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2019
Mathematics-I
Q.1. Choose the correct
answer (any seven):
(a) Determine the type of the following differential equation
d'y +Sin(x + y) = sin x
dx
(1) Linear, homogeneous
(ii) Non-linear, Homogeneous
(ii) Linear, non-homogeneous
(iv) Non-linear, non-homogeneous
Ans. (iv)
(b) The singular solution of p log(px- y)
=
is
(i) y=a(log x- 1) (ii)
(i)y = log x- 1
y=xlogx-1
(iv)y = xlog x

Ans. i)
Ify=o(x)is a particular solution ofy"+ (sinx)y' +2y=e" and
y= y(x) particular solution of y" +(sinx)y' +2y =cos2x,
os a

then particular of
y"+ (sinx)y' +2y= e +2sin*x is given by
(6)
o(x)-vx)+; (i) v(x)-p(x)+
ii) p(x)-y(x) +1 (iv) y(x)-P(x)+1
Ans. (iv)

(d)Let D-
Then the value of {1 / (xD +
1)} x' is
(i) logx (ii) (log x)/x
(ii)(log x)/x (iv)(log x)/x
Ans. (ii)
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(c) IfP, is the Legendre polynomial of first kind, then the value

of P,P,d i

2 2n 2 2n
(0
(1) 2n +|
2n+1 (2n+1 (iin) 2n+3 (iv) 2n+3
2n+3
Ans. (ii)
The partial differential eqúation
u-(1) u,y =0 is
i) parabolic in {(x, y):x <0} (i) hyperbolic in {(x, y): y> 0}
ii)elliptic in {(x, y) > 0, x2+y2<1}
(iv)parabolic in {(x, y):x>0}
Ans. (ii)

Ifux,t) satisfies the partial differential equation u =4

Ox
then u(x, t) can be of the form
(i) u(x, t)= f{x - 2t) + g(x +2t)

(i) u(x, t)- f(x2- 4t) +g(x2+4t)

(iii) u(x, t) = f(2x - 4t) + g (x + 2t)

(iv) u(x, 1) - f(2x-t) +g(2x + t)

Ans. (i)
(h) If J, is the Bessel's function of first kind, then the value of
2 is

) TCX
-sinxi) 2 sinx - cosx X

(i) Sin x (iv)

2 cOS x
10X

Ans. (ii)
(i) The region in which the following partial diferential equation
ayt3O
u70u, , 0'u
227U + 5u = 0
x Oxoy
acts as parabolie equation is
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(i) x>

1/3
(Gii)for all values of x (iv) x=

Ans. (iv)

G) Given =6x -0.5x2, y(0) =0, y(12) = 0 The value of

dx

a t y(4) using the finite difference method and a step size

ofh 4 can be approximated by

(i) Gii)6-2y(4)+y(0)
8 16

(ii) Y(12)-2y(8)+y(4) (iv) y4)-y0)

16 4
Ans. (ii)
Q.2. Solve the following differential equations

(a) (xy+e-l )dx- x?ydy = 0

Ans. xy+edx-xydy =0
xy+eAdx =x*ydy

=xy2xy+e-k*
dx
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Put y2= z then 2y dz

dx dx

I dz
i.e
2 dx -a
dz
dx
Which is linear in z

9 -e/x

Put y2= z then -2y-3 a_ dz

dx dx

dz
+ 2 x z = -2ex
dx
Which is iinear in z

PEx 9 2e

If d =e
The solution is

ze-2e.e dx+e
= 2ldx +e
ye-2x +e
Q.2. Solve the following differential equations:
(b) p+2py cotx = y*

Ans. Given p +2py cot x = y

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Given Equation can be written as

0
p+(2ycot x)p- y
=

Solving for p we get

-2ycot x t V4ys cot x+

4y
P
2

P = y cot x t y cos ecx

p=yl - sin x

p=y0Scos X* or P -coSX
sinx Sin x

2ysin -2ycos
or P
A

2sin.cos, 2sin.coS

P=ytan or p=-ycot

If p=y tan then

dy=ytan
X

dx

= tandx
y 2

on integrating

-uana
logy=2 log|sec+loge
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y=csec*

ycos
y(l+cos x) = 2c =c (say)

y(l+cos x)- c =0

If p=-y cot then

2

otdx
y

logy =-2 logl sin+ logc

y C2 vsin?
ysin" C2
sin
2
y(l-cosx) =2c2 =c (say)

y(1-cos x) -c =
0
Required solution is

(y(1+ cosx)-c) (y(l-cosx)- c) = 0

Q.3. (a) Solve the following differential equation:
e"(p-1) + p e =0
Ans. Given
e(p- 1) +p* e2y =0 )
Put eX X e= Y
eXdx = dx
edy dy
dydye x dy X .

