Differential Equations

Integrating both sides of (1), we have MODULE - V

Calculus
∫( x )
dy
= 2 + sin3x dx + C1 ,
dx
where C1 is an arbitrary constant

dy x 3 cos3x
or = − + C1 .....(2) Notes
dx 3 3

 x 3 cos3x 
or dy =  − + C1  dx
 3 3 
Further on integrating both sides of (2), we get
 x3 cos3x 
y = ∫ − + C1  dx + C2,
 3 3 
where C2 is another arbitrary constant.

x4 1
or y= − sin3x +C1x +C2 ....(3)
12 9
Now substituting y = 0, when x = 0 in ( 3 ), we have, 0 = C2
dy
Again substituting = 0 when x = 0 in (2) we have
dx
1 1
0=− + C1 or C1 =
3 3
Putting values of C1 and C2 in (3), we get

x4 1 x
y= − sin3x +
12 9 3
which is the required particular solution.
d 3y
28.7.2 Differential Equations of the Type = f(x)
dx 3
Consider the differential equation
d3 y
= f(x)
dx 3
This is a differential equation of order three. Its general solution will contain three arbitrary
constants. In order to find its particular solution we need three conditions [ see Example 28.21].
For solving the differential equation of order three of the form
d3 y
= f ( x ) , integrate the equation three times.
dx 3

MATHEMATICS 469
 Differential Equations
MODULE - V d3 y
Calculus Example 28.20 Solve 3
= x4
dx
Solution : The given differential equation is
d3 y
3
= x4
Notes dx

d  d2 y 
=x
4
or  .....(1)
dx  dx 2 

On integrating both sides of (1), we will get

d2 y x5
= + C1 ,
dx 2 5
where C1 is an arbitrary constat.

d  dy x5
or   = + C1 .....(2)
dx  dx  5
Further on integrating both sides of (2), we have
dy x 6
= + C1x +C2 ....(3)
dx 30

where C2 is another arbitrary constant.

On integrating both sides of (3), we have
x7 x2
y= +C1 +C2 x +C3 ,
7 ( 30 ) 2
where C3 is also an arbitrary constant.

x7 C x2
or y= + 1 +C2 x +C3
210 2
Example 28.21 Find the particular solution of the differential equation
d3 y
3
= xex
dx
dy
for which y = 1, = 2, when x = 0
dx
d 2y
and = 0 , when x = 1.
dx 2
Solution : The given differential equation is

470 MATHEMATICS
 Differential Equations
MODULE - V
d3 y
= xe x
.....(1) Calculus
dx3
On integrating both sides of (1), we get
d 2y
dx 2
= ∫ xe xdx + C1
Notes
where C1 is an arbitrary constant.

d 2y
or 2
= xe x − ∫ ex dx + C1
dx
d 2y
2
= xe x − ex + C1 .....(2)
dx
Again on integrating both sides of (2), we will get

∫ ( xe )
dy
= x − ex + C1 dx +C ,
dx 2

where C2 is another arbitrary constant.

dy
or = xex − ∫ e xdx − e x + C1x + C2
dx
dy
or = xex − 2e x + C1x + C2 ....(3)
dx
Further on integrating both sides of (3), we have
x2
y = xe x − ex − 2e x + C1 + C 2 x + C3 ,
2
where C3 is also an arbitrary constant.

x2
or y = xe x − 3ex + C1 + C2 x + C3 .....(4)
2
dy d 2y
Now we have y = 1, = 2 when x = 0 and = 0 , when x = 1
dx dx 2
d2 y
On substituting = 0,x = 1 in equation (2), we get
dx 2
0 = e1 − e1 + C1
∴ C1 = 0
dy
Now putting = 2 , when x = 0 in equation (3), we have
dx
2 = 0 − 2e 0 + C1 .0 + C2
⇒ 2 = − 2 + C2 ⇒ C2 = 4
MATHEMATICS 471
 Differential Equations
MODULE - V Now putting y = 1, when x = 0 in equation (4), we have
Calculus 1 = 0 − 3.e0 + C1 .0 + C2 .0 + C3
∴ C3 = 4
Further on substituting values of C1, C 2 and C3 in equation (4), we will get

Notes y = xe x −3ex + 4x +4
which is the required solution.

