1,a 2 ,…………,a n−1 , a n be the constants , X =f ( x )

Steps to solve: Equation (1) can be reduced to a linear differential equation with constant
Coefficients by using the following steps:
z a dz
Step 1: Put :a x +b=e so t h at log ⁡( a x +b)=z=≫ =
a x +b dx

dy dy dz a dy
Now = =
dx dz dx a x +b dz

dy dy d
∴ ( a x +b ) =a =a Dy w h ere D ≡
dx dz dz

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

2
2 d y 2 (
Similarly (a x +b) 2
=a D D−1 ) y ,
dx
3 n
3d y 3 ( nd y n
(a x +b) 3
=a D D−1 ) ( D−2 ) y∧so on (a x +b) n
=a D ( D−1 ) ( D−2 ) … . ( D−n+1 ) y
dx dx
Step 2 : Putting all the values in equation (1) then it will reduces into linear differential
equation with constant Coefficients in terms of z .
Step 3: Find the complete general solution is y=C . F .+ P . I
Step 4: Finally, we replace z by log ⁡(ax +b).

2
2 d y dy
Example-1 : Solvet h e equation x 2
−x + y =2logx
dx dx

Solution: Given equation is

2
2 d y dy
x 2
−x + y =2logx … …. (1)
dx dx

z 1 dz
Now to solve, Put x=e so t h at logx=z =≫∴ =
x dx

dy dy dz 1 dy
= =
dx dz dx x dz

dy dy d
∴x = =Dy w h ere D ≡
dx dz dz

2
d y
2
Similarly x 2
=D ( D−1 ) y ,
dx
Hence, the equation (1) becomes

D ( D−1 ) y−Dy+ y=2 z

Or (D¿¿ 2−2 D+1) y=2 z ¿---------------(2)

It is reduce into linear differential equation with constant Coefficients in terms of z .

Auxiliary equation is m2−2 m+1=0

( m−1 )2=0=¿ m=1 , 1

z
Hence C.F. = ( c 1 +c 2 z ) e

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Z 2z
∴P.I.¿ =
f ( D ) (D¿¿ 2−2 D+1)=2[1−(2 D−D2 )]−1 z ¿

=2 [ 1+ ( 2 D−D2 ) +… . ] z

= 2 [ z+ ( 2 Dz−D2 z ) +… . ]
= 2 [ z +2−0. ]

= 2 z+ 4

Hence the complete general solution is

z
y=C . F .+ P . I =( c1 +c 2 z ) e + 2 z + 4= ( c 1+ c 2 logx ) x +2logx +4

Example-2 : Solvet h e equation

2
2 d y dy
(2 x+ 1) 2
−2 (2 x +1 ) −12 y=6 x
dx dx

Solution: Given equation is

2
d y 2 dy
(2 x+ 1)
2
−2 (2 x +1 ) −12 y=6 x−−−(1)
dx dx
z 2 dz
Put :2 x+1=e so t h at log ⁡(2 x +1)=z=≫ =
2 x +1 dx

dy dy dz 2 dy
Now = =
dx dz dx 2 x +1 dz

dy dy d
∴ ( 2 x +1 ) =2 =2 Dy w h ere D ≡
dx dz dz

2
d y
=4 D ( D−1 ) y =4 ( D −D ) y
2 2
Similarly (2 x+ 1) 2
dx

Hence, the equation (1) becomes

4 ( D2−D ) y−22 Dy−12 y =3(e z −1)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
3 z
¿ ≫ ( D −2 D−3 ) y= (e −1)---------(2)
2
4
It is reduce into linear differential equation with constant Coefficients in terms of z .

Auxiliary equation is m2−2 m−3=0

( m+1 ) ( m−3 )=0=¿ m=−1 , 3

Hence C.F. = c 1 e−z +c 2 e 3 z

[ { ( )} ]
−1
Z 3 ( e −1 )
z
3 e
z
1
2
D 2
P.I.¿ = = − 1− − D 1
f ( D ) 4 ( D 2−2 D−3 ) 4 ( D2−2 D−3 ) (−3 ) −3 3

[ ]
z
3 e 1
‘’ = + .1
4 ( 1 −2.1−3 ) ( 3 )
2

‘’=
[
3 −e z 1
+ .
4 4 (3 ) ]
Hence the complete general solution is

[ ]
z
−z 3 −e 1
3z
y=C . F .+ P . I =c 1 e + c2 e + + .
4 4 ( 3)

−1 3 3 1
Or y=c1 ( 2 x+1 ) + c2 ( 2 x +1 ) − ( 2 x +1 ) +
16 4

2
2 d y ( dy
Example-3 : Solvet h e equation ( 1+ x ) 2
+ 1+ x ) + y=2 sin [log ( 1+ x ) ]
dx dx

2
2 d y ( dy
Solution: Given equation is( 1+ x ) 2
+ 1+ x ) + y=2 sin [log ( 1+ x ) ]… … .(1)
dx dx

2
dy dy 2d y
Put 1+ x=e , i.e. t=log (1+ x), so that
t
( 1+ x ) = =Dy and ( 1+ x ) =D ( D−1 ) y , where
dx dt dx
2

d
D≡ .
dt

Hence equation (1) becomes

D ( D−1 ) y + Dy+ y =2 sin(log t)⇒ ( D2 +1 ) y=2 sin t ……………..(2)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

It is reduce into linear differential equation with constant Coefficients in terms of z .

It’s auxiliary equation is m2 +1=0 ⇒ m=±i . Hence CF is

C.F.=c 1 cos t+ c 2 sin t

1 2
And PI = 2 sint=t sin t=−t cos t
2
D +1 2D

⇒ y =CF+ PI =c 1 cos t+ c2 sin t−t cos t=c1 cos [ log ( 1+ x ) ] +c 2 sin [ log ( 1+ x ) ] − [ log (1+ x ) ] cos [ log ( 1+ x ) ]

Part-V: Simultaneous linear differential equations

Important Ordinary Differential Equations arise at times in which there are two or more
dependent variables and a single independent variable, usually the time t. We consider here
only a system of linear differential equations with constant coefficients, the solutions of which
can be obtained by reducing to the solution of one or more separate equations,just analogous
to algebraic system of linear equations. The method is well explained by the following examples

Example-1 : Solvet h e simultaneous linear differential equation

dx dy
+ + 2 x + y=0
dt dt

dy
+5 x +3 y=0
dt
Also find the particular solution satisfying the conditions x=1 , y=2 w h en t=0.

Solution: Given equations can be written as in operator form

( D+2 ) x+ ( D+ 1 ) y =0 … … ..(1)

d
5 x+ ( D+ 3 ) y=0 … … ..(2)where D ≡
dt

Now operating equation (1) by ( D+3 ) and (2) by ( D+1 ) and then subtracting them, we get

( D2 +1 ) x=0………….(3)
Now the
Auxiliary equation is m2 +1=0=¿ m=± i

0
Hence C.F. = ( c 1 cost +c 2 sint ) and P.I.¿ =0
f ( D)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Hence the complete general solution is

x=C . F .+ P . I =c 1 cost+ c 2 sint …………….(4)
Now ,eliminating Dy∈ ( 1 )∧ ( 2 ) by subtracting (2) from (1).

1 3
We get , y= Dx+ x
2 2

1 3
¿> y= D ( c 1 cost +c 2 sint )+ ( c 1 cost+ c 2 sint )
2 2
1 3
¿> y= (−c 1 Sint +c 2 cost ) + ( c 1 cost+ c 2 sint )
2 2
−1 1
¿> y= ( c +3 c 2) Sint + 2 ( c 2−3 c 1 ) cost … … … .(5)
2 1

Now it is given that , x=1 , y=2 w h en t=0 ,

Hence from equation (4),c 1=1and by equation (5)c 2=7
Thus the required particular solution satisfying the given conditions is

x=cost +7 sint∧ y=−11 sint +2 cost

Example-2 : Solvet h e simultaneous equation

dx dy
+5 x−2 y=t ; +2 x + y=0
dt dt
Also find the particular solution satisfying the conditions x= y =0 w h en t=0.

Solution: Given equations can be written as in operator form

( D+5 ) x−2 y=t … … ..(1)

d
2 x+ ( D+ 1 ) y =0 … … .. ( 2 ) where D ≡
dt

Eliminate x as if D were an ordinary algebraic multiplier. Multiplying (1) by 2 and operating on

(2) by D+5 and then subtracting ,we get

( D2 +6 D+9 ) y=−2 t ………….(3)

Now the Auxiliary equation is m2 +6 D+9=0=¿ m=−3 ,−3

−3 t
Hence C.F. = ( c 1 +c 2 t ) e

32
 B.Tech , Sem- II, Mathematics -II (BT -202 )

( ) ( )
−2
−2 t −2t −2 D −2 D −2 4
∴P.I.¿ = = 1+ t= 1−2 + … . t= t+
( )
f D ( D+ 3 ) 2
9 3 9 3 9 27

Hence the complete general solution is

−3 t 2 4
y=C . F .+ P . I =( c1 +c 2 t ) e − t+ ……………………….(4)
9 27

Now to find x , either eliminate from (1) and (2) and solve the resulting equation or substitute
the value of y in (2).
From (4)

[ 2
Dy=D ( c 1 +c 2 t ) e−3 t− t+
9 27
4
]
= c2 e−3 t + ( c1 +c 2 t ) (−3 ) e−3 t −
2
9

∴ Substituting for y∧Dy in (2) ,we get

[( 1
2 ) ] t 1
x= c 1− c 2 + c 2 t e−3 t + +
9 27
………………….(5)

Since x= y =0 w h en t=0 , t h e equations ( 4 )∧ (5 ) give

−4 −2
c 1= andc 2=
27 9

Thus the required particular solution satisfying the given conditions is

−1 1 −2 2
x= ( 1+6 t ) e−3 t + ( 1+3 t ) ∧ y= ( 2+3 t ) e−3 t + ( 2−3 t )
27 27 27 27

Module 2 :Ordinary Differential Equations-II

Contents
Second order linear differential equations with variable coefficients, Method of variation of
parameters, Power series solutions; Legendre polynomials, Bessel functions of the first kind and
their properties.

Part-I : Second-Order Linear Differential Equations with Variable Coefficients

33
 B.Tech , Sem- II, Mathematics -II (BT -202 )

An equation is of the type

2
d y dy
2
+ P +Qy=R
dx dx
is called a Second-Order Linear Differential Equations with Variable Coefficients.
Where P∧Q be some functions of x only.
We shall study the following five methods for solving the Second-Order Linear Differential
Equations with Variable Coefficients.

Method-I: [Inspection Method] Complete solution in terms of one known integral belonging to the
complementary function :-

Step-1: Put the given equation in the standard form

2 2
d y dy d y
2
+ P +Qy=R … … … … … .(1) in which coefficient of 2 is unity.
dx dx dx

Step-2: Find an integral say u of C.F. by using the following table :

Sr. Condition satisfied An integral say u of C.Fis

No
.
1 1+ P+Q=0 u=e
x

2 1−P+Q=0 u=e
−x

3 2
a + aP+Q=0 u=e
ax

4 P+Qx=0 u=x
5 2
2+2 Px+Q x =0 u=x
2

6 2
m(m−1)+ Pmx +Q x =0 u=x where m=1 , 2 ,3 … … .
m

If an integral say uof C.F is given in the question, then this step is omitted .
Step-3: Now, assume that the complete solution of the given equation is
y=uv … … … … ( 2 )
Where u has been obtained in step (2). Then the given equation reduces to

( )
2
d v 2 du dv R
2
+ P+ = −−−−−−(3)
dx u dx dx u

2
dv d v dq
Step-4: Take =q so t h at 2
= ,∴Substituting this in (3),it will reduce to first order and
dx d x dx
first degree and we solve it for q as usual.

34
 B.Tech , Sem- II, Mathematics -II (BT -202 )
dv
Step-5: Now ,we replace q by and separate the variables v∧x . Integrate and determine
dx
v . Put this value of v in (2). We get the complete solution of the given equation.

Example-1 : Solvet h e differential equation

2
d y dy
x 2 −( 2 x−1 ) +(x−1) y =0
dx dx

Solution: Given equations can be written as

( ) ( )
2
d y 1 dy 1
2
− 2− + 1− y=0 … … … .(1)
dx x dx x

On comparing the above with

2
d y dy
2
+ P +Qy=R
dx dx

1 1
( )
∴ P=−(2− ) ,Q= 1− , R=0
x x

Here ,1+ P+Q=0 , S h owing t h at y =u=e x…………………..(2)

is a part of C.F. of solution of (1).

Let the complete solution of the given equation be

y=uv … … … … ( 3 )

( )
2
d v 2 du dv R
Now , we know that v is defined by 2
+ P+ =
dx u dx dx u

( )
2 x
d v 1 2 d e dv
∴ 2
+ −2+ + x =0
dx x e dx dx

()
2
d v 1 dv
¿> 2
+ =0 … … … … … ..(4 )
dx x dx

2
dv d v dq
Now to solve the above equation, Put =q so t h at 2
=
dx d x dx

dq q
()dq −q
+ =0=¿ =
dx x dx x

35
 B.Tech , Sem- II, Mathematics -II (BT -202 )
dq −dx
¿ =
q x
Integrate both sides

logq=−logx+ log c 1
Or
logq=−logx+ log c 1
log qx=log c 1

Or qx=c1

dv c 1
=
dx x
c1
¿> dv = dx
x
Integrate both sides,We get
v=c 1 logx+ c 2

Thus the complete general solution is

x
y=uv =e ( c 1 logx+ c2 )

Example-2 : Solvet h e differential equation

2
2 d y dy 1
x 2
+ x − y =0 , given t h at x + is one interal
dx dx x

Solution: Given equations can be written as

() ( )
2
d y 1 dy 1
2
+ − 2 y=0 … … … .(1)
dx x dx x

On comparing the above with

2
d y dy
2
+ P +Qy=R
dx dx

1 1
( )
∴ P=( ) ,Q=− 2 , R=0
x x

1
Here it is given that y=u=x + …………………..(2)
x
is a part of C.F. of solution of (1).

Let the complete solution of the given equation be

36
 B.Tech , Sem- II, Mathematics -II (BT -202 )

y=uv … … … … ( 3 )

( )
2
d v 2 du dv R
Now , we know that v is defined by 2
+ P+ =
dx u dx dx u

( ( )
x ) dx
d2 v 1 2 1 dv
∴ 2+ + 1− =0
( x)
2
dx x 1
x+

¿>
d2 v
dx
2
+
(
3 x 2−1 dv
x ( x 2+1 ) dx
=0
)
2
dv d v dq
Now to solve the above equation, Put =q so t h at 2
=
dx d x dx

+
(
dq 3 x 2−1
dx x ( x2 +1 )
q=0
)
Or
dq −1
q
+
(
+ 2
4x
x ( x +1 )
dx =0
)
Integrate both sides, We get
log q−logx+2 log ( x 2 +1 )=log c 1

c1 x
Or q= 2
( x 2+1 )

dv c1 x
∴ =
dx ( x 2 +1 )2
c1 x
¿> dv= 2
dx
( x 2+1 )
∴Integrate both sides,We get
−c1
v= + c2
2 ( x 2 +1 )

Thus the complete general solution is

y=uv = x + ( )( 2( x +1) +c )
1
x
−c 1
2 2

Method-II : [ Normal Form ] Complete solution by removal of first order derivative :-

37
 B.Tech , Sem- II, Mathematics -II (BT -202 )

If an integral included in the complementary function cannot be determined by Inspection

Method,it is sometimes useful to reduce the equation to normal form as follows in which the
term containing the first derivative is missing.

Step-1: Put the given equation in the standard form

2 2
d y dy d y
2
+ P +Qy=R … … … … … .(1) in which coefficient of 2 is unity.
dx dx dx

Step-2: To remove the first order derivative, we chooseu=e−∫ ( 2 ) Pdx

1

Step-3: Now, we choose complete solution of the given equation be

y=uv … … … … ( 2 )

Then the given equation reduces to normal form

2
d v
2
+ I v =S−−−−−−(3)
dx

1 2 1 dP R
Where I =Q− P − , S=
4 2 dx u
Now , the above equation can be solved by usual methods and we get v .

Step-4 : Thus the complete general solution is

y=uv .