Pdx e'dx y dx y
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dy
Where Pdx
Hence Equation (i)
Reduce too

p-1P'y-o
p-y+p-0 x0. y=0)
y= px +p
by constant e
Which is elqiraut's form replacing p
ie y Cx +c>
e ' = cex+ e3
to find the
Q.3. (b) Use the method of variation of parameters
solution of the given differential equation

y" =3y'+ 2y =cos(e

Ans. Given (D - 3D + 2 ) y = cos(e")

y2-3y + 2y = cos(e) ...)

Comparing (1) with y2 + p 9 y = R

Whose R cos(e) ..(in)

Auxilary Equation of (i) is
(D-3D+2) 0 giving =
D= 1.2
Hence C.F of (i)
CF = c,ex +c, e2x ...ii)
Where c, and c, being arbitrary constant
Let u = ex =e2x
Now

- 2e- =e #0
Than particular solution for integral of (i)
uf(x)+ vg(x)
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9.4. (a) Express x + 2xd+2x x-3
in terms of Legendre's polynomials.
Ans. Let
d+2x3+2x- x-3
ap,(x) + bp;(x) + cpi(x) + dp,(x)...(i)

+c(x)+ d(l)

ax-axbx-+Cxd 2

axb d
Equating the coefficients of like
Power of x we get

5a=l a
5

b2 b=
a+c=-1 c=-l-
e-1 =0-6-
1010

-d--3
d--3.-3 -3.14
Putting these values of a, b, c and d in Equation (i) we get

x+2x+2x -x -3
8
PCx)Pa(*)-P()Po)
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Q.4. (b) Prove that

P(cos)+P,(cos0)+.. =logl+cosec(0/2)
Ans. From the generating function

k=0

Integrating (i) w.r. to x from 0 to I

dz
SP.dz =- 22+2
n=00
i)
Replacing x by cosf on both side (ii) gives

dz

n=0 0 1-22 cos+z

dz

n=0
Vz-cos8 +sin e
Pn os6)=log{z- cos+y[z-cos6) +sin6
i
n+l
n=

log (1-cos0)+ -cos 0) +sin -log(1-cos)

=

log| (1-cos6)+ /2(1 - cos0)-log(1 -cos6)

log(-cos0)+v2/1-cos
1- cos6
= log V-cose1-cos+ 2/1-cos
VI-coseI-cos6

= log- V-cos+V2 2sin+N2

log
VI-cos0
2sin
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I+9Sin
= log-

Sin
2

Pocos)p,(cos8)+p(cos 6) +.

1+SIn
log-
Sin
2

1+pcos0)+Pa (cos 0) +..

1+Sin
= log 2
Sin
2

Since po (cos6) =

1| i)
ie
1P (cos0)+ P;(cos0)+. =log I+cosee
Q.5. (a) Solve the following differential equation :
z{x + y) p + z(x - y) q =x2+ y2

Ans. Given z(x + y) pt z(x - y)q=x +y2

The Lagrange's subsidiary equation are

dx
z(x + y) z(x -
dy y) x
dz
+y2

ChoosingX - Y1 - Z as multipliers we get

xdx - y d y - zdz
xdx -

ydy zdz
xz(x+ y)- yz(x -

y)-z(x* -

y^)
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x d x - y d y - zdz =0

2xdx-2ydy -2zdz =0
on integrating
y Z e
Where c, is an arbitrary constant
Again choosing y, X -z as multipliers

ydx +xdy-zdz
yz(x+y)+ xz(x - y) -z(x* - y*)

ydx+ xdy-zdz

y d x + xdy - zdz=0

2d(xy) -2zdz =0
on integrating
2xy z C where c, is an arbitrary constant

Hence Required Solution are

ox-y-, 2xy-z)=0
Q.5. (b) Find the complete integral of
zp?y +6zpxy + 2zqx2+ 4x?y =d
Aas. z<p'y + 6zpxy + 2zqx + 4x-y =0
f F z p í y + 6zpxy+ 2zqx- + 4x*y

6zpy+ 4zqx +8xy

OX

=z'p*+ 6zpx +4x

2zp y+ 6pxy 2qx*

+

d- 2pz?y + 6zxy

of 2zx
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Charpits Auxiliary Equations are

dq dz
dp
of

dx dy
of p
op cq

dp
6zpy + 4zqx +8xy + p(2zp°y +6pxy+ 2qx*)

zp +6zpx +4x +q(2zp'y +6pxy+2qx*)

dz
-P(2pz y +6zxy) - q(2zx<)

dx dy dp
-(2pz^y +6zxy) -2zx

From the first and last members we

get
dp =0 P

z ' a ' y +6zaxy + 2zqx* + 4x^y = 0

2zqx=-z'a'y + 6zaxy + 4x*y

(zay +6zaxy +4x y

2

dz= pdx + qdy

dz .za'y+ 6zaxy +4x*y

=adx + (-)
2 dy
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on integrating