CHECK YOUR PROGRESS 28.3

1. Find the particular solution of the differential equation

d 2y dy π
= sin 2 x given that = 0 , when x = ; y = 0 when x = 0
dx 2 dx 2
2. Solve the differential equation

d3 y dy d 2y
= x + sin x given that y = 0 , = −1 , = 2 when x = 0
dx 3 dx dx 2
3. Solve the following differential equations :

d 2y d2 y
(a) = xsinx (b) ( 1 − cos2x ) = 1 + cos2x
dx 2 dx 2

d 2y d 2y
(c) = sin 2 x (d) = cosx − s i n x
dx 2 dx 2
4. Solve the following initial value problems :

d2 y dy
(a) x = 1 , given that y = 1 , = 0 , when x = 1
dx 2 dx

d 2y dy
(b) = 1 + sinx, given that x = 0 , y = 0 , =0
2 dx
dx

d 2y
(c) = 2 , x = 1 when y = 3 and x = −1 when y = 1
dx 2
5. Solve the following differential equations :
d3 y d3 y d3 y
(a) = x 2 + e x (b) = ex + c o s x (c) =0
dx 3 dx 3 dx3

28.8 APPLICATIONS OF DIFFERENTIAL EQUATIONS

Here we consider the applications of differential equations in some of the problems of Physics,
Economics, Biology and other related fields.

472 MATHEMATICS
 Differential Equations
28.8.1 Problems of Velocity and Acceleration MODULE - V
We know that velocity 'v' and acceleration 'a' of a body at any instant are given by Calculus
velocity = rate of change of displacement
d
∴ v= ( s ) , where s is the displacement.
dt
Notes
And acceleration = rate of change of velocity
d
a = (v)
dt
d  d 
=  ( s) 
dt  dt 

d2
= (s )
dt 2
Example 28.22 A stone is dropped from the top of a tower 44.1 m high. Velocity 'v' at any
instant is given by v = 9.8 t m/s (metres per second)
If it is given that distance after one second is 4.9 m
(i) find the distance of the stone when t = 2 seconds
(ii) value of 't' when stone hits the ground.
Solution : We are given that v = 9.8 t
d
∴ ( s ) = 9.8 t
dt
On integrating both sides, we get

∫ ds = ∫ 9.8tdt
or s = 4.9t 2 + C , where C is an arbitrary constant .....(1)
It is given that s = 4.9 m when t = 1 second. Using this condition we get,
C=0
Substituting the value of C in (1), we will get

s = 4.9t2 .....(2)
(i) Putting t = 2 in (2), we get

s = 4.9 × 22 or s = 19.6 m
So it is clear that the distance of the stone in 2 seconds is 19.6 m.
(ii) Now when stone hits the ground the distance is 44.1 m. Thus, s = 44.1
Putting s = 44.1 in (2), the following will be the outcome.

44.1 = 4.9 t 2
MATHEMATICS 473
 Differential Equations
MODULE - V 44.1
Calculus or t2 = or t2 = 9 or t =3
4.9
Hence the stone hits the ground after 3 seconds.
28.8.2 Population Growth Problems
Rate of growth of population is proportional to the population. For example the bacteria
Notes
dx
population grows at a rate proportional to the population i.e. the growth rate is proportional
dt
to x where x = x (t) denotes the number of bacteria present at time t. This fact can be represented
by means of the differential equation
dx
= kx where k is positive constant of proportionality..
dt
Example 28.23 In a certain culture of bacteria the rate of increase is proportional to the
number present.
(a) If it is found that the number doubles in 4 hours, how may times is the number expected
at the end of 12 hours ?

(b) If there are 104 at the end of 3 hours and 4 × 10 4 at the end of 5 hours, how many
were there in the beginning ?
Solution : Let x denote the number of bacteria at time t hours. Then ,
dx
= kx , where k is an arbitrary constant
dt
dx
or = kdt .....(1)
x
On integrating both sides, we get
dx
∫ x
= ∫ k dt + logC,
where C is an arbitrary constant.
or log x = kt + log C
Note that the number x is positive.
x x
or log = kt or = e kt
C C
or x = Ce kt .....(2)
Assuming that x = x 0 at time t = 0, we get
x0 = C

Putting value of C in (2) we get

x = x 0e kt .....(3)
474 MATHEMATICS
 Differential Equations

(a) It is given that when t = 4, x = 2x 0 substituting these in (3), we get MODULE - V