Example-1 : Solvet h e differential equation

( )
2
d y dy
2
+ y cotx+2( + y tanx)=secx
dx dx
Solution: Given equations can be written as

2
d y dy
+2 tanx + ( 1+2 tan x ) y=secx tanx … … … .(1)
2
2
dx dx

On comparing the above with

2
d y dy
2
+ P +Qy=R
dx dx

∴ P=2 tanx ,Q=( 1+ 2 tan2 x ) , R=secx tanx

Now to remove the first order derivative, we chooseu=e−∫ ( 2 ) Pdx =e−∫ ( 2 ) 2tanx dx =cosx
1 1

38
 B.Tech , Sem- II, Mathematics -II (BT -202 )

∴we choosecomplete solution of the given equation be

y=uv … … … … ( 2 )

Then the given equation reduces to normal form

2
d v
2
+ I v =S−−−−−−(3)
dx

1 2 1 dP 1 1
=( 1+ 2 tan x ) − 4 tan x − 2 sec x=0 ,
2 2 2
Where I =Q− P −
4 2 dx 4 2
R 2
S= =sec xtanx
u
Hence the equation (3) becomes

2
d v 2
2
=sec xtanx
dx

Integrate both sides, We get

dv 1 2
= sec x+ c 1
dx 2
Again Integrate both sides

1
v= tanx+c 1 x + c2
2
Thus the complete general solution is
1
y=uv =cosx ( tanx+ c 1 x +c 2).
2
Example-2 : Solvet h e differential equation

( ) ( )( )
2
d y 2 dy 2 x
2
− + 1+ 2 y=x e …………………..(1)
dx x dx x
Solution: On comparing the above with
2
d y dy
2
+ P +Qy=R
dx dx

−2 Q=1+ 2 , R=x e x
∴ P= , 2
x x

Now to remove the first order derivative, we chooseu=e−∫ ( 2 ) Pdx =e−∫ ( 2 ) x dx=x
1 1 2

∴we choosecomplete solution of the given equation be

y=uv … … … … ( 2 )

39
 B.Tech , Sem- II, Mathematics -II (BT -202 )

Then the given equation reduces to normal form

2
d v
2
+ I v =S−−−−−−(3)
dx

1 2 1 dP
Where I =Q− P −
4 2 dx
2
x( ) ( ) ( )
= 1+ 2 −
1 4
4 x 2
−
1 2
2 x2
=1 ,

R x
S= =e
u
Hence the equation (3) becomes

2
d v
+ v=e =¿ ( D +1 ) v=e …………..(4)
x 2 x
2
dx

Now the auxiliary equation is m2 +1=0=¿ m=± i

Hence C.F. = ( c 1 cosx+c 2 sinx )

x x
X e e
∴P.I.¿ = 2 = ,
f ( D ) D +1 2
The, complete general solution of (4) is

x
e
v=C . F .+ P . I .=c1 cosx +c 2 sinx +
2

Thus the complete general solution of given equation is

( )
x
e
y=uv =x c 1 cosx + c 2 sinx+ .
2

Method-III: Method of change of Independent Variable i.e. Transformation of the equation by

changing of Independent Variable:--Some times by the change of Independent Variable, the
differential equation may become easily integrable.

Step-1: Put the given equation in the standard form

2 2
d y dy d y
2
+ P +Qy=R … … … … … .(1) in which coefficient of 2 is unity.
dx dx dx

Step-2: Now ,we assume a following relation between the new independent variable z and the
old independent variable x as

( )
2
dz
=+Q
dx
Note: Carefully that ,we omit -ve sign of Q

40
 B.Tech , Sem- II, Mathematics -II (BT -202 )

Now , we solve the above

dz
=√ Q
dx
Integrate both sides,
z=∫ √ Q dx ………………………………(2)
we omit the constant of Integration ,since ,we are interested in finding just a relation between
z and x .

Step-3: With the help of above relation (2),We transform the equation (1) as
2
d y dy
2
+ P1 +Q 1 y =R 1 … … … … … .¿)
dz dz
Where
2
d z dz
+P Q R
dx
2
dx Q1= R 1=
( ) ( )
2 2
P 1= , dz , dz
( )
2
dz
dx dx
dx

Step-4: After solving equation (3) by usual methods,the variable z is replaced by relation(2) .
Example-1 : Solvet h e differential equation

( ) ( )
2
d y dy 2 4
2
+ tanx −2cos x y=2 cos x …………………..(1)
dx dx
Solution: On comparing the above with
2
d y dy
2
+ P +Qy=R
dx dx

∴ P=tanx ,Q=−2cos 2 x , R=2 cos 4 x

Now , we assume a following relation between the new independent variable z and the old
independent variable x as

( ) ( )
2 2
dz dz dz
=2 cos x=¿ =√ 2 cosx
2
=+Q 0r
dx dx dx
¿> dz=√ 2 cosx dx
Integrate both sides, We get

z=√ 2 sinx ……………………………… (2)

Now , with this z ,we know that

2
d y dy
2
+ P1 +Q 1 y =R 1 … … … … … .¿)
dz dz

41
 B.Tech , Sem- II, Mathematics -II (BT -202 )

Where
2
d z dz
+P Q R z
2
dx
2
dx Q 1= =−1 R = =1−
( )
1

( )
2
P 1= =0, dz , dz
2
2 =
( )
2
dz
dx dx
dx

Hence the equation (3) becomes,

2 2
d y dy z
2
+0. − y=1−
dz dz 2

2 2
d y z
¿> 2
− y=1−
dz 2
2
z
¿> ( D −1 ) y=1− ……………….(4)
2
2

Now the auxiliary equation is m2−1=0=¿ m=±1

Hence C.F. = ( c1 e + c2 e )
z −z

2
z
1−
∴P.I. Z 2
¿ = 2
f ( D ) D −1

( )
2
2 −1 z
¿−( 1−D ) 1−
2

¿−[ 1+ D + D +… ] (1− )
2
2 4 z
2

¿−[ 1− + D (1− )+0+ 0 … 0 ]

2 2
z 2z
2 2

[ ]
2 2
z ( z
P . I .=− 1− + 0−1 ) =
2 2
The complete general solution of (4) is
2
z
2
( √ 2 sinx )
y=C . F .+ P . I .=( c1 e +c 2 e ) + =( c1 e
z −z √2 sinx
+ c2 e− √ 2 sinx
)+ 2
2
Example-2 : Solvet h e differential equation

( ) ( )
2
d y 1 dy 2 2 2
2
− −4 x y=8 x sin x …………………..(1)
dx x dx
Solution: On comparing the above with
2
d y dy
2
+ P +Qy=R
dx dx

42
 B.Tech , Sem- II, Mathematics -II (BT -202 )
−1
∴ P= ,Q=−4 x 2 , R=8 x 2 sin x 2
x
Now , we assume a following relation between the new independent variable z and the old
independent variable x as

( ) ( )
2 2
dz dz 2 dz
=+Q 0r =4 x =¿ =2 x
dx dx dx
¿> dz=2 x dx
Integrate both sides, We get

z=x ……………………………… (2)

2

Now , with this z , we know that

2
d y dy
2
+ P1 +Q1 y =R 1 … … … … … .¿ )
dz dz
Where
2
d z dz
+P Q R
dx
2
dx Q 1= =−1 R1= =2 sinz
( ) ( )
2 2
P 1= =0, dz , dz
( )
2
dz
dx dx
dx
Hence the equation (3) becomes,
2
d y dy
2
+0. − y=2 sinz
dz dz

2
d y
¿> 2 − y=2 sinz
dz
¿> ( D2−1 ) y=2 sinz ……………….(4)

Now the auxiliary equation is m2−1=0=¿ m=±1

Hence C.F. = ( c1 e + c2 e )
z −z

Z 2 sinz 2 sinz
∴P.I.¿ = 2 = =−sinz
f ( D ) D −1 −1−1
The complete general solution of (4) is

y=C . F .+ P . I =¿ ( c 1 e + c2 e )−sinz=( c1 e x +c 2 e−x )−sin x 2

2 2
z −z

Method-IV: Method of Variation of Parameters:-The Method of variation of parameters is used for

finding the complete solution of the equation
2
d y dy
2
+ P +Qy=R … … … … … .(1)
dx dx

43
 B.Tech , Sem- II, Mathematics -II (BT -202 )

,when the complementary function [C.F.] is known.Where , P , Q∧R be the some functions of
independent variable x
Let the C.F. be C.F.=( c 1 y 1+ c 2 y 2 ) … … … … …(2)
Where, y 1∧ y 2be the some functions of independent variable x .
Now according to this P.I. = A y 1 +B y 2 …………(3)
Where, A∧B be the some functions of independent variable x are given by

| |
y1 y2
y2 y1
A=−∫ R dx and B=∫ R dx with W = d y 1 d y2 ≠ 0
W W
dx dx

Thus the complete solution is

y=C . F .+ P . I .=c 1 y 1 +c 2 y 2+ A y 1+ B y 2

−1
Example -1: Solve by variation of parameters method ( D2−1 ) y =2 ( 1−e−2 x ) 2 .

Sol.: The auxiliary equation m2−1=0⇒ m=1 ,−1

Hence C.F.¿ c 1 e x + c 2 e− x. And

y2 X y1 X −1
P.I.¿− y 1∫ dx + y 2∫ dx , where y =e x , y =e− x , X =2 ( 1−e−2 x ) 2 and
W W 1 2

| |
x −x
e e
W= x −x =−2
e −e
−1 −1
e 2 ( 1−e ) e 2 ( 1−e )
−x −2 x 2 x −2 x 2 x −x
∫ ∫ dx = e I 1 +e I 2
x −x
⟹ P.I.=−e dx +e
−2 −2
−1

Where I 1=∫ e 2 ( 1−e )

−x −2 x 2 −x
e −x −x
dx=−∫ dx put e =t ⟹−e dx=dt
−2 √ 1−e−2 x

dt
⟹ I 1 =∫ =sin t=sin ( e )
−1 −1 −x

√ 1−t 2

−1

And I 2=∫ e 2 ( 1−e )

x −2 x 2 2x
e 2x 2x
dx=−∫ dx put e −1=t ⇒ 2 e dx=dt
−2 √ e −1
2x

44
 B.Tech , Sem- II, Mathematics -II (BT -202 )
1 dt
∫ =√ t=√ e −1. Hence P.I.=e x ⋅sin−1 ( e− x )+ e−x ⋅ √ e2 x −1
2x
⇒ I 2=
2 √t

y=CF+PI=c 1 e x + c 2 e− x + e x ⋅sin−1 ( e−x ) +e−x ⋅ √ e2 x −1

Example -2: Solve by variation of parameters method

2
( x−1 ) d y2 −x dy + y=(x−1)2
dx dx

Solution: Given equations can be written as

2
d y x dy 1
− + y=( x−1 ) … … … (1)
d x (x−1) dx (x−1)
2

On comparing the above with

2
d y dy
2
+ P +Qy=R
dx dx

−x 1
∴ P= ,Q= , R=( x−1 )
( x−1 ) ( x−1 )

∴ 1+ P+Q=0 P+Qx=0 ,Hence , y 1=e x , y 2=x be the integrals of C.F. of equation (1).
x
i.e. C.F. =c 1 y 1+ c2 y 2=c 1 e + c 2 x

| ||
y1 y2
We have W = d y 1
dx
ex x
d y2 = x
dx
e 1 |
=e x −x e x =(1−x)e x ≠0

Now P.I. = A y 1 +B y 2 …………(2) , where

y2 x
A=−∫ R dx=−∫ x
( x−1 ) dx=∫ x e−x dx=[−x e−x −e−x ]
W (1−x)e

and
x
e
B=∫ x
( x−1 ) dx=−∫ 1 dx=−x
(1−x)e

Hence the P.I. = A y 1 +B y 2=¿]e x −x 2=− ( x 2+ x +1 )

Thus the complete solution is

45
 B.Tech , Sem- II, Mathematics -II (BT -202 )

y=C . F .+ P . I =c 1 e x + c 2 x−( x 2 + x +1 ) .

Part-II : Power series solutions

1 Power Series Method

1.1 Power Series

am ( x  x0 )m = a0 + a1 ( x  x0 ) + a2 ( x  x0 )2 + . . .

where a0, a1 , . . ., are constants (coefficients)

x0 is a constant (center)

Taylor's Formula

f(x) = m + RN (x  x0)

If (x  x0) is sufficiently small, RN (x  x0)  0 as N , then, we say f(x) is analytic at x0, and

f(x) = m Taylor Series

When x0 = 0  Maclaurin Series

Examples: ex = = 1 + x + + + . . .

sin x =

cos x =

1.2 Basic Idea of the Power Series Method

In the previous discussion, the linear differential equations with constant coefficients were
solved and shown to have solution for

They can be anyone of the following 3 forms:

46
 B.Tech , Sem- II, Mathematics -II (BT -202 )
But, exponential, sine and cosine functions can be expressed in terms of Maclaurin series or
Taylor series expanded around zero.

Example. y'' + y = 0

Solution. Assume

y = am xm = a0 + a1 x + a2 x2 + . . .

y' = m am xm-1 = a1 + 2 a2 x + 3 a3 x2 + . . .

y'' = m ( m - 1 ) am xm-2 = 2a2 + 6 a3 x + 12 a4 x2 + . . .

Since y'' + y = 0

 (2a2 + 6a3 x + . . .) + (a0 + a1 x + a2 x2 + . . .) = 0

or (2a2 + a0) + (6a3 + a1) x + (12a4 + a2) x2 + . . . = 0

Since 1, x, x2, . . ., xn are linearly independent functions, we have

2a2 + a0 = 0 coefficients of x0

6a3 + a1 = 0 coefficients of x1

12a4 + a2 = 0 coefficients of x2

or a2, a4, a6, . . ., can be expressed in terms of a0

a3, a5, a7, . . ., can be expressed in terms of a1

where a0 and a1 are arbitrary constants. After solving the above simultaneous
equations, we have

a2 =  = 

a3 =  = 

a4 = . . . = ; ...

thus y = a0 + a1

= a0 cos x + a1 sin x

47
 B.Tech , Sem- II, Mathematics -II (BT -202 )

 Since every linear differential equation with constant coefficients always possesses a
valid series solution, it is natural to expect the linear differential equations with
variable coefficients to have series solutions too.

 Also, since the majority of series cannot be summed and written in a function form,
it is to be expected that some solutions must be left in series form.
y'' + p(x) y' + q(x) y = 0

where p(x) and q(x) are expressed in polynomials.

We assume

y = am xm = a0 + a1 x + a2 x2 + . . .

y' = m am xm-1 = a1 + 2 a2 x + . . .

y'' = m ( m  1 ) am xm-2 = 2 a2 + 3 2a3 x + . . .

(1) Put y, y' and y'' into the differential equation

(2) Collect terms of x0, x1, x2, . . .,

(3) Solve a set of simultaneous equations of a0, a1, a2, ....

2 Theory of Power Series Method

2.1 Introduction

Power Series:

S(x) = am ( x  x0 )m = a0 + a1 ( x  x0 ) + a2 ( x  x0 )2 + . . . (1)

Partial Sum:

Sn(x) = a0 + a1 ( x  x0 ) + a2 ( x  x0 )2 + . . . + an ( x  x0 )n (2)

Remainder:

Rn(x) = an+1 ( x  x0 )n+1 + an+2 ( x  x0 )n+2 + . . . (3)

Note that Rn = S  Sn or | Sn  S | = | R n |

Convergence:

48
 B.Tech , Sem- II, Mathematics -II (BT -202 )

Definition 1

If limSn(x1) = S(x1), then the series (1) converges at x = x1 and

Definition 2

If the series converges, then for every given positive number  (no matter how small,
but not zero), we can find a number N such that

| Sn  S | <  for every n  N

2.2 Radius of Convergence

We know that for the series xm = 1 + x + x2 + . . .

|x| > 1 divergent

|x| < 1 convergent

but for the series = 1 + x + + . . . ( = ex )

Convergent for all x.

If a series converges for all x in | x  x0 | < R

and diverges for | x  x0 | > R (0 < R < )

then R = radius of convergence

R = if series converges for all x.

R can be calculated by the following formula:

R =

or R = (Ratio Test)

Ratio Test

 = lim

= lim

if  > 1 divergent

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

 < 1 convergent

 = 1 test fails (i.e., inconclusive)

Note that radius of convergence R can be 'calculated' by

since  < 1 : convergence, we need

lim< 1

 | x - x0 | < = R (radius of convergence)

Example: ex = 1 + x + + . . . =

 = lim= lim

= lim= 0 < 1

 The series converges, i.e.,

R = , i.e.,

converges for all x.

Example: xm = 1 + x + x 2 + x3 + . . .

 = lim= lim| x | = | x |

thus, converges for | x | < 1 , diverges for |x|>1

and for x = 1, ratio test fails.

|x|< R =

i.e., converges for all x in | x | < 1.

In fact, this series converges to for 1 < x < 1.

Example: m! xm = 1 + x + 2x2 + 6x3 + . . .

 = lim= = x ( m + 1 ) =   1

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

thus, this series diverges for all x  0.

Example: x3m

This is a series in powers of t = x3 with coefficients am = , so that

 = lim=

thus, converges for

1 or |t| 8 or

i.e., |x|2

2.3 Properties of Power Series

(1) A power series may be differentiated term by term (Term-wise Differentiation).

i.e., if y(x) = am ( x  x0 )m, |x  x0| < R and R > 0

then y'(x) = m am ( x  x0 )m-1 =

(2) Two power series may be added term by term (Term-wise Addition).

i.e., if f(x) = am ( x  x0 )m; g(x) = bm ( x  x0 )m

then f(x) + g(x) = m

(3) Two power series may be multiplied term by term (Term-wise Multiplication).

f(x) g(x) = m

(4) Vanishing of all Coefficients (Linearly Independence).

if f(x) = am ( x  x0 )m = 0

for all x in |x  x0| < R, then

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

am = 0 for all m.