Z a
za'y 3axy 2x y
4 2z
differential equations:
Q.6. Find the solution of following partial
(a) r - t = tan' x tan y- tanx tany
Ans. Given r - t = tan' x tany- tanx tan'y

tan'x tan y - tan x tan" y

y
Auxiliary Equation is m2-1 =0 m=+|
CF y+x)+ f,y-)

PI tan x tan
y| sec x
-

sec y

D- D
D+DD_nitan xsec x tan y

tan x tan y secs y]

tan xsec xtan(c-x)dx

D+D'L.

tan x tan(c-x)seci(c x)dx

-

Where y c-x

D.D
D+D'|
tan(c -x) see x(c-x)

tan dx +tan xtan2 X

2

e* xtan? d«
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tan(cx)tan' x + tan s
tan'(c -y)

see (e-xXsecx -1)dx

-ssee-*)-1]ak
1 tan(c-x)
2/0.Dtan(c x)tan* x tan
-

secx -se e-x)

. t a n ( c - x ) tan* x+ tan xtan (c-x)
2D+D')
+tan x +tan(c x)]

[tan xsec(c-x)+ tan(c-x)sec- x]

2(D+D)

tanx secy+ tan ysecs x]

2(D+D')

where cy+z
- tan x sec (b+ xdk +| tan(b+ x)sec? dk

where y =b+x

tan(b + x)dx
tan x. tan(b x)-| sec
+ x

+tan(b+x)sec*xdx

=tan x tan(b + x)
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t a n x tan y
where b y- x

Hence complete solution is

z
=
CF +
pf f,(y+x)+ f,(y
-

x) +
, tan xtan y
differential equations
Q.6. Find the solution of following partial
(b) (2D -3D' + 7) (D2 +3D') z 0
=

(2D -3D'+7} (D2 +3D') z 0

=

Ans. Given
Put D m D' 1
Auxiliary Equation are

(2m-3+7) (m+3) 0 =

(2m+4) (2m +4) (m3 +3) =0

= 2m +4 0 m=-2

2m +4 0 m=-2

m2+3 0 m=+ V3i

CF f(4-2x)+x f,(4-2x) +f,(4 t 3 ix)
+4-3 ix)
Q.7. (a) Classify the partial differential equation

axet*2

Ans. u +t +x +2 6u = 0
o ôx
Let us consider the general form of second order linear partial
differential equation
autbuy
Now
t cu,
b=t,
du, t eu, +fu= yx, u) +

a =
1, c =

x, d =2, e =
1, f=6
Cannot all be zero.
Itsdiscriminate to beb- 4ac
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Now Ifb:-4ac > 0 then the equation is called haperbola, If b2.

4ac 0, then the equation is called parabolic, Ifb-4ac <0 then

the equation is called elliptic
In general elliptic equation describe processes in equilibrium
while the hyperbolic and parabolic equations model processes
which evolve over time
Q.7. (b) ind the solution of Laplace's equation in cylindrical
coordinates.
Ans. Laplace equation in cylindrical coordinates is

u1u 1 8u 1)
Let u(r,0,z) =
R(r) F(0)z(2) ...(2)
or u= RFZ be solution
a
of (1)
Putting it in (1) we get

RFZ+R'FZ+RF°Z + RFZ" =
0

Dividing by RFZ

1 d'R 1 dR1 dF id'z_0

R dr? rdrPF do 2 i ..3)

F
Assuming = -n^F and =k'z .(4)
de dz
Equation (3) reduces to

d'R,dK
Rdr
+k0
dR dR
(k-n
dr dr )R =0
This is Bessels Equation
It solution is R
=C/, (kr) C^ Y, (kr)
+

The solution
of equation (3) are
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F=C3 COS ne + C4 sin ne

Z=ce +c,e-kz
Hence equation (1) is
u=
[ei, (kr) +c2 y, (kr))
cosn+c sin 0] (,e" +c,e
Which is known as a
cylindrical harmonic
Q.8. The points of trisection ofa string are pulled aside
through
the same distance on opposite sides of the
position of
equilibrium and the string is released from rest. Derive an
expression for thedisplacement of the string at subsequent
time and show that the midpoint of the
string always remains
at rest.
Ans. The Equation for the vibration of the
string is

ot .1)
The solution of Equation (1) is
y(x. t)= (c,cos cpt + c,sin cpt) (c, c sin
L be the length of
cos pxt px) .2)
Equation of OB is
string

y- h-0 x-0)
-0 3n
..3)