Calculus
2x 0 = x 0 e4k or e 4k = 2
When t = 12

( )
3
x = x 0e12k or x = x 0 e4k
Notes
x = x0 ( 2 ) x = 8x 0
or 3 or
Hence the growth of bacteria is 8 times the original number at the end of 12 hours.
(b) Now it is given that at t = 3, x = 104 and at t = 5, x = 4 × 10 4
Substituting these values in (2), we have
104 = Ce3k .....(4)
and 4 × 10 4 = Ce5k .....(5)
From (4) and (5), we get
4 × Ce3k = Ce5k or 4 = e 2k
1
or log 4 = 2k or k = log4
2
or k = log 4 = l o g 2
Substituting value of k in equation (4), we get
3
104 = Ce3 l o g 2 or 104 = Ce log2
or 104 = Ce log8 or 104 = 8C
104
or =C
8
Substituting values of C and k in equation (2), we get
104 t(log 2 )
x= e .....(6)
8
Substituting t = 0 in equation (6), we will get the number of bacteria in beginning.
104 104 × 1
∴ x= × e0 or x=
8 8
25 × 10 2
or x= or x = 25 × 50 = 1250
2

28.8.3 Newton's Law of Cooling Problems

Newton's law of cooling states that the rate at which the temperature T = T(t) changes in a
cooling body is proportional to the difference between the instantaneous temperature T of the
body and the constant temperature T0 of the surrounding,

MATHEMATICS 475
 Differential Equations
MODULE - V dT dT
i.e. = −k ( T − T0 ) , where k > 0. Negative sign indicates that if T > T0 , is negative,
Calculus dt dt
that is the temperature decreases with time.
Example 28.24 A body whose temperature is initially 100°C is allowed to cool in air, whose
temperature remains at a constant 20°C. It is given that after 10 minutes the body has cooled to
Notes 40°C. Find the temperature of the body after half an hour.
Solution : Let T denote the instantaneous temperature of body and t denote the time. Then
according to the Newton's Law of Cooling
dT
= −k ( T − 20 ) , where k > 0, is a constant.
dt
dT
= −kdt .....(1)
T − 20
dT
On integrating both sides of (1), we get ∫ T − 20 = −∫ kdt
log ( T − 20 ) = −kt + logC , where C is an arbitrary constant
or T − 20 = Ce− kt
or T = 20 + Ce −kt .....(2)

Now substituting t = 0, T =100 in equation (2), we get

100 = 20 +C
or C = 80
On substituting value of C in equation (2), we get
T = 20 + 80e −kt .....(3)

It is given that when t = 10, T = 40 and on substituting these in (3), we get

40 = 20 + 80e−10k or 80e−10k = 20
1
e −10k
1  1 10
or = or e −k = 
4 4
t
 1 10
Thus T = 20 + 80  
4
Now after half an hour i.e. when t = 30, we have
1 3
T = 20 + 80  
4
1 1 1
or T = 20 + 80 × × ×
4 4 4
or T = 20 + 1.25 or T = 21.25°C
476 MATHEMATICS
 Differential Equations
MODULE - V
CHECK YOUR PROGRESS 28.4 Calculus
1. The population of a certain country is known to increase at the rate of 10 % per year.
How long does it take for the population to double ?
2. A body at temperature 80°F is placed at time t = 0 in a medium, the temperature of which
is maintained at 50°F. At the end of 5 minutes, the body has cooled to a temperature of Notes
70°F. When will the temperature of the body be 60°F ?

3. A particle is moving in a straight line, the velocity v (m/s) at any instant is given by v = 4t 2 .
It is given that the distance of the particle at the end of 1 second is 2 metres. Find the
distance of the particle at the end of 3 seconds.
4. The number of bacteria in a yeast culture grows at a rate which is proportional to the
number present. If the population of a colony of yeast bacteria triples in 1 hour, how
many may be expected at the end of 5 hours ?

Example 28.25 Verify if y = emsin −1x is a solution of

( 1 − x 2 ) ddx y2 − dx
2 dy
− m2 y = 0

Solution : We have,
−1 x
y = e msin .....(1)
Differentiating (1) w.r.t. x , we get
−1
dy memsin x my
= =
dx 1 − x2 1 − x2
dy
or 1 − x2 = my
dx
Squaring both sides, we get
2
(1 − )x2  dy  = m 2 y2
 
 dx 
Differentiating both sides, we get
dy 2
( )
2
−2x   + 2 1 − x 2 dy . d y = 2m 2 y dy
 dx  dx dx 2 dx

or −x
dy
dx
(
+ 1 − x2
d 2y
dx 2 )
= m2 y

( 1 − x 2 ) ddx y2 − x dx
2 dy
or − m2 y = 0

MATHEMATICS 477
 Differential Equations
MODULE - V −1
Hence y = e msin x is the solution of
Calculus
( 1 − x2 ) dx 2 − x dx
d2 y dy
− m2 y = 0