Let's ask ourselves a question: Can all linear second-order variable coefficient
differential equations be solved by power series method? Let us answer this question
by the following illustration:

Example: Solve the equations

x2 y'' + a x y' + b y = 0

where (i) a =  2, b = 2

(ii) a =  1, b = 1

(iii) a = 1, b = 1

Solution: we assume

y = cm xm ; y' = m cm xm-1;

y'' = m ( m  1 ) cm xm-2

 x2 y'' + a x y' + b y =

[m ( m  1 ) + a m + b ] cm xm = 0

Note that m (m - 1 ) + a m + b = 0 is the characteristic equation for Euler

equation.

Case (i) a =  2, b = 2

 ( m2  3 m + 2 ) cm xm = 0

or ( m2  3 m + 2 ) c m = 0

or ( m  2 ) ( m  1 ) cm = 0

 cm = 0 for all m  1 or 2

 y = c1 x + c2 x2

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Case (ii) a =  1, b = 1

 ( m2  2 m + 1 ) cm xm = 0

 ( m  1 )2 c m = 0

 cm = 0 for all m  1

 y = c1 x

In this case, power series method yields only one solution: y = c1 x.

We need another linearly independent solution to get the general

solution of the differential equation.

 Reduction of order: let y2 = x u

 x3 u'' + x2 u' = 0

 u = c ln|x|

 y = A x + B x ln|x|

Case (iii) a = 1, b = 1

 ( m2 + 1 ) c m = 0

 cm = 0 for all m

i.e., the power series method fails completely, but why??

By the way, the general solution of Case (iii ) is

y = A cos(ln|x|) + B sin(ln|x|)

2.4 Regular Point and Singular Point

Analytic Function: If g is a function defined on an interval I, containing a point x0, we say

that g is analytic at x0 if g can be expanded in a power series about x0 which has a positive
radius of convergence.
Any polynomial in x is analytic for all x.
Any rational function (ratio of polynomials) is analytic for all values of x which are not zeros
of the denominator polynomial.
Question: Are ex, , and analytic at x = 0?

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Theorem (Existence of Power Series Solutions)

If the function p, q, r in

y'' + p(x) y' + q(x) y = r(x)

are analytic at , then every solution y(x) of the above equation is analytic at
and can be represented by a power series of x - x0 with radius of
convergence , i.e. y = am ( x  x0 )m

Definition: Regular Point and Singular Point

We call x = 0 a regular point (or ordinary point) of the differential equation

y'' + p(x) y' + q(x) y = 0

when both p(x) and q(x) are analytic at x = 0.

If x = 0 is not a regular point, it is called a singular point of the differential

equation.

Example: x y'' + 2 y' + x y = 0

y'' + y' + y = 0

 x = 0 is a singular point!

 may give some trouble in power series method.

If we nonetheless assume (This is inappropriate!)

y = c m xm

the differential equation becomes

m ( m  1 ) cm xm-1 + 2 m cm xm-1 + cm xm+1 = 0

m ( m  1 ) cm xm-1 = ( k + 1 ) k ck+1 xk

2 m cm xm-1 = 2 ( k + 1 ) ck+1 xk

= 2 c1 + 2 ( k + 1 ) ck+1 xk

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

cm xm+1 = ck-1 xk

Thus we have

2 c1 + {[ ( k + 1) k + 2 ( k + 1 )] ck+1 + ck-1 } xk = 0

or 2 c1 + { ( k + 1 ) ( k + 2 ) ck+1 + ck-1 } xk = 0

 c1 = 0

ck+1 = for k  1

 c3 = c5 = c7 = . . . = 0

c2 =  c4 =

∴ y = c 0 = c0

Only one solution is obtained! The other linearly independent solution can be
obtained by the method of reduction of order:

 y2 = u

 y2 = (Exercise!)

 y = A+B

Note that

= x-1

This suggests that we may try y = xr (c0 + c1 + c2 x2 + . . .) in the first place to

obtain the second linearly independent solution.

3 Frobenius Method

3.1 General Concepts

y'' + p(x) y' + q(x) y = 0

If p(x), q(x) are analytic at x = 0

 x = 0 is a regular point, two linearly independent exist.

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 y = am x m

If p(x), q(x) are not analytic at x = 0  singular point

For x = 0 is a singular point, rewrite the differential equation in the following form:

y'' + y' + y = 0

If b(x), c(x) analytic at x =0

 regular singular point, at least one solution exist with the following form

 y = xr am xm
where r is a parameter which need to be determined. It can be positive or negative.

If b(x), c(x) not analytic at x=0

 irregular singular point, a non-trivial solution may or may not exist.

Theorem 1 (Frobenius Method)

Any differential equation of the form

y'' + y' + y = 0

where b(x) and c(x) are analytic at x = 0 ( a regular singular point)

has at least one solution of the form

y = xr amxm = xr ( a0 + a1x + a2 x2 + ... ),

where a0  0 and r may be any number ( real or complex ).

x=0 regular singular point!!!

3.2 Indicial Equation

y'' + y' + y = 0

or x2 y'' + x b(x) y' + c(x) y = 0

Since b(x) and c(x) are analytic, i.e.,

b(x) = b0 + b1 x + b2 x2 + . . .

c(x) = c0 + c1 x + c2 x2 + . . .

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We let y = xr am xm =

y' = ( m + r ) am xm+r-1 = xr-1 [ r a0 + ( r + 1 ) a1 x + . . . ]

y'' = ( m + r ) ( m + r  1 ) am xm+r-2

= xr-2 [ r ( r  1 ) a0 + ( r + 1 ) r a1 x + . . . ]

Put y, y', y'', b(x), c(x) into the differential equation and collect terms of x p, we have (for
xr terms)

[ r ( r  1 ) + b 0 r + c 0 ] a0 = 0

Since a0  0, we have

r ( r  1 ) + b 0 r + c0 = 0 Indicial Equation !!!

Two roots for r:

one r for y1 = xr am xm

another r  Theorem 2 for y2

Theorem 2 Form of the Second Solution

Case 1: r1 and r2 differ but not by an integer

y1 = x

y2 = x

Case 2: r1 = r2 = r, r =

y 1 = x r ( a 0 + a1 x + a 2 x 2 + . . . )

y2 = y1 ln x + xr (A1 x + A2 x2 + . . . )

Case 3: r1 and r2 differ by a nonzero integer, where r1 > r2

y1 = x

y2 = k y1 ln x + x

where r1  r2 > 0 and k may or may not be zero!!!

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Note that in Case 2 and Case 3,

the second linearly independent solution y2 can also be obtained by reduction

of order method ( i.e., by assuming y2 = u y1 ).

Case 1: r1 and r2 differ but not by an integer

Example: y'' + y' + y = 0, x>0 (Euler Equation)

Solution. y = xr ( a0 + a1 x + a2 x2 + . . . ) = xr am xm = am xm+r

y' = . . .

y'' = . . .

 am xr+m-2 = 0

For m = 0, am  0 ( )

thus,we have the indicial equation:

r(r1)+ r + = 0

or r1 = 1/4 and r2 = 1/2

Note that in this case, r1  r2 and r1  r2 is not an integer.

For r =, we have

y1 = x1/4 (a0 + a1 x + a2 x2 + . . . )

or am x= 0

or amx= 0

which is valid for all x  0.

Thus, we have

am m = 0 for all m (=0, 1, 2, …)

for m = 0 a0 = arbitrary constant

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but for m = 1, 2, … am = 0

 y1 = a0 x1/4

Similarly, for r = 1/2, we have

( by setting y2 = x1/2 ( A0 + A1 x + A2 x2 + A3 x3 + . . . , )

 y2 = A0 x1/2

Hence, the general solution is

y = a0 x1/4 + A0 x1/2

Example: y'' + y' + y = 0, x0

Solution. Letting y = xr am xm , we have

am xr+m-2 = 0

The indicial equation is ( )

r ( r  1 ) + r + 1 = r2 + 1 = 0

or r1 = i, r2 =  i, r1  r2 =2i is not an integer.

For r = i

am xi+m-2 = 0

or am [ ( i + m ) 2 + 1 ] = 0

or am m ( m + 2 i ) = 0

 m=0 am  0, i.e., a0 is an arbitrary constant

m0 am = 0

 y1 = a0 xi = a0 [ cos(lnx) + i sin(lnx) ]

= cos(lnx) + i sin(lnx) take a0 = 1

Similarly, for r =  i, we have (Exercise!)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

y2 = x-i = cos(lnx)  i sin(lnx)

Since the linear combinations of solutions are also solutions of the linear
differential equation, thus,

y1* = = cos(lnx)

y2* = = sin(lnx)

 y = c1 cos(lnx) + c2 sin(lnx)

Case 2, r1 = r2 = r , Double Roots

Example : y'' + y' + y = 0, x0

Solution: Letting y = xr am xm, we have

am ( r + m ) ( r + m - 1 ) xr+m-2 + am ( r + m ) xr+m-1

+ am xr+m-2 = 0

Since am ( r + m ) xr+m-1 = ak-1 ( r + k  1 ) xr+k-2

= am-1 ( r + m  1 ) xr+m-2

The differential equation becomes

a0 xr-2 +

xr+m-2 = 0

The indicial equation is

r ( r  1 ) + = 0 or = 0

or r1 = r 2 = r =

For r =

x= 0

or x= 0

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 am = – for m  1 (recurrence equation)

Hence

a1 = , a 2 = ,...

and y1 = a0 x

= x1/2 , x0

Note that we have set a0 = 1 in the above equation.

Approach 1

Since r = r1 = r2, another solution can be obtained by directly letting

y2 = y1 lnx + xr (A1 x + A2 x2 + ... )

 y2 = y1 lnx + x

Approach 2

We can also use the method of reduction of order to produce the second linearly
independent solution, y2, by letting

y2 = u y 1

Put into the differential equation,

u'' y1 + u' ( 2 y1' + y1 ) = 0

 =  2  1 = . . . (long division) = –  + . . .

i.e., ln u' =  ln x  + . . .

or u' = exp = . . .

By expanding the exponential function in Taylor series and then integrating

u = ln x  + . . .

 y2 = y1 u = y 1

= y1 ln x +

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Exercise : x y'' + ( 1  x ) y'  y = 0, x0

Case 3: r1 and r2 differ by an nonzero integer, r1 > r2

Example: x2 y'' + x y' + y = 0 (Bessel's equation of order 1/2)

Solution . Put y = xr am xm = am xm+r

the differential equation becomes

am ( r + m ) ( r + m  1 ) xr+m-2 + am ( r + m ) xr+m-2

 am xr+m-2 + am xr+m = 0

After substituting am-2 xr+m-2 for the last term of the lhs of the above equation,
we have

a0 xr-2

+ a1 xr-1

+ xr+m-2

= 0

Thus, we have the indicial equation:

r(r1)+r= 0

or r1 = r2 = 

Note that r1  r2 = 1 is an integer!!!

Note also that, when m=1, !

For r1 =

2 a1 x-1/2 + x = 0

 a1 = 0 and am =  for m  2

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 y1 =

For r2 =  , both a0 and a1 are arbitrary!

x= 0

 am = –

or a2 =  a0/2! a4 = a0/4! a6 =  a0/6!

a3 =  a1/3! a5 = a1/5! ...

 y2 =

Thus, the linearly independent solution is

(Alternatively, the linearly independent solution y2 can also be obtained by

reduction of order method.)

 y = A + B

Note that k=0 in this case!

Example: x2 y'' + x y' + (x2  1) y = 0 (Bessel's equation of order 1)

Solution. Letting y = xr am xm

we have

a0 [ r ( r  1 ) + r  1 ] xr-2 + a1[ r ( r + 1 ) + ( r + 1 )  1] xr-1

+ ( am ( ( r + m ) ( r + m  1 ) + ( r + m )  1 )  am-2 ) xr+m-2

= 0

 The indicial equation is

r ( r  1 ) + r  1 = r2  1 = 0

or

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

For r = 1, we have

3 a1 + ( am ( m2 + 2 m ) + am-2 ) xm-1 = 0

 a1 = 0 and

am =  for m  2

 y1(x) = x

For r =  1, we have

 a1 x-2 + { am ( m2  2 m ) + am-2 } xm-3 = 0

 a1 = 0 and

m ( m  2 ) am = - am-2 for m 2

but for m = 2, we have 0 = a0 which is not true.

Thus we can not obtain the second linearly independent solution by setting

y = xr am xm

with r =  1.

Approach 1

From the theorem, we need to directly assume that the second solution is of the
form:

y2 = k y1 lnx + x

= y1 ln x  x-1 + + . . . (Exercise!)

Approach 2

Note that the second linearly independent solution can also be obtained by the
method of reduction of order (Exercise!):

y2 = u y 1

=  = + + ...

or ln u' =  3 ln x + + . . .

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

u' = x-3 exp = x-3 + x-1 + . . .

or u = – x- 2 + ln x + . . .

Exercise: x y'' + (x  1) y'  2 y = 0

(A) J0(0) = 1, J1(0) = J2(0) = . . . = 0 ,J0() = J1() = . . . = 0

Plots of J0, J1, are given in the following figures:

(B) Bessel Function of Half Integer Order

J = sin x , J = cos x
1/2 -1/2

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4 Legendre's Equation

4.1 Legendre's Differential Equation :

( 1  x2 ) y''  2 x y' + n ( n + 1 ) y = 0

where n is any non-negative real number. Since n(n+1) is unchanged when n is

replaced by –(n+1), then (1) solution of n=n’ (where ) is the same as n=-(n’+1); (2)
solution of n=-n” (where ) is the same as n=n”-1. The above equation can be
written as

y''  y' + y = 0

But = 1 + x2 + x 4 + . . .

Which is analytic at x = 0 (regular point).

 We can solve the above equation by assuming

y = am xm

 Recurrence formula

am+2 =  am , m = 0, 1, . . .

or a2 =  a 0 a3 =  a 1

a4 = a 0

a5 = a 1

...

 The general solution is y = a0 y1 + a1 y2

where

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If n = 0, 1, 2, . . . (non-negative integer), then

y = c1 Pn(x) + c2 Qn(x)

where Pn(x) = Legendre polynomials [It is desirable that Pn(1) = 1]

Qn(x) = Legendre functions of the second kind converges in -1<x<1, but

Qn (1) = unbounded (This is due to the fact that the Legendre equation is not
analytic at x=+1 and x=-1)

4.2 Legendre Polynomials Pn(x)

Since for m = 0, 1, . . .

when n = non-negative integer,

am+2 = 0 for m = n

i.e., an+2 = an+4 = an+6 = . . . = 0

when n = even, y1  polynomial of degree n

n = odd, y2  polynomial of degree n

These polynomials, multiplied by an appropriate constant, are called the Legendre polynomials
Pn(x), which have the value Pn(1) = 1. In other words, let

Specifically, we have to choose 1

an = 1 if n = 0

an = ; if n = 1, 2, . . .

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Then, we can obtain the other coefficients in Pn(x)

an-2 = =

...

an-2m = (1)m

Then, the Legendre polynomial of degree n, Pn(x) is given by

Pn(x) = (  1 )m xn-2m

where M =

Example:

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In all cases, Pn(1) = 1, and Pn(1) = (1)n.

4.3 Some Important Properties of Pn(x)

(1) Values of Pn(x)

Pn(1) = 1

Pn(1) = (1)n

Pn(x) = (1)n Pn(x)

n : even, Pn(x) : even function†

n : odd, Pn(x) : odd function

Pn'(x) = (1)n+1 Pn'(x)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

(2) Rodrigues' Formula

Pn(x) = [ ( x2  1 )n ]

Exercise: Show that P2(x) =

(3) Generating Function for Legendre Polynomials

= Pn(x) tn

(4) Recurrence Formulas

(i) ( n + 1 ) Pn+1(x) = ( 2 n + 1 ) x Pn(x)  n Pn-1(x), n = 1, 2, . . .

(ii) Pn+1'(x)  Pn-1'(x) = ( 2 n + 1 ) Pn(x)

Exercise: starting with P0 = 1, P1 = x, calculate P2, P3, P4, . . .

(5) Integrating Formulas : Orthogonal Property

(i) = n = 0, 1, . . .

(ii) Pm(x) Pn(x) dx = 0, m  n, m, n  N

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Module 3: Partial Differential Equations

Contents
Formulation of Partial Differential equations, Linear and Non-Linear Partial Differential
Equations, Homogeneous Linear Partial Differential Equations with Constants Coefficients.

Basic concepts and definitions

An equation containing the dependent and independent variables and one or more

partial derivatives of the dependent variable is called a partial differential equation. In general

it may be written in the form

F (x,y,.....,u,ux, uy,......,uxx, uyy,......)=0

involving several independent variables x,y,.....,an unknown function u of these variables and

the partial derivatives ux, uy,......,uxx, uxy, uyy ........of the function. is considered in a suitable

subset D of Rn. For the sake of convenience we confine our discussion for n=2. However

extension of properties discussed here to higher values of n is possible.

Here as in the case of ordinary differential equations, we define the order of a partial

differential equation to be the order of the derivative of highest order occurring in the

equation. The power of the highest order derivative in a differential equation is called the

degree of the partial differential equation.