(2)
7u.o
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Equation of BC is

y-h= -2h

y-h=-oh

y= 3h ..4)
Equation of CA is

y+h-
3hx 2h

y= 3 5)

Hence boundary conditions are

y(0, t)= 0
y, t)=0

Cy=0 whent =
0

3h
0sxs

and y(x, o) -2x)xs

3h

From (2)
y(0, t)= (c,cos cpt t c, sin cpt) (c3)
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From (2)
y(x, t)= (c, cos cpt + c, sin cpt)c sin px

.(6)
yl.t)=0=(c, coscpt+c, sin cpt) c sin pl
sin p/ = 0 sin na (n e J)

rom (6) y(x,t) =C, S nCL Sin nzctin nzx

...(7)
sin ntct cosTnct C nTX
+C2
ot

At t=0 -0-e,Sin
C0
ntct nTx
From (7) yx, t)= Cq.C4COSsin

nTx
=
b, cos C -Sin

The most general solution is

nict nTX
y (x, ) =

, cosSin
n=

nTX
y (x, 0) =
2b, sin- where

y x.0)sindx
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dx -2x)sindx
/3

x-)sindx
2//3

-2h nT 6h
nT-cOs 22 S

2hcos
cos 2nt,2
coscos
nT

12h 2nT
n?2S1n- s i n

l
18h 18h 2nT
2 Sin-3 n 2 - Sin
3

18h
3 nun
18h Sinar 18h nit
n?2SIn
Sn coSnTT
3
n
18h
n22Si
+cos n7)

36h nt
n2 3
When n
is even
0 when n is odd
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sincos sinx
y(x,t) = n
n

2mTs S 2mrx

ml m
Where n= 2m ..(9)
Which is the required expression for
y(x. t)

Putting x =; in equation (6) we get

2mt 1 2mtct
CoS-
Sin ma = 0

m=
Hence mid point of the string is always at rest.
Q.9. (a) Solve the initial value problem yy'=x, y(0)= 1, using the
Eulermethod in 0Sxs0.8,with h 0.2 and h= 0.1. Compare
the results with the exact solution at x= 0.8. Extrapolate the
result.
Ans. We have y' = f(x, y) = (wly)

Euler method gives

Y+, + hf [x, y] =
y, +

Initial condition gives x% =0 Y%=

When h 0.2 We get

Yi Y+
0.2x
yi
Now

y(x)= y(0.2) = yi = Yot 0.2x0 10

Yo
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y(x2)= y(0.4) = y2 = y+ 0.2x

= 1.0+0.2X0.2)

1.0
1.04

Yx3)= y(0.6) = yy = y2 + 0.2x2

Y2

=1.04+20.4) 1.11
1.04
y(x) = y(0.8)

y4y3 +
0.2(x3)
y2

=1.11+.2(0.6)
1.11
=1.22
When h 0.1 we get

Yi+Yi
0.Ix1
y1

y(x)= y(O.1) = Y1 = Yot-

0.1X01.0
Yo

y(x2)= y(0.2) = y2 = y i t 0.1X1

= 1.0
1.0

y(x)= y(0.3) = ys = Y2t- 0.1x2

y2
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= 1.01+ o.I(0.2) - 1.02
1.01

0. Ix3
+
y(x)= y(0.4) =y4
=
Ys
Y3

=1.02+ .I(0.3)=
1.02
|.05
y(xs)= y(0.5) = ys = Y4 + 0.1x4
Y4

= 1.05+(0.4)=1.09
1.05

t 0.1xs
yx)= y(0.6) =
Ys =
Ys
Ys

1.09+.0.1(0.3)1.14
1.09

0. Ixg
y(x7)= y(0.7) = y7 = Ys +
Y6

= 1.14+ 0.10.6) 1.19

1.14

0.1x7
y(xg)=y(0.8) =ys =
y7 +
y7

=1.19+ 21.25
=1.19*1.19
The exact solution is y =vx2 +I
Atx = 0.8 the exact value is y(0.8) = VI.64=1.28
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Organizer
The magnitudes of the errors in the solution the
h 0.2 |1.280- 1.2241 =0.05
are
following
h =0.1 |1.28-1.25|
=0.02
Now extrapolated result as
y(0.8) = 12 yp (h/2)- y (h)

= 2(1.25)-1.22 1.28
The magnitude of the error in Extrapolated result is givenn by
|1.280 1.282] = 0.001
Q.9. (b) Write the bound on the truncation error of the Taylor
series method.
Ans. The n terms of the approximation are simply the first n terms of
the exact expansion

x
e = 1+x+
2
So the function f(x) can be written as the Taylor's series
approximation plus an error (Truncation) term
f(x) = f, (x) + E, (x)

Where E, (x) =
(E)
*
(x-c
n+