Example 28.26 Find the particular solution of

Notes dy
e dx = x + 1 , given that y = 2 when x = 0
Solution : We are given that
dy
dy
e dx = x + 1 or = log ( x + 1 )
dx
or dy = log ( x + 1 ) dx .....(1)
On integrating both sides of equation (1), we get
y = ∫ log ( x + 1 ) .1dx + C
where C is an arbitrary constant
x
or y = x log ( x + 1 ) − ∫ x + 1 dx + C
x +1 − 1
or y = x log ( x + 1 ) − ∫ dx + C
x +1
or y = x log ( x + 1 ) − [ x − log ( x + 1 ) ] + C
or y = ( x + 1 ) log ( x + 1 ) − x + C .....(2)

Substituting y = 2 and x = 0 in equation (2), we get

2 = log1 − 0 + C or C=2
Putting the value of C in equation (2), we get
y = ( x + 1 ) log ( x + 1 ) − x + 2
which is the required solution.
Example 28.27 Find the equation of the curve represented by
( y − yx ) dx + ( x + xy ) dy = 0
and passing through the point ( 1, 1 ).
Solution : The given differential equation is
( y − yx ) dx + ( x + xy ) dy = 0
or ( x + xy ) dy = ( yx − y ) dx
or x ( 1 + y ) dy = y ( x − 1 ) dx

(1 + y ) x −1
or dy = dx .....(1)
y x
478 MATHEMATICS
 Differential Equations
Integrating both sides of equation (1), we get MODULE - V
Calculus
 1 + y x −1
∫   dy = ∫   dx
y   x 

1   −  dx
1
or ∫  y + 1 dy =∫  1 x
.....(2) Notes

or log y + y = x − logx + C
Since the curve is passing through the point (1,1), therefore,
substituting x = 1, y = 1 in equation (2), we get
1=1+C
or C=0
Thus, the equation of the required curve is
log y + y = x − l o g x

or log ( xy ) = x − y

dy 3e2x + 3e 4x
Example 28.28 Solve = x
dx e + e− x

dy 3e2x + 3e 4x
Solution : We have = x
dx e + e− x

or
dy 3e
=
3x e−x + ex
( ) or
dy
= 3e 3x
dx ex + e− x dx

or dy = 3e3x dx .....(1)
Integrating both sides of (1), we get

y = ∫ 3e 3x dx + C

where C is an arbitrary constant.

e3x
or y=3 +C or y = e3x + C
3
which is required solution.
Example 28.29 Find the solution of the following differential equation :

dy  dy 
y−x = a  y2 + x2 
dx  dx 

satisfying x = a, y = a

MATHEMATICS 479
 Differential Equations
MODULE - V Solution : The given differential equation is
Calculus dy  2 2 dy 
y−x = a y + x 
dx  dx 

or y − ay 2 = x
dy
dx
+ ax 2
dy
dx
or ( y − ay 2 ) = ( x + ax 2 ) dy
dx
Notes
dy dx dy dx
= =
y ( 1 − ay ) x ( 1 + ax )
or or
y − ay 2
x + ax 2

1 a  1 a  dx
or  y + 1 − ay  dy =  x − 1 + ax  .....(1)
 

Integrating both sides of (1), we get

1 a   1 − a  dx +C
∫  y + 1 − aydy = ∫  x 
1 + ax 

where C is an arbitrary constant

a a
or log y − log ( 1 − ay ) = log x − log ( 1 + ax ) + logC1
a a
where C = logC1 ( x, y > 0,) ( 1 − ay ) > 0, (1 + ax ) > 0
or log y − log ( 1 − ay ) = logx − log ( 1 + ax ) + logC1

 y 
= log  1 
Cx
or log  
 1 − ay   1 + ax 
or y ( 1 + ax ) = C1 x ( 1 − ay ) .....(2)

Now using the condition that y = a, where x = a in equation (2), we get

1 + a2
a 1+( a2 ) = C1a ( 1 − ) a2 or C1 =
1 − a2
Substituting value of C1 in equation (2), we get

(
y ( 1 + ax ) 1 − a 2 ) = x ( 1 − ay ) ( 1 + a 2 )
which is the required solution.