∂u ∂u
Example: (a) x ∂ x + y ∂ y =0 is a first-order equation in two variables with variable coefficients.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

∂u ∂u
(b) a ∂ x + b ∂ y =c; where x,y are independent variables, a and b are constants; is partial

differential equation of first-order with constant coefficients.

∂u ∂u
(c) ∂ x + ∂ y -(x+y) u=0 is a partial differential equation of first-order.

∂2 u ∂2 u ∂2 u ∂u ∂u
2
a(x) ∂ x +2b(x) ∂ x ∂ y +c(x) ∂ y =x+y+u+ ∂ x + ∂ y is a partial differential equation of
2
(d)

second-order.

∂2 u ∂2 u ∂2 u ∂u ∂u
2
a(x) ∂ x +2b(x) ∂ x ∂ y +c(x) ∂ y = f(x,y,u, ∂ x , ∂ y )
2
(e)

∂u ∂u
where a(x), b(x) and c(x) are functions of x and f(..,..,.,.,.) is a function of x,y,u, ∂ x and ∂ y

, is a partial differential equation of second order.

∂2 u ∂u
(f) u ∂ x ∂ y + ∂ x =y is a partial differential equation of second-order.

∂2 u ∂2 u ∂2 u
(g) ∂ x 2 + 2y ∂ x ∂ y + 3x ∂ y 2 = 4 sin x is a partial differential equation of second-order and

degree one.

∂2 u ∂2 u
(h) ∂ x 2 = ∂ y 2 is a partial differential of second-order.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

( ) ( )
2
∂ u 2 ∂u
(i) ∂ x + ∂ y = 1 is a partial differential equation of first-order and second degree.

By a solution of a partial differential equation of the type

F (x,y,u, ux,uy,uxx,uyy, uxy) =0

we understand functions u=(x,y) which satisfy identically in D, that is, if we put values of

quantities on the left hand side we get right hand side.

The general solution of a linear partial differential equation is a linear combination of

all linearly independent solutions of the equation with as many arbitrary functions as the order

of the equation; a partial differential equation of order 2 has 2 arbitrary functions. A particular

solution of a differential equation is one that does not contain arbitrary functions or constants.

Homogeneous linear partial differential equation has an interesting property that if u is its

solution then a scalar multiple of u, that is, cu, where c is a constant, is also its solution. Any

equation of the type F(x,y,u,c1,c2)=0, where c1 and c2 are arbitrary constants, which is a solution

of a partial differential equation of first-order is called a complete solution or a complete

integral of that equation. An equation F(,)=0 involving arbitrary function. F connecting two

known functions  and  of x, y and u, and providing a solution of a first order differential

equation is called a general solution or general integral of that equation.

Formation of Partial Differential Equations:-

(i) By eliminating arbitrary constants :

Let φ ( x , y . z . a . b )=0…(i)

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Where a and b are arbitrary constants, x and y are independent variables and z is

dependent variable.

Differentiate φ ( x , y . z . a . b )=0 w. r. t. x, we get

∂φ ∂φ ∂ z
+ . =0
∂ x ∂ z ∂x

∂φ ∂φ
Or + . p=0 … (ii)
∂x ∂z

Differentiate φ ( x , y . z . a . b )=0 w. r. t. y, we get

∂φ ∂φ ∂ z
+ . =0
∂ y ∂z ∂ y

∂φ ∂φ
Or + .q=0 … (iii)
∂ y ∂z

Eliminating a and b from (i), (ii) and (iii), we get

Fx , y , z , p ,q ¿=0

Which is required Partial Differential Equation of first order.

(ii) By eliminating arbitrary functions :

Let u and v be two given functions of x,y z connected by the relation

f ( u , v )=0

Where f is an arbitrary function

The linear PDE is Pp+Qq=R

Where,

| |
∂u ∂u
∂(u , v ) ∂ u ∂ v ∂u ∂ v ∂ y ∂z
P= = . − =
∂( y , z ) ∂ y ∂ z ∂ z ∂ y ∂ v ∂v
∂y ∂z

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| |
∂u ∂u
∂(u , v) ∂ u ∂ v ∂ u ∂ v ∂ z ∂x
Q= = . − . =
∂ (z , x) ∂ z ∂ x ∂ x ∂ z ∂ v ∂v
∂z ∂x

| |
∂u ∂u
∂ (u , v ) ∂u ∂ v ∂ u ∂ v ∂ x ∂y
R= = . − . =
∂ (x , y ) ∂ x ∂ y ∂ y ∂ x ∂ v ∂v
∂x ∂y

Example-1: Form partial differential equations from the following equations by eliminating

arbitrary constants

(i) z=ax+by + ab (ii) 2

z=ax+ a y +b
2

Solution:

(i) Given z=ax+by + ab

Differentiating partially differentiating w.r.t. x and y, we get

∂z ∂z
=aand =b
∂x ∂y

Putting the values of a and b in the given equation, we get

z= px +qy + pq

which is required PDE.

(ii) Given z=ax+ a2 y 2 +b

Differentiating partially differentiating w.r.t. x and y, we get

∂z ∂z 2
=aand =2 a y
∂x ∂y

Eliminating a from the above equations, we get

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( )y
2
∂z ∂z
=2
∂y ∂x

2
q=2 p y

Example-2: Form partial differential equations from the following equations by eliminating

functions constants (i) z=f (x 2− y 2 ) (ii) z= y 2 +2 f ( 1x + logy )

Solution:

(i) Given z=f (x 2− y 2 )

Differentiating partially differentiating w. r.t. x and y, we get

∂z ' 2
=f ( x − y ) .2 x
2
∂x

∂z
=−f ( x − y ) .2 y
' 2 2
∂y

From the above equations, we get

p −x
=
q y

py+ qx=0

Which is required PDE.

(ii)
2
Given z= y +2 f ( 1x + logy )
Differentiating partially differentiating w.r.t. x and y, we get

∂z
∂x
=2 f (
' 1
x
+logy
−1
x
2 )( )
∂z
∂y
1
(
=2 y +2 f ' +logy
x )( 1y )
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Eliminating a from the above equations, we get

−1
p x
=
q−2 y 1
y

p −y
=
q−2 y x

px+ y ( q−2 y )=0

Which is required PDE.

Lagrange's Method :

The general form of first-order linear partial differential equations with variable

coefficients is

P(x,y)ux+Q(x,y)uy+f(x,y)u=R(x,y)

We can eliminate the term in u from by substituting u=ve -(x,y), where (x,y) satisfies the

equation

P(x,y) x(x,y)+ Q (x,y) y(x,y)=f(x,y)

Hence, Eq is reduced to

P(x,y)ux+Q (x,y) uy =R(x,y)

where P,Q,R in are not the same as in The following theorem provides a method for solving

often called Lagrange's Method.

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Theorem : The general solution of the linear partial differential equation of first order

Pp+Qq=R;

∂u ∂u
,q =
where p= ∂ x ∂ y , P, Q and R are functions of x y and u

is F(, ) = 0

where F is an arbitrary function and  (x,y,u) =c1 and  (x,y,u)=c2 form a solution of the

auxiliary system of equations

dx dy du
= =
P Q R

Proof: Let  (x,y,u)=c1 and  (x,y,u)=c2 satisfy then equations

xdx+y dy +udu=0

and

dx dy du
= =
P Q R

must be compatible, that is, we must have P x+Qy+Ru=0

Similarly we must have

Px+Qy+Ru=0

Solving these equations for P,Q, and R, we have

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P Q R
= =
∂(ϕ, ψ )/∂( y , u) ∂(ϕ ,ψ )/∂(u , x) ∂(ϕ ,ψ )/∂( x , y)

where  (,)/(y,u)= yu- yu0 denotes the Jacobian.

Let F(,)=0. By differentiating this equation with respect to x and y, respectively, we

obtain the equations

{
∂F ∂ϕ ∂ϕ
+
∂ϕ ∂ x ∂u
p + } {
∂ F ∂ ψ ∂ψ
+
∂ψ ∂ x ∂ u
p =0 }
{
∂ F ∂ϕ ∂ ϕ
+ q +
∂ϕ ∂ y ∂u } {
∂ F ∂ψ ∂ψ
+ q =0
∂ψ ∂ y ∂u }
∂F ∂F
and if we now eliminate ∂ ϕ and ∂ψ from these equations, we obtain the equation p
∂(ϕ, ψ ) ∂(ϕ, ψ ) ∂(ϕ, ψ )
∂( y , u ) +q ∂(u , x) = ∂( x , y )

Substituting from equations into equation we see that F(,)=0 is a general solution of

The solution can also be written as

 =g() or =h(),

Example: Find the general solution of the partial differential equation y2up + x2uq = y2x

Solution: The auxiliary system of equations is

dx dy du
2
= 2 = 2
y u x u xy

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Taking the first two members we have x 2dx = y2dy which on integration given x3-y3 = c1.

Again taking the first and third members,

we have x dx = u du

which on integration given x2-u2 = c2

Hence, the general solution is

F(x3-y3,x2-u2) = 0

Char pit’s Method for solving nonlinear Partial Differential Equation of First-Order :

We present here a general method for solving non-linear partial differential equations.

This is known as Charpit's method.

Let F(x,y,u, p.q)=0

be a general non linear partial differential equation of first-order. Since u depends on x

and y, we have

du=uxdx+uydy = pdx+qdy

∂u ∂u
where p=ux= ∂ x , q = uy= ∂ y

If we can find another relation between x,y,u,p,q such that

f(x,y,u,p,q)=0

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then we can solve for p and q and substitute them in equation This will give the solution

provided is integrable.

To determine f, differentiate and w.r.t. x and y so that

∂F ∂F ∂F ∂ p ∂F ∂q
+ p+ + =0
∂ x ∂u ∂ p ∂ x ∂q ∂ x

∂f ∂f ∂ f ∂ p ∂ f ∂q
+ p+ + =0
∂ x ∂u ∂ p ∂ x ∂q ∂ x

∂ F ∂ F ∂ F ∂ p ∂ F ∂q
+ q+ + =0
∂ y ∂u ∂ p ∂ y ∂q ∂ y

∂f ∂f ∂f ∂ p ∂f ∂q
+ q+ + =0
∂ y ∂ u ∂ p ∂ y ∂q ∂ y

∂p ∂q
Eliminating ∂ x from, equations and ∂ y from equations and we obtain

( ∂∂ Fx ∂∂ fp − ∂∂ fx ∂∂ Fp )+( ∂∂uF ∂∂ fp − ∂∂fu ∂∂ Fp ) p+( ∂∂ Fq ∂∂ fp − ∂∂qf ∂∂ Fp ) ∂dxq =0

( ∂∂ Fy ∂∂qf −∂∂ fy ∂∂ Fq )+( ∂∂uF ∂∂ qf − ∂∂uf ∂∂qF ) q +( ∂∂ Fp ∂∂ fq − ∂∂ fp ∂∂ Fq ) ∂dyp =0
Adding these two equations and using

∂q ∂2 u ∂ p
= =
∂ x ∂ x∂ y ∂ y

and rearranging the terms, we get

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(−∂∂ Fp ) ∂∂ fx +(−∂∂ Fq ) ∂f∂ y +(− p ∂∂ Fp −q ∂∂qF ) ∂∂uf +(∂∂ Fx + p ∂∂ uF ) ∂∂ fp

+ ( +q ) =0
∂ F ∂f ∂f
∂ y ∂u ∂ q

Following arguments in the proof of Theorem we get the auxiliary system of equations

dx dy du dp dq df
= = = = =
−∂ F −∂ F ∂F ∂F ∂ F ∂ F ∂F ∂F 0
− p −q +p +q
∂p ∂q ∂ p ∂ q ∂ x ∂u ∂ y ∂ u

An Integral of these equations, involving. p or q or both, can be taken as the required

equation. p and q determined will make integer able.

Example: Find the general solution of the partial differential equation.

( ) ( )
∂u 2
∂x
x+
∂u 2
∂y
y −u =0
………………… (1)

∂u ∂u
Solution: Let p = ∂ x , q = ∂ y

The Char pit’s auxiliary system of equations is

dx dy du dp dq
= = = =
2 px 2 qy 2( p x+q y ) p−p q−q 2
2 2 2

Which, we obtain from (1) by putting values of

∂F ∂F ∂F ∂F ∂F
=2 px , =2qy , = p2 , = −1, =q2
∂p ∂q ∂x ∂u ∂y

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and multiplying by -1 throughout the auxiliary system. From first and 4 th expression we get

p 2 dx +2 pxdp
dx = py . From second and 5th expression

q2 dy+2qydq
dy= qy

Using these values of dx and dy in 1st and 2nd expression of Char pit’s auxiliary system ,

we get

p 2 dx +2 pxdp q2 dy+ 2qydq

p2 x = q2 y

dx 2 dy 2 dq
+ dp = +
or x p y q

Taking integral of all terms we get

ln|x| + 2ln|p| = ln|y|+2ln|q|+lnc

or ln|x| p2 = ln|y|q2c

or p2x=cq2y, where c is an arbitrary constant.

for p and q we get cq2y+q2y -u=0

(c+1)q2y=u

{ } { }
1 1
u 2 cu 2

q=
(c+1) y p=
(c+1) x

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Therefore

du =p dx +q dy

{ } { }
1 1
cu 2 dx + u 2 dy

du=
(c+1) x (c +1) y

( ) () ()
1 1 1
1+c 2 du= c 2 dx+ i 2 dy
or u x y

1 1 1

By integrating this equation, we obtain (( 1+ c )u ) 2 =( cx ) 2

+( y ) 2 + c 1

This is a complete solution.

Solutions of special type of nonlinear Partial Differential Equation of First-Order :-

(i) Equations containing p and q only

Let us consider a partial differential equation of the type

F(p,q)=0

The auxiliary system of equations of Charpit's method takes the form

dx dy du dp dq
= = = =
F p F q pF p +qF q 0 0

It is clear that p=c is a solution of these equations. Putting value of p in we have

F(c,q)=0

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So that q=G(c) where c is a constant

Then observing that

du= cdx +G(c) dy

we get the solution u=cx +G(c) y+c1,

where c1 is another constant.

Example: Solve p2+q2=1

Solution: The auxiliary system of equation is

dx dy du dp dq
= = = =
- −2 p 2 q −2 p 2
−2q 2 0 0

dx dy du dp dq
= = 2 2= =
or p q p +q 0 0

Using dp =0, we get p=c and q= √ 1−c 2 , and these two combined with
du =pdx+qdy yield

u=cx+y √ 1−c 2 + c which is a complete solution.

1

dx dx
Using du = p , we get du = c where p= c

x
Integrating the equation we get u = c + c1

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dy
2
√
Also du = q , where q = 1− p = 1−c
2
√
dy 1
or du = √1−c 2
. Integrating this equation we get u = √1−c 2 y +c 2

This cu = x+cc1 and u √ 1−c 2 = y + c √ 1−c 2

2

Replacing cc1 and c2 √ 1−c 2 by -  and - respectively, and eliminating c, we get

u2 = (x-)2 + (y-)2

This is another complete solution.

This is another complete solution.

(ii) Clairaut’s equations

An equation of the form

u=px+qy+f(p,q) or F=px+qy+f(p,q)-u=0

is known as Clairaut’s equation.

The auxiliary system of equations for Clairaut’s equation takes the form

dx dy du dp dq
x+ f p = y+ f q = px +qy + pf p +qf q = 0 = 0

From here we find that

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dp=0, dq=o implying

p=c1, q=c2

If we put these values of p and q in we get

u = c1 x +c2y +f (c1, c2)

Therefore, F(x,y,u,c1,c2) = c1x + c2y + f (c1,c2) -u=0 is a complete solution of

(iii) Equations not containing x and y

Consider a partial differential equation of the type

F(u,p,q) = 0

The auxiliary system of equations take the form

dx dy du dp dq
F p = F q = pF q +qF q = −pF u = −qF u

dp dq
The last two terms yield p = q

i.e. p = a2q where a2 is an arbitrary constant

This equation together with 11.43 can be solved for p and q and we proceed as in

previous cases.

Example: Solve u2+pq – 4 = 0

Solution: The auxiliary system of equations is

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dx dy du dp dq
q = p = 2 pq = −2up = −2uq

The last two equations yield p = a2q.

Substituting in u2+pq – 4 = 0 gives

1
± √ 4−u2 2
√
q= a and p = + a 4−u

Then du = pdx+qdy yields

√ 4−u2 ( adx+ 1a dy )
du = +

du 1
adx + dy
or √ 4−u = +
2
a

u
Integrating we get sin--1 2 = +
1
(
adx+ y +c
a )

or u = + 2 sin
( ax+
1
a
y +c )
which is the required complete solution.

(iv) Equations of the type

f(x,p) = g(y,q)

Then each of these functions must be constant, that is

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f(x, p) = g(y, q) = C

Solving for p and q, and using du=pdx+qdy we can obtain the solution

Example: Solve p2(1-x2)-q2(4-y2) = 0

Solution: Let p2(1-x2) = q2 (4-y2) = a2

a a
This gives p = √ 1−x and q = √ 4− y
2 2

( neglecting the negative sign).

Substituting in du = pdx + q dy we have

a a
du = √ 1−x dx + √ 4− y dy
2 2

Integration gives u = a
(sin'x + sin' 2y ) + c.

This is the required complete solution.

Partial Differential Equations Of Higher Order With Constant Coefficients:-

Partial differential equations (PDEs) are equations involving functions of more than one variable
and their partial derivatives with respect to those variables. Most (but not all) physical models
in engineering that result in partial differential equations are of at most second order and are
often linear. (Some problems such as elastic stresses and bending moments of a beam can be
of fourth order). In this course we shall have time to look at only a very small subset of second
order linear partial differential equations.

Homogeneous Linear Equations with constant Coefficients.

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A homogeneous linear partial differential equation of the nth order is of the form

homogeneous because all its terms contain derivatives of the same order. Equation (1) can be
expressed as

As in the case of ordinary linear equations with constant coefficients the complete solution of
(1) consists of two parts, namely, the complementary function and the particular integral.

The complementary function is the complete solution of f (D,D') z = 0-------(3)

Which must contain n arbitrary functions as the degree of the polynomial f(D,D'). The particular
integral is the particular solution of equation (2).

Finding the complementary function:

Let us now consider the equation f(D,D') z = F (x,y). The auxiliary equation of (3) is obtained by
replacing D by m and D' by 1.

Solving equation (4) for „m‟, we get „n‟ roots. Depending upon the nature of the roots, the
Complementary function is written as given below:

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Finding the particular Integral

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Expand [ f (D,D')]-1 in ascending powers of D or D' and operate on xm yn term by term.

Case (iv) : When F(x,y) is any function of x and y.

into partial fractions considering f (D,D') as a function of D alone.

Then operate each partial fraction on F(x,y) in such a way that

where c is replaced by y+mx after integration

Example 1. Solve(D3 – 3 D2 D'+ 4 D ' 3)z =e x+2 y

Solution: The auxillary equation is m3 – 3 m2+ 4=0

The roots are m = -1,2,2

Therefore the C . F=f 1( y−x)+ f 2 ( y +2 x)+ x f 3 ( y +2 x) .

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Hence, the solution is z = C.F. + P.I

Example2. Solve (D2 – 4 DD '+ 4 D ' 2 ) z=cos(x – 2 y )

Solution: The auxiliary equation is m2 – 4 m+ 4=0

Solving, we get m = 2, 2.

Therefore the C . F=f 1( y+ 2 x )+ x f 2 ( y +2 x )

cos ⁡( x−2 y )
The solution is z=f 1 ( y +2 x ) + x f 2 ( y +2 x ) +
25

Example 3. Solve ( D 2 – 2 DD ' ) z=x 3 y +e 5 x

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Solution: The auxiliary equation is m2 – 2 m=0.

Solving, we get m = 0, 2.

Hence the C . F=f 1( y)+ f 2 ( y +2 x).

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Example 4. Solve ( D2 + D D' −6 D'2 ) z = ycosx

Solution: The auxiliary equation is m2 +m – 6=0

Therefore, m = –3, 2.

Hence the C . F=f 1( y−3 x)+ f 2 ( y +2 x)

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¿(c+ 3 x )(– cosx) – (3)(−sinx)– 2 sinx

¿ – y cosx+ sinx

Hence the complete solution is

z=f 1 ( y−3 x )+ f 2 ( y+ 2 x ) – y cosx + sinx

Example 5. Solve r – 4 s+ 4 t=e 2 x+ y

Solution: The auxiliary equation is m2 – 4 m+ 4=0.

Therefore, m = 2, 2

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Hence the C . F is f 1 ( y+ 2 x )+ x f 2( y+ 2 x ).

Since D2 – 4 D D' + 4 D'2=0 for D = 2 and D' = 1, we have to apply the general rule.

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Module 4: Functions of Complex Variable

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Contents
Functions of Complex Variables: Analytic Functions, Harmonic Conjugate, Cauchy-Riemann Equations
(without proof), Line Integral, Cauchy-Goursat theorem (without proof), Cauchy Integral formula
(without proof), Singular Points, Poles & Residues, Residue Theorem, Application of Residues theorem
for Evaluation of Real Integral (Unit Circle).

COMPLEX VARIABLES

If z=x +iy , then Z is called a complex variable. Also x and y are respectively called real and
imaginary parts of z. sometimes we express z as z=x +iy=(x , y).

We also write R(z) = x and I(z) = y.

The complex conjugate, or briefly conjugate of z=x−iy .

For example, conjugate z=−3−5 i is z=−3+5 i

z+ z z+z
It is easy to verify that: R ( z )=x= , and I ( z )= y= .
2 2i

POLAR FORMS OF A COMPLEX NUMBER

The polar form of a complex number is another way to represent a complex number. The form
z=a+bi is called the rectangular coordinate form of a complex number.

The horizontal axis is the real axis and the vertical axis is the
imaginary axis. We find the real and complex components in
terms of r and θ where r is the length of the vector and θ is the
angle made with the real axis.

From Pythagorean Theorem:

2 2 2
r =a +b

By using the basic trigonometric ratios:

cosθ=a/r and sinθ=b /r

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Multiplying each side by r:

rcosθ=a and rsinθ=b

The rectangular form of a complex number is given by z=a+bi

Substitute the values of a and b in z=a+bi, Therefore

z=rcosθ+ ( rsinθ ) i

i.e. z=r ( cosθ +isinθ ) or z=ℜiθ

In the case of a complex number, r represents the absolute value or modulus and the angle θ is
called the argument of the complex number.

This can be summarized as follows:

The polar form of a complex number z=a+bi is
z=r ( cosθ+isinθ ) , where ¿∨z∨¿ √ a 2+b 2 , a=rcosθ and b=rsinθ , and θ=tan−1(b /a) for a>0
and
θ=tan−1(b /a)+ π
Or θ=tan−1(b /a)+180 ° for a< 0.

ANALYTIC FUNCTION

A function f (z) is said to be analytic in a region R of z-plane, if it is uniquely differentiable at

every point of R .
Necessary condition or Cauchy-Riemann equations for analytic function :
The necessary condition for the function w = f(z) = u(x,y) + i v(x,y) to be analytic in a region R is that
u(x,y) and v(x,y) satisfy the following equations
∂u ∂v ∂u −∂ v
= and =
∂x ∂ y ∂ y ∂x
u x =v y and u y =−v x
Which are known as Cauchy-Riemann equations.
Sufficient condition for analytic function:
If w = f(z) = u(x,y) + i v(x,y) be a function of complex variable defined in the region R such that

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
∂ u ∂ v ∂u ∂ v
I) , , , are continuous functions of x and y in the region R and
∂ x ∂ y ∂ y ∂x

∂ u ∂ v ∂u −∂ v
II) = , =
∂x ∂ y ∂ y ∂x

Then the function w = f(z) = u(x,y) + i v(x,y) is analytic and

∂u ∂v
f '(z) = ¿ +i
∂x ∂ x
CAUCHY-RIEMANN EQUATIONS [ In Polar Form ]:
Let w=f ( z ) =u+iv=f (x +iy)
Put x=rcosθ and y=rsinθ, then
u+iv=f (rcosθ+irsinθ)
u+iv=f ( r eiθ ) …(i)
Differentiate equation (i) partially w.r.t. r
∂ u ∂ v ' iθ iθ
+i =f ( r e ) e … (ii)
∂r ∂r
Differentiate equation (i) partially w.r.t. θ
∂ u ∂ v ' iθ
=f ( r e ) ir e …(ii)
iθ
+i
∂ θ ∂θ
∂u ∂v
+i
∂ θ ∂θ
=r i − (
∂u ∂v
∂r ∂r )
Equating real and imaginary parts, we get
∂u ∂v
=−r
∂θ ∂r
∂v ∂u
=r
∂θ ∂r
Which is the polar form of C-R equations.
Example 1: Test the analytic behavior of logz
Solution: let f ( z )=logz
u+iv=log ( x +iy )
u+iv=log ⁡(rcosθ+irsinθ)
iθ
u+iv=log ⁡(ℜ )
u+iv=log r +iθ

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Equating real and imaginary part

u=logr and v=θ

Differentiating partially w.r.t r and θ

∂u 1 ∂v
= and =0
∂r r ∂r
∂u ∂v
=0 and =1
∂θ ∂θ
∂u 1 ∂ v ∂ v −1 ∂u
It is clear that = and =
∂r r ∂θ ∂r r ∂θ
The function logz is analytic except at origin.
Example 2 : Show that f ( z )=z 2 is an analytic function.

Solution: Since z=x +iy

2 2
Therefore z =( x+iy)
2 2 2
z =x − y +2 ixy

Hence, f ( z )=u +iv=x 2− y2 +2 ixy

Equating the real and imaginary parts,

2 2
∴ u=x − y , v =2 xy

∂u ∂v
=2 x , =2 y ,
∂x ∂x

∂u ∂v
= - 2 y, =2 x ,
∂y ∂y

∂ u ∂ v ∂u −∂ v
Hence, = , = ie ,u∧v satisfy Cauc h y – Riemann equations .
∂x ∂ y ∂ y ∂x
Therefore, f(z) is analytic.
HARMONIC FUNCTION
A real valued function u=u(x , y ) is called harmonic function, if
2 2
∂ u ∂ u
2
+ 2 =0
∂x ∂y
i.e. u satisfies Laplace’s equation

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2 2
2 ∂ u ∂ u
∇ u= 2
+ 2 =0
∂x ∂ y
Example : Prove that, if f(z)=u+iv is analytic function in domain D, then u and v are harmonic.
Solution: Since f(z)=u+iv is analytic,it means it is satisfies C-R equations.

∂u ∂v ∂u −∂ v
i.e. = and =
∂x ∂ y ∂ y ∂x
Differentiate w.r.t. x and y, we get
2 2 2 2
∂ u ∂ v ∂ u −∂ v
2
= and 2
=
∂ x ∂ x∂ y ∂ y ∂ y∂x
By adding both the equation, we get
2 2
∂ u ∂ u
2
+ 2 =0
∂x ∂y
Here, u is harmonic
Similarly,
2 2
∂ v ∂ v
2
+ 2 =0
∂x ∂y
Here, v is harmonic.
METHODS FOR CONSTRUCTING AN ANALYTIC FUNCTION:-
Method-I
If u is given function then to find v:
∂v ∂v
dv = dx + dy
∂x ∂y
−∂u ∂u
dv = dx + dy
∂y ∂x
dv =Mdx + Ndy …(i)

−∂ u ∂u
Where, M = and N=
∂y ∂x
2 2
∂ M −∂ u ∂N ∂ u
Differentiate = 2 and
=
∂y ∂y ∂ x ∂ x2
Since u is harmonic, therefore

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2 2
∂ u ∂ u
2
+ 2 =0
∂x ∂y
2 2
∂ u −∂ u ∂ M ∂ N
2
= 2
⇒ =
∂x ∂y ∂ y ∂x
Hence equation (i) is exact differential equation. dv can be solve to get v.
Method-II: Milne Thomson Method
Type-I: To construct analytic function f(z) in terms of z, when real part u is given by the
following formula,
f ( z )=∫ [∅ 1 ( z , 0 ) −i∅ 2 (z , 0) ] dz +C
∂u ∂u
Where ∅ 1 ( x , y )= and ∅ 2 ( x , y )=
∂x ∂y
Type-II: To construct analytic function f(z) in terms of z, when imaginary part v is given by the
following formula,
f ( z )=∫ [∅ 1 ( z , 0 ) +i ∅ 2 (z , 0) ] dz+C
∂v ∂v
Where ∅ 1 ( x , y )= and ∅ 2 ( x , y )=
∂x ∂y
Type-III: To construct analytic function f(z) in terms of z, when U = u – v is given by the
following formula,
(1+i)f ( z )=∫ [∅ 1 ( z , 0 ) −i∅ 2 (z , 0) ] dz +C
∂U ∂U
Where ∅ 1 ( x , y )= and ∅ 2 ( x , y )=
∂x ∂y
Type-IV: To construct analytic function f(z) in terms of z, when V = u + v is given by the
following formula,
(1+i)f ( z )=∫ [∅ 1 ( z , 0 ) +i ∅ 2 (z , 0) ] dz+C
∂V ∂V
Where ∅ 1 ( x , y )= and ∅ 2 ( x , y )=
∂x ∂y
Example: Use C-R equation to find v , where u=3 x 2 y− y 3.
Solution: Here u=3 x 2 y− y 3
Differentiating u w.r.t x and y
We have

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∂u
=6 xy =∅ 1 ( x , y )
∂x
∂u 2 2
=3 x −3 y =∅ 2 (x , y)
∂y
Also
2 2
∂ u ∂ u
2
+ 2 =6 y + (−6 y )=0
∂x ∂y
Here u is harmonic
Now putting x = z, y = 0

∅ 1 ( x , y )=0 and ∅ 2 ( x , y )=3 z 2

Hence by Milne Thomson method,

f ( z )=∫ [∅ 1 ( z , 0 ) −i∅ 2 (z , 0) ] dz +C

f ( z )=∫ [ 0−3 z i ¿ ] dz +C
2

3
f ( z )=−z i+ C
3
f ( z )=−( x +iy) i +C

f ( z )=u +iv=( 3 x 2 y− y 3 ) +i(3 xy 2−x 3 +C)

Here v=(3 xy 2−x 3 +C).
INTEGRATION IN THE COMPLEX PLANE
Complex Line Integrals and Some Integral Theorems
b
a ≦t ≦ b ∫ f ( z)dz=∫a f ( z(t ))⋅z' (t )dt
For smooth curve C: z=z(t) for , then c
Special case 1C: |z-z0|=r⇔ z(t)=z0+reit, dz = z’(t)dt=ireitdt, and 0 ≦t ≦ 2 π
Special case 2C: |z|=r⇔ z(t)=reit, dz=z’(t)dt=ireitdt, and 0 ≦t ≦ 2 π

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

1
∮ z dz
Example 1. Find c , C: |z|=1.
ieit
∮ dzz =∫0
2π
¿ z∨¿ 1 ¿
z=e 0 ≤ t ≤ 2 π z ’ (t )=ie
¿ dt=2 πi
⇔ c eit
Solution: , , ,
dz
∮ z−3 i 1
Example 2: Evaluate: C
, C: |z-3i|= 3 .
Solution:
¿ ¿
z (t)=3i+e , z ’ (t )=ie ,
2π 1 i
dz =∫ ⋅ eit dt=2 πi
∮ z−3 i 0 1
e it
3
C 3
∮ z dz
Example 3: Evaluate c , C: |z|=1.
2 π −it
z (t)=e
¿ −¿ ¿ ∮ z dz=∫0 it
e ⋅ie dt=2 πi
Solution:
z e z ’ (t )=ie , c
, = ,

∮ [ z−R e ( z)]dz
Example 4: Evaluate c , C: |z|=2.
⇔
Solution : |z|=2 z(t)=2eit, z’(t)= i2eit
1 1 1 1
Re (z )= ( z + z )= (2 eit +2 e−it ) z−R e ( z )= ( z−z )= (2 e it −2 e−it )
2 2 ,
2 2
2π
∮ [ z−R e ( z)]dz=∫0 [ z(t )−Re (z(t ))]⋅z' (t )dt =∫ 2 (e it−e−it )⋅2ieit dt=2 i ∫ ( e2 it −1)dt
2π 2π

c
0 2 0 =−4 πi

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CAUCHY’S INTEGRAL THEOREM

Let f(z) be analytic in a simply-connected domain D, C is a simple closed curve in D, then
∮ f ( z)dz=0
c
.
Example 1: Integrate z2 along the straight line OA and also along the path OBA consisting of two
straight lines OB & BA where O is the origin , B is the point z=3 and A is the point z=3+i.
Hence show that the integral of z2 along the closed path OBAO is zero.
Solution : We have z=x+iy , dz=dx+idy
Along the curve C , We have

A (z=3+i)

O B(z=3) x

………………………..(1)

The point A is z=3+I i.e. A(3,1)

The equation of the line OA is
1−0
y−0= (x −0)
3−0

i.e. x=3 y
Now, On the line OA, x=3y
dx=3dy y:0 to 1

Therefore from (1) we have,

1

∮ z dz=∫ ( 8+6 i ) ( 3+i ) y dy

2 2

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
26
=6+i
3
Again,

3 1
¿ ∫ x dx +∫ ( 9− y +6 iy ) i dy
2 2

0 0

26
=6+i
3
Lastly , The integral of z2 along the closed path OBAO is given by

i.e Cauchy Integral theorem is verified.

Application of Cauchy Integral theorem :
1
∮ z dz
Example 1. Evaluate c , C: |z-2|=1.
dz
1 ∮ z =0
Solution : f(z)= z is analytic except z=0. No poles are within C, ∴ c
dz
∮ z 2−4
|z|=1
Example 2. Evaluate
dz
2
1
z 2−4
∮
Solution : f(z)= z −4 is analytic except z=±2. No poles are within C, ∴|z|=1 =0
z
C ∮ sin( z)( z−2 i)3 dz
Example 3 . Evaluate , C: |z-8i|=1.
z
f ( z )=
Solution:
sin (z )( z−2i )3 is analytic except z=2i, nπ (n=0,±1, ±2, …)

No poles are within C, ∴

∮ f ( z)dz=0
CAUCHY’S INTEGRAL FORMULAE
Let f(z) be analytic in a simply-connected region D, and let C be a simple curve enclosing z 0 in D,
f ( z) f ( z )dz 2 πi
∮ z−z dz ∮ ( z−z )n = (n−1 )!
⋅f
(n−1)
( z0 )
c 0 c 0
then =2πi f( ) and .

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e 2 z dz
∮
|z|=3 z−2
Example 1 : Evaluate .
e2 z
∮
|z|=3 z−2
Solution : =2πie2∙2=2πie4

e iz
∮ z3 dz
c
Example 2: Evaluate , C: |z|=3.
iz iz ' '
(e )
∮ ez3 dz=i2 π⋅(3−1 )! =−iπ
c |z0 =0
Solution :

f ( z) f ( z)
∮C dz=2 πi ∮C ( z−1)2 dz=4 πi
Example 3: Given z and . If let f(z)=a+bz, find a and b

Solution: 2πi=2πi．f(0)=2πai and 4πi=2πi．f'(0)=2πbi, ∴a=1 and b=2.

sin 6 ( z ) sin 6 ( z )
∮ π dz ∮ dz

Example 4: Evaluate
c z−
6 and
c
z−
π 3
6
if
π
6 ( )
C :|z − |=δ>0
.
sin 6 ( z )
∮ π dz =2 πi⋅sin6 π6 =32
c z−
πi
()
Solution: Let f(z)=sin6(z) and n=3 , 6 ,
sin 6 ( z )
∮ π 3 dz= 22πi! ⋅[sin6 ( z )]|z' ' = π =21 πi
c
( )
z−
6
0 6
16

dz dz
∮ z−z ∮ ( z−z n n ≧2
c 0 c 0)
Example 5: Let be within C, find and , .
dz dz
⇒∮
z−z 0
=2 πi ∮ ( z−z n
=0
c 0)
Solution: Let f(z)=1, f(n-1)(z0)=0 and .
2 sin( z 2 )
∮ (z−1)4 dz
c
Example 6: Evaluate , C is a closed curve not passing 1.

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2sin ( z2 ) 2 sin( z 2 )
( z−1 )4
∮ (z −1)4
dz=0
c
Solution: If C does not enclose 1, is analytic within C,∴
2 sin( z 2 ) f ( z)
f ( z )=2 sin( z 2 )⇒ =
( z−1 )4 ( z−1) 4
If C encloses 1, let ,n=4, n-1=3
( 3) 2 3 2
f ( z )=−24 z sin( z )−16 z cos( z )
2 sin( z 2 )
∮ (z−1)4 dz= 23!πi [−24 sin(1 )−16 cos(1)] = πi [−24 sin(1)−16 cos(1 )]
∴ c 3
CAUCHY’S RESIDUE THEOREM
Let f(z) be analytic in D except z1, z2, …,zn and C encloses z1, z2, …,zn within D. Then we have
n

∮ f ( z )dz=2 πi⋅∑ Rez s (f )

c j=1 j

Method of finding Residue:

1 d m−1 m
Re s ( f )= lim m−1 [( z−z j ) ¿ f ( z)]
zj (m−1)! z → z j dz
, where m is the order of a pole z=zj.
Re s ( f )=lim [ ( z−z j )⋅f ( z )]
z z→zj
In case of m=1, j .

z
Example 1: Find the residues of f(z)= .
z−1
z
Re s ( f )=lim [( z−1)⋅ ]
1 z→ 1 z−1
Solution: At z=1, m=1, =1

z
Example 2: Find the residues of f(z)= 2.
( z−1)(z +1)
z
Re s ( f )=lim [( z−1)⋅ 2
] 1
1 z→ 1 (z −1)( z+1) 4
Solution : At z=1, m=1, =
1 d 2 z ( z−1)−z 1
Re s ( f ) (2−1 )! lim [( z+1) ⋅ 2
]= |
2 z=−1 −
dz ( z+1 ) ( z−1) (z−1)
= 4
z→−1
At z=-1, m=2, −1 =

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
1
Example 3: Find the residues of f(z)= 2 2.
( z−1) (z+ 1)
1 d 2 1 −2( z+1) 1
Re s ( f ) (2−1 )! lim [( z−1 ) ⋅ 2 2
]= |
4 z=1 −
z→1 dz ( z+1) ( z−1) ( z+1 ) 4
Solution: At z=1, m=2, 1 = =
1 d 2 1 −2( z−1) 1
Re s ( f ) (2−1 )! lim [( z+1) ⋅ 2 2
]= |
4 z=−1
−1 z→−1 dz ( z+1 ) ( z−1) ( z−1) 4
At z=-1, m=2, = =
z 2−2 z
( z +1)2 ( z 2 + 4 )
Example 4 : Find the residues of f(z)= .
z 2−2 z z 2−2 z
( z +1)2 (z 2 + 4 ) ( z +1)2 (z +2 i)( z−2 i)
Solution: f(z)= =
At z=-1, m=2,
1 d z2 −2 z (2z−2)( z 2 +4 )−( z 2 −2 z )(2z)
lim [( z+1)2⋅ ]= |z=−1 −14
(2−1 )! z→−1 dz ( z+1 )2 ( z 2 +4) (z 2 +4 )2 25
= =

z 2 −2 z
Re s ( f )= lim [( z−2 i)⋅ 2
] 7+i
2i z→ 2i ( z−2 i)( z +2 i)( z+1) 25
At z=2i, m=1, =
2
z −2 z
Re s ( f )= lim [( z +2i )⋅ ] 7−i
−2i z→−2i ( z−2 i)( z+2i)( z +1)2 25
At z=-2i, m=1, =

cot (z )coth (z )
Example 5: Find the residues of f(z)= z3 .
m−1
1 d m
Re s ( f )= lim m−1 [( z−z j ) ¿ f ( z)]
z (m−1)! z→ z j dz
Solution: It is difficult to compute the residue at 0 by j . We
∞ a−2 a−1
f ( z )= ∑ a n ( z−z 0 )n =⋯+ +
n=−∞ ( z−z 0 )2 ( z−z 0 ) + a0 + a1 ( z−z 0 )+⋯
utilize
cot (z )coth (z ) cos ( z )cosh ( z )
z3 z3 sin( z )sinh( z )
f(z)= =

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( 1−
z2 z 4 z2 z4
+ −+⋯ 1+ + +⋯
2! 4! 2! 4 ! ) ()( ) 1−
z4
6
+⋯

z ( z− +−⋯) ( z + +⋯) ( z − +−⋯)

3 3 9
3 z z z 5 1 7 1
= 5 − ⋅ +⋯
3! 3! 90 45 z
= = = z
Re s −7
0
∴ (f)= 45
z 2 +1
∮ z2−1 dz
c
Example 6: Evaluate , C: |z-1|=1.
Solution: There is only one pole1 within C.
2
+1
z 2 +1
Re s ( f ) lim [( z−1 )⋅ 2 ] lim
z2 +1 ∮ zz2−1 dz
1
= z→1 z −1 = z→1
z +1 =1,∴ c =2πi．1 =2πi
cos( z) 1 1
∮ z2( z−1 ) dz
Example 7: Evaluate c
3 3
for (a) C: |z|= , (b) C: |z-1|= , (c) C: |z|=2.[
Solution : (a)There is only one pole 0 within C.
1 d 2 cos( z) −(z−1)sin( z)−cos( z) cos( z)
Re s ( f ) (2−1)! lim
z →0 dz
[ z ⋅
z 2( z−1)
]=
( z−1)2
|z=0 ∮ z2( z−1 ) dz
0
= =-1, c =-2πi
(b) There is only one pole1 within C.
cos( z ) cos( z)
Re s ( f ) lim [( z−1 )⋅ 2
z→1 z ( z−1)
] ∮ z2( z−1 ) dz
c
1
= =cos(1), =2πi．cos(1)
cos( z)
∮ z2( z−1 ) dz
c
(c) There are two poles 0 and1 within C. =2πi．[-1+cos(1)]

sin ( z ) sin (z )
f ( z )= ∮ dz
( z−i )3 and evaluate c ( z−i)3 , C:|z-i|=2.
Example 8: Find the residue of

1 d2 sin( z ) −1 1
Re s ( f )= lim 2 [( z−i)3⋅ ]= sin(i)=− i sinh(1)
i (3−1)! z →i dz ( z−i) 3 2 2
Solution: m=3,

∴
∮
sin (z )
2
( z−i )
1
( )
dz=2 πi⋅ − isinh (1) =π sinh(1)
2

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∮ tan zdz
Example 9: Evaluate c , C: |z|=2.
Solution: There are two poles ±π/2 within C.
π
z sin( z )− sin( z )
π π sin( z ) 2
Re s ( f ) lim [( z− )⋅tan( z )] lim [( z− )⋅ ] lim
π z→
π 2 π 2 cos( z ) π cos( z )
z→ z→
2
= 2 = 2 = 2
π
sin( z )+ z cos( z )− cos ( z )
2
lim
π −sin ( z )
z→
= 2 =-1
π
z sin ( z )+ sin( z )
π π sin( z ) 2
Re s ( f ) lim [( z+ )⋅tan( z )] lim [( z+ )⋅ ] lim
π z→−
π 2 π 2 cos ( z ) π cos( z )
− z→− z→−
2
= 2 = 2 = 2
π
sin ( z )+ z cos ( z )+ cos (z )
2
lim
π −sin( z )
z→−
= 2
=-1
∮ tan zdz
c
=2πi．[(-1)+(-1)]= -4πi

sin (z )
∮ z 2( z 2+4 ) dz
Example 10: Evaluate c , C is any piecewise-smooth curve enclosing 0, 2i, and –2i.

sin (z )/ z
f ( z )= lim [ sin( z )/z ] =1
z (z +2 i)( z−2 i) z→0
Solution : ,∵ ,∴f(z) has a removable singularity at 0⇒
sin( z ) 1
Re s ( f )=lim z⋅ 2 2 =
0 z→ 0 z ( z +4) 4
m of the pole z=0 in f(z) is 1.
sin ( z ) i 1
Re s ( f )= lim ( z−2 i)⋅ 2 = sin(2 i)=− sinh(2)
2i z→ 2i z ( z +2 i)( z−2i ) 16 16
sin( z ) i 1
Re s ( f )= lim ( z +2 i)⋅ 2 = sin (2i)=− sinh(2 )
−2i z→−2i z ( z+2 i)( z−2 i) 16 16

∮ f ( z)dz=2 πi
c
[ 1 1 1 πi πi
]
− sinh(2 )− sinh (2) = − sinh(2 )
4 16 16 2 4

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1
3 z
∮ zz3+1
e
dz
c
Example 11: Evaluate , C: |z|=3.
Solution: Let t=1/z, z=1/t, dz=-dt/t2, C: |t|=1/3
1
( )3 et
1 t −dt
z e 3 z∮ 1 3 ⋅( t 2 ) −e t
∮ z 3+1 dz c ( ) +1 ∮ t2 (t3 +1) dt
c t c
= = .
There is only one pole 0 within C.
1
t 3 z
1 d −e
Re s ( f )=
0
lim [t 2⋅ 2 3
1 ! t→ 0 dt t (t +1)
]= ∮ zz3+1
e
dz
c
-1, ∴ =2πi∙(-1)= -2πi

e− z
∮ cos( z) dz
c
Example 12: Evaluate ,C: |z|=2.
Solution: cos(z) has two zeros at z=±π/2 within |z|=2.
π z π
z e ze z −
e z + ze z− e z
π e 2 2
Re s ( f ) lim [( z− )⋅ ] lim lim π
π π 2 cos( z ) π cos( z ) π −sin( z ) 2
z→ z→ z→
2
= 2
= 2
= 2
=−e
π π z
z ze z + e z e z + ze z + e
π e 2 2
Re s ( f ) lim [( z+ )⋅ ] lim lim π
π π 2 cos ( z ) π cos( z ) π −sin ( z ) −
2
− z→− z→− z→−
2
= 2
= 2
= 2
=e
e− z
∮ cos( z) dz −
π
2
π
c e e2
∴ =2πi∙( - )
1
z
e
∮ z+1 dz
c
Example 13: Evaluate , C: |z|=0.5.
Solution: Let t=1/z, z=1/t, dz= -dt/t2, C: |t|=2

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1 et −dt
∮ ⋅( 2 ) t
∮ t(t−e+1) dt
z
e 1
∮ z+1 dz c ( )+1 t
c t c
= = . There are twopoles0and -1 within C.

−et −et
Re s ( f )=lim [ t⋅ ]=−1 Re s ( f )= lim [(t +1)⋅ ]=e−1
0 t →0 t (t+1 ) −1 t→−1 t (t+1 )
, ,
1
3 z
∮ zz3+1
e
dz
c
∴ =2πi∙(-1+e-1)
Evaluation of real Integral by Cauchy Residue Theorem i.e. Integration round the unit circle
2π

Consider the integral ∫f ¿¿ ……………………… (1)

0

Where f ¿ is a rational function of cosθ and sinθ .

Such integrals can be reduced to complex line integrals by the substitution

iθ
z=e
iθ
∴ dz=e idθ
iθ −iθ
e +e 1 1
We also know that , cos θ= = (z+ )
2 2 z

and
iθ −iθ
e −e 1 1
sin θ= = ( z− )
2i 2i z

As θ varies from 0 to 2π ,moves once round the unit circle in the anti-clockwise direction
where C is the unit circle |z| = 1.

Hence, after putting the above values in the given real integral (1) ,then it will be reduce in to
complex integral and the we, sovle by Cauchy Residue Theorem .
2π dθ
∫0 5−3 cosθ .
Example 1: Evaluate

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
2π dθ dz /iz 2idz 2idz
∫0 =∮ =∮ 2 ∮c
Solution:
5−3 cos θ c 1
( )
5−3⋅ z+
2
1
z
c 3 z −10 z+3

=
1
3( z− )( z−3)
3
1
There is only one pole within |z|=1.
3
1 2i 2i π
2 πi⋅Re s (f )=2 πi⋅lim ( z− )⋅ =2 πi⋅ =
2π dθ 3 1 −8 2
∫0 5−3 cosθ =
1
3
z→
1
3
3( z− )( z−3)
3
∴ .

2π dθ
∫0 3−2cos θ+sin θ .
Example 2: Evaluate
2π dθ dz /iz 2dz
∫0 =∮ =∮

Solution:
3−2cos θ+sin θ c 1
( ) ( )
1 1
3−2⋅ z+ +
2 z 2i
z−
1 c (1−2 i) z 2 +6 iz−1−2 i
z
2 dz
∮c 2−i
(1−2i )( z− )[ z−(2−i)]
=
5 .
2−i
There is only one pole within |z|=1.
5
2π dθ
∫0 3−2cos θ+sin θ
∴
2−i 2
2 πi⋅Re s (f )=2 πi⋅ lim {(z− )⋅ } 1
2−i 2−i 5 2−i
z→ (1−2i )( z− )[ z−(2−i)] 2i
=
5 5 5 =2πi∙( )=π.
2π dθ 2π dθ 2π
∫0 a+b sin θ
=∫0 = 2 2
a+b cos θ √a −b if a>|b|.
Example 3: Show that

2π dθ dz /iz 2 dz
∫0 =∮ =∮ 2

Proof:
a+ b sin θ c 1
a+b⋅ z−
2i ( )1
z
c bz +2 iaz−b

=
2 dz
∮c b ( z−z )( z−z )
1 2

i i
z 1=(−a+ √ a2 −b2 ) z 2 = (−a− √ a 2−b2 )
Poles:
b is within C: |z|=1, but
b is not.
2 2 1 2 b 1
Re s ( f )=lim ( z−z 1 )⋅ = ⋅ = ⋅ =
z1 z→z
1
b( z−z 1 )( z−z 2 ) b z 1−z 2 b 2i √ a −b i √ a2 −b 2
2 2

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
2π dθ 1 2π
∫0 =2 πi⋅ 2 2 = 2 2
∴
a+b sin θ i √ a −b √ a −b

Module 5: Vector Calculus:

Contents

Differentiation of Vectors, Scalar and vector point function, Gradient, Geometrical meaning of
gradient, Directional Derivative, Divergence and Curl, Line Integral, Surface Integral and Volume
Integral, Gauss Divergence, Stokes and Green theorems.

Vector Differentiation
^ ^ ^
One-variable vector function: R(t )= x(t ) i+ y(t ) j+z(t ) k

Multi-variable vector function: F ( x, y , z )=F1 ( x, y , z ) ^i +F2 ( x, y , z) ^j+F 3 ( x, y , z ) k^

d R (t ) d R(t ) dt dR(t ) df (t )
= ⋅ = /
The derivatives of vector functions: df (t ) dt df (t ) dt dt

∂ F ( x, y ,z ) ∂ F ( x , y , z) ∂x ∂ F ( x , y , z) ∂y ∂ F ( x, y , z) ∂z
= ⋅ + ⋅ + ⋅
∂ g(x , y ,z) ∂x ∂ g( x , y , z) ∂y ∂ g( x , y , z) ∂z ∂ g ( x, y ,z )

^ ^ 3^
Eg. For R(t )=2t i−cos(3t ) j+t k , 0¿ t¿ 1, find R '(t )=d R (t )/dt .

Let s(t)= 0 ∫ 1 √ 4+9sin 2(3 t )+9 t 4⋅dt , then find d R(t )/ds(t ) .
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 B.Tech , Sem- II, Mathematics -II (BT -202 )

^ ^ 2^
Sol: d R(t )/dt=2 i+3sin (3t ) j+3t k

d R (t ) d R(t ) dt 2 ^i +3 sin(3 t ) ^j+3 t 2 k^

= ⋅ =
∴ ds(t ) dt ds(t ) √ 4+9 sin2 (3 t )+9 t 4

Some theorems of derivatives of vector functions:

1. ( F⋅G)'=F '⋅G+F⋅G '

2. ( F×G)'=F '×G+F ×G'

3. ( F×F ' )'=F×F ''

Proof: (F×F')'=F'×F'+F×F= { {F}} times { {F}}

^ ^ ^
4. For R(t )= x(t ) i+ y(t ) j+z(t ) k , if R(t ) does not change direction, then R(t )×R ' (t )=0 , and
vice versa.

5. Let R(t ) denote the position of a particle at time t. If the particle moves so that equal areas
are swept out in equal times, then we have R(t)×R \( t \) =0} {¿, and vice versa. (Kepler's law)

Proof:

1
= R 2θ
Area 2 and |R(t +Δt )−R(t )|≈Rθ
∴ 2 area=R θ=R (t )×[ R(t+ Δt )−R (t )]
2

If area=0 ⇔ R(t )×[ R (t+ Δt )−R (t )]=0 ⇔ R(t )×R ' (t )⋅Δt =0
⇔ R(t )×R ' (t )=0

Equal area in equal time

⇔ R(t )×R ' (t )= Constant⇔ [ R (t )×R '(t )]'=0 ⇔ R(t)×R \( t \) =0} {¿

Differential Geometry
^ ^ ^
Position vector: R (t )=x (t ) i + y (t ) j+z (t ) k

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

' dx (t ) ^ dy (t ) ^ dz (t ) ^
v (t )=R (t )= i+ j+ k
Velocity: dt dt dt
t2 ds
s (t )=∫t |R (t )|dt , =|R (t )|=|v (t )|
' '

Arc length: 1 dt
' d 2 x ( t ) ^ d 2 y (t ) ^ d 2 z ( t ) ^
a ( t )=v ( t )= i+ j+ k
Acceleration: dt 2 dt 2 dt 2
d T^ '
R (t ) v (t ) d R(t )
κ=| | T^ = ' = =
Curvature: ds , where |R (t )| |v (t )| ds(t )
^ ^ ^
Eg. C: R(t )=t i+(t−2) j+(3t−1 ) k is a straight line.

v ^i+ ^j+ 3 k^ κ=| d T^ |=| d T^ ⋅dt |=0

T^ = =
|v| √ 11 , ds dt ds .
^i+2sin (t ) ^j+4 k^
Eg. C : R (t )=2cos(t ) is a circle of radius 2 at z=4.
^ ^ ^
^T = −2sin(t ) i +2cos (t ) j ,κ=| d T ⋅dt |= 1 = 1
2 dt ds 2 r .

d|v| ^ |v|2 ^
a= T+ N=
Theorem : dt ρ tangential acceleration+centripetal acceleration
d v (t ) d ^
d|v (t )| ^ d T^
a (t )= = [|v (t )|T ] = ⋅T +|v (t )|
⋅
Proof : dt dt dt dt

=
d|v (t )| ^
dt (
ds d T^
⋅T +|v (t )|⋅ ⋅
dt ds = )
d|v (t )| ^
dt
⋅T +|v (t )|2
d T^
ds . Define
N=
^
^ ρdT
ds
d|v (t )| ^ |v (t )|2 ^ ^ ^ ^ ^ ^ T^ )
d ( T⋅
⇒ a(t )= T+ N ( N ⊥ T ⇐ T⋅T =1) =0
dt ρ , ds
^ ^ 2^
Eg. : For R(t )=[ cos(t )+t sin(t )] i +[sin(t )−t cos(t )] j+t k , t>0, we have
v(t )=t cos(t ) ^i +t sin (t ) ^j+2t k^
^ sin(t )+t cos(t )] ^j+2 k^
a (t )=[ cos(t )−t sin (t )] i+[
v (t ) 1 1 2
T^ = = cos(t ) ^i+ sin (t ) ^j+ k^
|v (t )| √5 √5 √5
1 1 1
ρ= = = =5 t
κ d T^ dt d T^ 1
| ⋅ | | ⋅ |
dt ds dt |v (t )|

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
^ ^ ^
^ ρ d T =ρ dt ⋅d T = ρ ⋅d T =−sin (t ) ^i +cos (t ) ^j
N=
ds ds dt |v (t )| dt
∴a (t )=√ 5 T^ +t N^ √5
, at= , an=t
^
Binormal vector: B=T × N
^ ^

Fernet formulae:
{ {
dT^ ^ N^ d N^ ^ ^
=κN= (1)¿ =−κT+τB (2)¿ ¿
ds ρ ds
Gradient, Divergence and Curl: the Basics

We first consider the position vector, r:

r = xi+yj+zK

Where x, y, and z are rectangular unit vectors. Since the unit vectors for rectangular
coordinates are constants, we have for dr:

dr = dx i + dy j + dz k .

The operator, del: Ñ is defined to be as:

Ñ = ¶/¶x i + ¶/¶y j + ¶/¶z k ,

This operator operates as a vector.

1. Gradient

If the del operator, Ñ operates on a scalar function, f(x,y,z), we get the gradient:

Ñf = (¶f/¶x) i+ (¶f/¶y) j + (¶f/¶z) k .

We can interpret this gradient as a vector with the magnitude and direction of the
maximum change of the function in space. We can relate the gradient to the differential
change in the function:

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

df = (¶f/¶x) dx + (¶f/¶y) dy + (¶f/¶z) dz = Ñf · dr = df .

Since the del operator should be treated as a vector, there are two ways for a vector to
multiply another vector: dot product and cross product. We first consider the dot product:

2. Divergence

The divergence of a vector is defined to be:

Ñ·A = [¶/¶x i + ¶/¶y j + ¶/¶z k] · [Ax x + Ay y + Az z]

= (¶Ax /¶x) + (¶Ay /¶y) + (¶Az /¶z) ;

since the rectangular unit vectors are constant, ¶x/¶x = 0 (etc.). This will not necessarily
be true for other unit vectors in other coordinate systems. We'll see examples of this soon.

3. Curl

The curl of a vector is defined to be:

Ñ´A = [¶/¶x i + ¶/¶y j + ¶/¶z k] ´ [Ax i + Ay j + Az k]

Where we have used the fact that the unit vectors do not change with position (¶x/¶x = 0)
and the fact that (x´x=0 and x´y=z, etc.). For other coordinate systems, unit vectors may
change with position.

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Remark: i.e. Properties:

Example 1 : Determine if
is a conservative vector field.

Solution:

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

So, the curl isn’t the zero vector and so this vector field is not conservative.

Example 2 : Compute for

Solution: There really isn’t much to do here other than compute the divergence.

Example 3: Verify the above fact for the vector field

.

Solution: Let’s first compute the curl.

Now compute the divergence of this.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Solution: The divergence of F(x, y) is given by ∇•F(x, y) which is a dot product.

Example 4 : Compute the divergence of F(x, y) = 3x2i + 2yj.

Example 5: Find the directional derivative of the function f ( x , y )=3 x 2 y at a point (-2,1) along
^ ^j.
the direction 4 i+3

^ ^
^ ^j is u^ = 4 i+ 3 j .
Solution: The unit vector along 4 i+3
5

^ x 2 ^j .
Gradient of the given function is 6 xy i+3

Thus the directional derivative at (x,y) is ( 245 ) xy +( 95 ) x .

2

At (-2,1) its value is -12/5.

Example 6: Find the directional derivative of the function f ( x , y , z )=3 x 3 +2 x y 2 + xyz along the
^ 4 ^j+ 4 k^ at a point (1,1,1)
direction 2 i+

^ ^j+2 k^
i+2
^ 4 ^j+ 4 k^ is u^ =
Solution: The unit vector along 2 i+ .
5

^
Gradient of the given function is ( 9 x 2 +2 y 2+ yz ) i+(4 xy + xz ) ^j+ xy k^ .

^ ^j+ k^
At (1,1,1) the gradient is 12 i+5

Thus the directional derivative at (1,1,1) is 8.

Example 7: Find the equation to the tangent plane to the surface x 2+ y 2−z 2=7 at the point
(2,2,1).

Solution: The level surface is x 2+ y 2−z 2=7.

^ 4 ^j−2 k^ .
Normal to the surface is in the direction of gradient which is 4 i+

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

The equation to the tangent plane is given by f x ( x−x 0 ) +f y ( y − y 0 ) + f z ( z−z 0)=0 , where f x , f y
and f z are the partial derivative of the function f(x,y,z)= constant at the point (x 0 , y 0 , z 0).

In this case f x =2 x 0=4 , f y =2 y 0=4 , f z=−2 z 0=−2.

Thus the equation is 4 ( x−2 ) +4 ( y−2 )−2 ( z−1 ) =0.

2
x − y2
Example 8: Find the directional derivative of the function f ( x , y )=e at the point

P = (1, 1) in the direction of the point Q = (0, 0).

Solution: We first need to find a unit vector u that travels in the same direction is the vector with
an initial point P and terminal point Q. First, we see that
→
v=PQ=¿0−1,0−1>¿ <−1,−1>¿−i−j

Then, since ||v||= √(1)2 +(1 )2=√ 2 , we can find the unit vector u to be
v 1 1
u= =¿−1 ,−1> ¿ =− i− j=a i+b j ¿
||v|| √ 2 √2 √2
1 1
a=− b=− x2 − y2
Hence, √2 and √2 . For f ( x , y )=e , we have that
2
− y2 2 2
f x ( x , y )=2 xe x f ( x , y )=−2 ye x − y
and x .

Hence,

Therefore,

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Example 9 : Find the constants a, b, c so that

F = (x + 2y + az) i + (bx - 3y - z) j + (4x + cy + 2z) k
is irrotational.
Solution: We have a vector F is irrotational, if curl F=0. Here
i j k
∂ ∂ ∂
curl F=∇ ×F=| |=0
∂x ∂y ∂z
x +2 y +az bx−3 y −z ax+cy+2 z ,
implies
(c+1)i + (a - 4)j + (b - 2)k = 0 i + 0 i + 0 i ,
implies that
a=4, b=2, c= -1
and hence
F = (x + 2y + 4z) i + (2x - 3y - z) j + (4x - y + 2z) k.
LINE INTEGRAL:

Line integral ¿ ∫ F .
c
( ⃗
dr
ds)ds=∫ F . dr
c

Note:
1) Work: If F represents the variable force acting on a particle along arc AB, then the total
B

work done ¿ ∫ F . dr
A

2) Circulation: If V represents the velocity of a liquid then ∮ V . dr is called the circulation

c

of V round the closed curve c .

If the circulation of V round every closed curve is zero then V is said to be irrotational
there.
3) When the path of integration is a closed curve then notation of integration is ∮ in place
of∫ .

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
B

Note: If ∫ F . dr is to be proved to be independent of path, then F=∇ ∅

A

here F is called Conservative (irrotational) vector field and ∅ is called the Scalar potential.
And ∇ × F=∇ × ∇ ∅ =0

Example 1: Evaluate ∫ F . dr where F=x 2 i+

^ xy ^j and C is the boundary of the square in the
c

plane z=0 and bounded by the lines x=0 , y=0 , x=a∧ y =a.

Solution: ∫ F . dr=∫ F . dr + ∫ F . dr + ∫ F . dr + ∫ F . dr
c OA AB BC CO

Here ^ y ^j , dr=dx i+
r =x i+ ^ dy ^j , F=x 2 i+
^ xy ^j

2
F . dr =x dx + xydy _______ (i)

 On OA , y=0
2
∴ F . dr=x dx (From (i))

[ ]
a 3
x3 a3
∫ F . dr=∫ x dx=
2
=
3 0 3
_______ (ii)
OA 0

 On AB , x=a
∴ dx=0
∴ F . dr=ay dy (From (i))

[ ]
a a
y2 a3
∫ F . dr=∫ ay dy =a =
2 0 2
_______ (iii)
AB 0

 On BC , y=a
∴ dy=0
∴ F . dr=x dx
2
(From (i))

[ ]
0 0
x3 −a3
∫ F . dr=∫ x dx=
2
3 a
=
3
_______ (iv)
BC a

 On CO , x=0
∴ F . dr=0 (From (i))

∫ F . dr=0 _______ (v)

CO

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

On adding (ii), (iii), (iv) and (v), we get

3 3 3 3
a a a a
∫ F . dr= + − +0=
3 2 3 2
C

Example 2: A vector field is given by

^ ( xz ) ^j+( yz−x ) k^ . Evaluate ∫ F . dr along the path c is

F=( 2 y +3 ) i+
C
3
x=2 t , y=t , z =t ¿ t=0 ¿ t=1.

Solution:

∫ F . dr=∫ ( 2 y +3 ) dx+ ( xz ) dy +( yz−x )dz

C C

[ ]
3
since x=2 t y=t z=t
dx dy dz 2
∴ =2 =1 =3 t
dt dt dt
1
¿ ∫ ( 2 t+3 )( 2 dt )+ ( 2t ) ( t ) dt + ( t −2t ) (3 t dt )
3 4 2

1
¿ ∫ ( 4 t+6 +2 t + 3t −6 t ) dt
4 6 3

[ ]
1
t2 2 5 3 7 6 4
¿ 4 + 6 t+ t + t − t
2 5 7 4 0

[ ]
1
2 3 3
¿ 2 t 2+6 t + t 5+ t 7 − t 4
5 7 2 0

2 3 3
¿ 2+6+ + −
5 7 2

¿ 7.32857

y ^j+ z k^ is the force field. Find the work done by ⃗

3^
Example 3: Suppose ⃗ F ( x , y , z )=x i+ F along
the line from the (1, 2, 3) to (3, 5, 7).

Solution: Work done ¿ ∫ F . dr

c

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
(3 , 5 ,7)

¿ ∫ ( x3 i+
^ y ^j+ z k^ ) . ( i^ dx + ^j dy + k^ dz )
(1 , 2 ,3)

(3 , 5 ,7)

¿ ∫ ( x3 dx + ydy + zdz )
(1 , 2 ,3)

3 5 7
¿ ∫ x dx+∫ y dy +∫ z dz
3

1 2 3

[ ][ ][ ]
3 5 7
x4 y2 z2
¿ + +
4 1 2 2 2 3

¿
[ 81 1
− +
4 4
25 4
− +
2 2
−
][
49 9
2 2 ][ ]
80 21 40
¿ + +
4 2 2

202
¿
4

¿ 50.5 units

Exercise:

^ xy ^j displaces a particle in the xy -plane from (0, 0) to (1,

1) If a force F=2 x2 y i+3
4) along a curve y=4 x 2. Find the work done.
2) If⃗ ^
A=( 3 x2 +6 y ) i−14 yz ^j+ 20 x z 2 k^ , evaluate the line integral ∮ ⃗
A⃗dr from (0, 0,
0) to (1, 1, 1) along the curveC .
(3 ,4 )

3) Show that the integral ∫ ( x y 2+ y 3 ) dx +(x 2 y +3 x y 2 )dy is independent of the

(1 ,2)

path joining the points (1, 2) and (3, 4). Hence, evaluate the integral.

SURFACE INTEGRAL:

Let F be a vector function and S be the given surface.

Surface integral of a vector function F over the surface S is defined as the integral of the
components of F along the normal to the surface.

Component of F along the normal¿ F . n^

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Where n = unit normal vector to an element ds and

grad f dx dy
n^ = ds=
|grad f | ( n^ . k^ )

Surface integral of F over S

¿ ∑ F . n^ ¿ ∬ ( F . n^ ) ds
S

Note:

1) Flux ¿ ∬ ( F . n^ ) ds where, F represents the velocity of a liquid.

S

If∬ ( F . n^ ) ds=0, then F is said to be a Solenoidal vector point function.

S

VOLUME INTEGRAL:

Let F be a vector point function and volume V enclosed by a closed surface.

The volume integral ¿ ∭ F dv

V

Example 1: Evaluate ∬ ( yz i+
^ zx ^j+ xy k^ ) . ds where S the surface of the sphere is
S

x + y + z =a in the first octant.

2 2 2 2

Solution: Here, ∅ =x 2 + y 2+ z2 −a2

Vector normal to the surface ¿ ∇ ∅

∂∅ ^ ∂∅ ^ ∂∅
¿ i^ +j +k
∂x ∂y ∂z

(
¿ i^
∂ ^ ∂ ^ ∂
∂x
+j
∂y
+k
∂z )
( x 2 + y 2 + z 2−a 2)

^
¿ 2 x i+2 y ^j+ 2 z k^

^ y ^j+2 z k^
∇ ∅ 2 x i+2
n^ = =
|∇ ∅ | √ 4 x 2+ 4 y 2 + 4 z 2

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
^ y ^j+ z k^
x i+
¿
√ x 2+ y 2+ z 2
^ y ^j+ z k^
x i+
¿ [ ∵ x 2+ y2 + z 2=a2 ]
a

Here, ⃗ ^ zx ^j+ xy k^
F = yz i+

⃗
^ ^ ^
^ zx ^j+ xy k^ ) . x i+ y j+ z k = 3 xyz
F . n^ =( yz i+
a a ( )
Now, ∬ ⃗F . n^ ds=∬ ( ⃗F . n^ ) dx^ dy
S S |k . n^|
a √ a2−x 2
3 xyz dx dy
¿∫ ∫
0 0
a
z
a ()
a √ a2−x 2
¿ 3∫ ∫ xy dy dx
0 0

y √ a −x
( )
a
2 2 2

¿ 3∫ x dx
0 2 0
a
3
¿ ∫ x (a −x )dx
2 2
20

( )
2 2 4 a
3 a x x
¿ −
2 2 4 0

( )
4 4
3 a a
¿ −
2 2 4
4
3a
¿
8

^
Example 2: If F=2 z i−x ^j+ y k^ , evaluate ∭ F dv where, v is the region bounded by the
V
2
surfaces x=0 , y=0 , x=2 , y=4 , z=x , z=2.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Solution: ∭ F dv=∭ ( 2 z i−x

^ ^j+ y k^ ) dx dy dz
V

2 4 2
¿ ∫ dx ∫ dy ∫ ( 2 z i−x
^ ^j+ y k^ ) dz
0 0 x
2

2 4
2
¿ ∫ dx ∫ dy [ z i−xz ^j+ yz k^ ]
2^
2
x
0 0

2 4
¿ ∫ dx ∫ dy [ 4 i−2
^ x ^j+ 2 y k^ −x i+ x ^j−x y k^ ]
4^ 3 2

0 0

[ ]
2 2 2 4

¿ ∫ dx 4 y i−2
^ 2^
xy ^j+ y k−x
4 ^ x3 y ^j− x y k^
y i+
0
2 0

2
¿ ∫ ( 16 i−8
^ x ^j+16 k^ −4 x i+4
4^
x ^j−8 x k^ ) dx
3 2

[ ]
5 3 2
^
¿ 16 x i−4 ^ 4 x i+
x ^j+16 x k−
2 ^ x 4 ^j− 8 x k^
5 3 0

^ 128 i+16
¿ 32 i^ −16 ^j +32 k− ^ ^j− 64 k^
5 3

32 i^ 32 k^
¿ +
5 3

32 ^ ^
¿ ( 3 i +5 k )
15

Exercise:

1) Evaluate∬ ( F . n^ ) ds, where, ⃗ ^

F =18 z i−12 ^j+3 y k^ and S is the surface of the plane
S

2 x+3 y +6 z =12 in the first octant.

^
F =( 2 x 2−3 z ) i−2
2) If⃗ xy ^j−4 x k^ , then evaluate∭ ∇ ⃗
F dv , where V is bounded by the
V

plane x=0 , y=0 , z=0 and2 x+ 2 y + z=4 .

GREEN’S THEOREM: (Without proof)

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
∂ϕ
∧∂ Ψ
If ∂y be continuous functions over a region R bounded by
∅ ( x , y ) ,Ψ ( x , y ) ,
∂x
simple closed curve C in x− y plane, then

∮ ( ϕ dx +Ψ dy )=∬ ( ∂∂Ψx − ∂∂ ϕy ) dx dy
C R

Note: Green’s theorem in vector form

∫ F . dr=∬ ( ∇ × F ) . k^ dR
c R
^
Where, F=∅ i+Ψ ^ y ^j, k^ is a unit vector along z -axis and dR=dx dy .
^j , r=x i+

Example 1: Using green’s theorem, evaluate∫ (x y dx + x dy ), where c is the boundary

2 2

described counter clockwise of the triangle with vertices( 0 , 0 ) , ( 1 ,0 ) ,(1, 1).

Solution: By green’s theorem, we have

∮ ( ϕ dx +Ψ dy )=∬ ( ∂∂Ψx − ∂∂ ϕy ) dx dy
C R

∫ ( x 2 y dx+ x 2 dy )=∬ ( 2 x−x 2 ) dx dy

c R
1 x
¿ ∫ ( 2 x−x ) dx ∫ dy
2

0 0
1 x
¿ ∫ ( 2 x−x ) dx [ y ] ¿0 ¿
2

0
1
¿ ∫ (2 x −x )dx
2 3

( )
3 4 1
2x x
¿ −
3 4 0

¿ ( 23 − 14 )
5
¿
12

Example 2: Use green’s theorem to evaluate

∫ ( x 2 + xy ) dx +( x2 + y 2 )dy, where c is the square formed by the lines y=±1 , x=±1 .

c

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Solution: By green’s theorem, we have

∮ ( ϕ dx +Ψ dy )=∬ ( ∂∂Ψx − ∂∂ ϕy ) dx dy
C R

[ ]
1 1
∂ 2 2
¿∫ ∫ ( x + y )− ∂ (x 2+ xy ) dx dy
−1 −1 ∂x ∂y
1 1
¿ ∫ ∫ ( 2 x−x ) dxdy
−1 −1

1 1
¿ ∫ ∫ x dxdy
−1 −1

1 1
¿ ∫ x dx ∫ dy
−1 −1

1 1
¿ ∫ x dx ( y )−1
¿
¿
−1

1
¿ ∫ x dx (1+1)
−1

1
¿ ∫ 2 x dx
−1

1¿
2
¿( x )−1
¿

¿ 1−1

¿0

Exercise:

1) Apply Green’s theorem to evaluate

∫ [ ( 2 x 2− y 2 ) dx +(x2 + y 2 )dy ], where C is the boundary of the area enclosed by the x -

C

axis and the upper half of circle x 2+ y 2=a2 .

^ x (1+cos y) ^j .
2) A vector field F is given by F=sin y i+

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Evaluate the line integral ∫ F . dr where C is the circular path given by x 2+ y 2=a2 .
C

STOKE’S THEOREM: (Relation between Line integral and Surface integral)

(Without Proof)

Surface integral of the component of curl F along the normal to the surface S, taken
over the surface S bounded by curve C is equal to the line integral of the vector point
function F taken along the closed curveC .

Mathematically

∮ F . dr =∬ curl F . n^ ds
S

^
Where n^ =cos ∝ i+cos β ^j+ cos γ k^

is a unit external normal to any surface ds .

OR

The circulation of vector F around a closed curve C is equal to the flux of the curve of
the vector through the surface S bounded by the curveC .

∮ F . dr =∬ curl F . n^ ds=∬ curl F . d S

S S

Example 1: Apply Stoke’s theorem to find the value of

∫ ( y dx+ z dy + x dz)
c

Where c is the curve of intersection of x 2+ y 2+ z 2=a2 and x + z=a .

Solution: ∫ ( y dx+ z dy + x dz)

c

¿ ∫ ( y i+
^ z ^j+ x k^ ) .( i^ dx + ^j dy + k^ dz )
c

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

¿ ∫ ( y i+
^ z ^j+ x k^ ) . d r
c

¿ ∬ curl ( y i+
^ z ^j+ x k^ ) . n^ ds (By Stoke’s theorem)
S

¿ ∬ i^
S ∂ x
+j
∂ (
∂ ^ ∂ ^ ∂
y
+k
∂ z )
^ z ^j+ x k^ ) . n^ ds
× ( y i+

¿ ∬ – ( i+
^ ^j+ k^ ). n^ ds _______ (i)
S

Where S is the circle formed by the integration of x 2+ y 2+ z 2=a2 and

x + z=a .

∇∅
n^ =
|∇ ∅ |

(¿ i^ ∂∂x + ^j ∂∂y + k^ ∂∂z )( x + z−a)

|∇ ∅|
^ k^
i+
¿
√1+1
i^ ^k
¿ +
√2 √2
Putting the value of n^ in (i), we have

¿ ∬ – ( i+
^ ^j+ k^ ).
S
( i^
+
√2 √2
k^
)
ds

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

( ) [ ]
2 2
1 1 a a
¿ ∬−¿ + ds ¿ 2
Use r =R − p =a −
2 2 2
=
S √2 √2 2 2

( )
2 2
2 −2 a −π a
¿− ∬
√2 S
ds= π
√2 √2
=
√2

Example 2: Evaluate ∮ F . dr by stoke’s theorem, where

C

x ^j−(x+ z ) k^ and C is the boundary of triangle with vertices at ( 0 , 0 , 0 ) ,(1 , 0 , 0)

2^ 2
F= y i+
and (1 , 1, 0).

Solution: We have, curl F=∇ × F

| |
i^ ^j k^
∂ ∂ ∂
¿
∂x ∂y ∂z
2 2
y x −(x+ z )

^ ^j+2(x− y ) k^
¿ 0. i+

We observe that z co-ordinate of each vertex of the triangle is zero.

Therefore, the triangle lies in the xy -plane.

∴ n^ = k^

∴ curl F . n^ =[ ^j+2( x− y) k^ ] . k=2

^ ( x− y ) .

In the figure, only xy -plane is considered.

The equation of the line OB is y=x

By Stoke’s theorem, we have

∮ F . dr =∬ (curl F . n^ )ds
C S

1 x
¿∫ ∫ 2 ( x− y ) dxdy
x=0 y=0

[ ]
1 2
x
¿ 2∫ x −
2
dx
0 2

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
1 2
x
¿ 2∫ dx
0 2
1
¿ ∫ x dx
2

[ ]
1
x3
¿
3 0

1
¿
3

Exercise:

1) Use the Stoke’s theorem to evaluate ∫ [ ( x +2 y ) dx+ ( x −z ) dy +( y−z )dz ] where C

C

is the boundary of the triangle with vertices ( 2 , 0 , 0 ) , ( 0 , 3 , 0 )∧(0 , 0 , 6) oriented

in the anti-clockwise direction.

2) Apply Stoke’s theorem to calculate ∫ 4 y dx +2 z dy +6 y dz

c

Where c is the curve of intersection of x 2+ y 2+ z 2=6 z and z=x +3

3) Use the Stoke’s theorem to evaluate∫ y dx + xy dy + xz dz, where C is the

2

bounding curve of the hemisphere x + y + z 2=1 , z ≥ 0, oriented in the positive

2 2

direction.

GAUSS’S THEOREM OF DIVERGENCE: (Without Proof)

The surface integral of the normal component of a vector function F taken around a
closed surface S is equal to the integral of the divergence of F taken over the volume V
enclosed by the surface S.
Mathematically

∬ ⃗F . n^ ds=∭ ¿ ⃗F dv
S V

Example 1: Evaluate ∬ ⃗
F . n^ ds where ⃗ ^ y 2 ^j+ yz k^ and S is the surface of the
F =4 xz i−
S

cube bounded by x=0 , x=1 , y=0 , y=1 , z=0 , z=1.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Solution: By Gauss’s divergence theorem,

∬ ⃗F . n^ ds=∭ ( ∇ . ⃗F ) dv
S V

¿ ∭ i^
v
∂ ^ ∂ ^ ∂
∂x
+j
∂y (
+k
∂z
^ y 2 ^j+ yz k^ ) dv
. ( 4 xz i− )
¿∭
v
[ ∂
∂x
( 4 xz ) +
∂
∂y
2 ∂
]
(− y )+ ( yz ) dx dy dz
∂z

¿ ∭ ( 4 z −2 y + y ) dx dy dz
v

¿ ∭ (4 z− y )dx dy dz
v

( )
1 1 2 1
4z
¿ ∫∫
¿
− yz dx dy ¿
0 0 2 0
1 1 1
¿ ∫ ∫ (2 z − yz )¿0 dx dy ¿
2

0 0
1 1
¿ ∫ ∫ ( 2− y ) dx dy
0 0

( ) dx
1 2 1
y
¿∫ 2 y −
0 2 0
1
3
¿
20
∫ dx
3 1¿
¿ [ x ]0¿
2

3
¿ (1)
2

3
¿
2

Example 2: Evaluate surface integral where ⃗ ∬ ⃗F . n^ ds ,

^ ^j+ k^ ) , S
F =( x 2 + y 2+ z2 ) ( i+
is the surface of the tetrahedron x=0 , y=0 , z=0 , x + y + z=2 and n is the
unit normal in the outward direction to the closed surface S.

Solution: By Gauss’s divergence theorem,

∬ ⃗F . n^ ds=∭ ¿ ⃗F . dv
S V

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 B.Tech , Sem- II, Mathematics -II (BT -202 )

Where S is the surface of tetrahedron x=0 , y=0 , z=0 , x + y + z=2

¿ ∭ i^
V ∂ x
+j(
∂ ^ ∂ ^ ∂
∂ y
+k
∂ z
2 2 2 ^ ^ ^
. ( x + y + z ) ( i+ j+ k ) dv)

¿ ∭ (2 x +2 y+ 2 z )dv
V

¿ 2∭ ( x + y + z ) dx dy dz
V

2 2− x 2− x− y
¿ 2∫ dx ∫ dy ∫ ( x+ y+ z ) dz
0 0 0

( )
2 2− x 2 2−x− y
z
¿ 2∫ dx ∫ dy ¿
xz + yz + ¿
0 0 2 0

[ ]
2 2− x 2
(2−x − y)
¿ 2∫ dx ∫ dy 2 x−x 2−xy + 2 y −xy− y 2 +
0 0 2

[ ]
2 3 2− x
y3 ( 2−x− y ) ¿
¿ 2∫ dx 2 xy −x y−x y + y − − 2 2 2
¿
0
3 6 0

[ ]
2 3 3
(2−x) (2−x )
¿ 2∫ dx 2 x ( 2−x )−x 2 ( 2−x )−x (2−x )2+(2−x)2− +
0 3 6

[ ]
2 3 3
(2−x ) (2−x)
¿ 2∫ 2 2 3
4 x −2 x −2 x + x −4 x + 4 x −x +(2−x ) − + 2 3 2

0 3 6

[ ]
3 4 4 2
4 x x (2−x ) (2−x) (2−x)
3 4 3 4
4x x 2
¿2 2 x − + −2 x 2 + − − + −
3 4 3 4 3 12 24 0

[ ]
2
−(2−x)3 ( 2−x) 4 (2−x )4
¿2 + −
3 12 24 0

¿2
[ 8 16 16
− +
3 12 24 ]
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 B.Tech , Sem- II, Mathematics -II (BT -202 )
¿4

Example 3: Verify Stokes' theorem for F = (x2 - y2) i + 2xy j in the rectangular region in the xy plane,
the vertices of the rectangle being given by (0, 0), (a, 0), (b, 0) and (a, b).

Solution : Here the bounded surface S is the rectangle R given by

Given F = (x2 - y2) i + 2xy j .

i j k
∂ ∂ ∂
∴ ∇×F =| ∂ x ∂ y ∂ z |=i ( 0 )− j ( 0 ) +k ( 4 y )=4 y k .
x2 − y2 2 xy 0
The surface is the rectangular region in the xy plane. Hence the outward normal to the surface is k.

 n dS = k dx dy.

____ (1)

Now ____ (2)

(0, b) B C(a, b)

(0,0) O A (a, 0) X

F.dr = [ (x2  y2) i + 2xy j ] . [ dx i + dy j + dz k ]

= (x2  y2)dx + 2xy dy

(i) In OA, y = 0 , dy = 0 and x varies from 0 to a.

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 B.Tech , Sem- II, Mathematics -II (BT -202 )
a a
∫ F .dr=∫ ( x −0 )dx+2 x.0.0=∫ x 2 dx= 13 a3 .
2 2

OA 0 0

Similarly,

b
( ii ) ∫ F . dr=∫ 2 aydy=ab 2 . (∵ x=a dx =0 .)
AC 0

0 a
1
( iii ) ∫ F .dr =∫ ( x −b )dx=−∫ ( x2 −b 2 )dx=ab 2 − a3 (∵ ¿ ¿
2 2

CB a 0 3

( iv ) ∫ F .dr=0. (∵ x=0 dx=0.)

BO

Using the above values, (2) becomes

____ (3)

From (1) and (3), we get

∬ ( ∇× F ) . n dS=∮ F . dr
S C ,

So that Stokes’ theorem is verified.

Example 4: Verify Stokes' theorem for F = (2x – y) i  yz2 j  y2z k, where S is the upper half surface of the
unit sphere x2 + y2 + z2 = 1 and C is its boundary.

Solution

i j k
∂ ∂ ∂
∇×F =| ∂ x ∂y ∂ z |=k
2 2
2 x− y − yz − y z

142
 B.Tech , Sem- II, Mathematics -II (BT -202 )

where R is the orthogonal projection of S on the xy-plane. i.e., R is

the region enclosed by the circle

Now is the area of the region enclosed by the circle and is hence, we
obtain

____ (1)

The boundary C of S is a circle in the xy plane of radius 1 and center at the origin. Suppose x = cos t,
y = sin t, z = 0, 0  t  2 be the parametric form of C.

Then

= (2)

From (1) and (2), we get

∬ ( ∇× F ) . n dS=∮ F . dr
S C

Hence Stokes' theorem is verified.

******************************************************************************

143