CHECK YOUR PROGRESS 28.5

1. (a) If y = tan −1 x , prove that ( 1 + x ) dx 2 + 2x dy

2 d2 y
dx
=0

d 2y dy
(b) y= ex s i n x , prove that 2
−2 + 2y = 0
dx dx
480 MATHEMATICS
 Differential Equations
2. Find the particular solution of the differential equation MODULE - V
Calculus
 dy 
log   = 3x + 4y , given that y = 0 when x = 0
 dx 
3. (a) Find the equation of the curve represented by
dy
= xy + x + y + 1 and passing through the point (2, 0) Notes
dx
(b) Find the equation of the curve represented by
dy π 
+ y c o t x = 5ec o s x and passing through the point  , 2 
dx 2 

dy 4e3x + 4e5x
4. Solve : =
dx ex + e−x
5. Solve the following differential equations :
(a) dx + xdy = e − y sec2 ydy

(b) ( 1 + x 2 ) dx
dy
− 4x = 3cot −1 x

(c) ( 1 + y ) xydy = ( 1 − x 2 ) ( 1 − y ) dx

LET US SUM UP
• A differential equation is an equation involving independent variable, dependent variable
and the derivatives of dependent variable (and differentials) with respect to independent
variable.
• The order of a differential equation is the order of the highest derivative occurring in it.
• The degree of a differential equation is the degree of the highest derivative after it has
been freed from radicals and fractions as far as the derivatives are concerned.
• A differential equation in which the dependent variable and its differential coefficients
occur only in the first degree and are not multiplied together is called a linear differential
equation.
• A linear differential equation is always of the first degree.
• A general solution of a differential equation is that solution which contains as many as the
number of arbitrary constants as the order of the differential equation.
• A general solution becomes a particular solution when particular values of the arbitrary
constants are determined satisfying the given conditions.
dy
• The solution of the differential equation of the type = f ( x ) is obtained by integrating
dx
both sides.
dy
• The solution of the differential equation of the type = f ( x ) g (y) is obtained after
dx
separating the variables and integrating both sides.
MATHEMATICS 481
 Differential Equations
MODULE - V • The differential equation M (x, y) dx + N ( x , y ) dy = 0 is called homogeneous if
Calculus M (x , y ) and N ( x, y ) are homogeneous and are of the same degree.
• The solution of a homogeneous differential equation is obtained by substituting y = vx or
x = vy and then separating the variables.
dy
• The solution of the first order linear equation + Py = Q is
dx

∫ Q ( e∫ ) dx + C ,
Notes
ye ∫
Pdx Pdx
= where C is an arbitrary constant.

The expression e ∫ Pdx is called the integrating factor of the differential equation and is
written as I.F. in short.
d 2y
• The solution of a second order differential equation = f ( x ) is obtained simply by
dx 2
integrating it twice with respect to x and its general solution contains two arbitrary constants.

SUPPORTIVE WEB SITES

l http://www.wikipedia.org
l http://mathworld.wolfram.com

TERMINAL EXERCISE

1. Find the order and degree of the differential equation :

2
 d2 y  4
2  dy  = 0
(a)  2  + x   (b) xdx + ydy = 0
 dx   dx 

d 4y dy dy
(c) −4 + 4y = 5cos3x (d) = cosx
4 dx dx
dx
d 2y dy d 2y
(e) x2
2
− xy =y (f) +y=0
dx dx dx 2
3
dy  dy 
2   dy 
2 2
d 2y
(g) y=x +a 1+  (h) 1 +    =a
dx  dx    dx   dx 2

2. Find which of the following equations are linear and which are non-linear
dy dy y
(a) = cosx (b) + = y2 l o g x
dx dx x

482 MATHEMATICS
 Differential Equations

3
MODULE - V
 d2 y  2
2  dy  = 0 dy Calculus
(c)  2  + x   (d) x −4=x
 dx   dx  dx

(e) dx + dy = 0

3. Form the differential equation corresponding to y 2 − 2ay + x 2 = a 2 by eliminating a. Notes

4. Find the differential equation by eliminating a, b, c from

y = ax 2 + bx + c . Write its order and degree.

5. How many constants are contained in the general solution of
(a) Second order differential equation.
(b) Differential equation of order three.
(c) Differential equation of order five.
6. Show that y = acos ( log x ) + bsin ( log x ) is a solution of the differential equation

d 2y dy
x 2
+x +y=0
dx 2 dx
7. Solve the following differential equations:
dy dy
(a) sin 2 x = 3cosx + 4 (b) = e x − y + x 2 e− y
dx dx
dy c o s x s i n y
(c) + =0 (d) dy + xydx = xdx
dx cosy

(e)
dy
dx
+ y tan x = x m cosmx (f) ( 1 + y 2 ) dx
dy
= tan −1 y − x

d2 y d2 y
(g) 2
= cos x
2
(h) 2
= sin −1 x
dx dx
d2 y d3 y
(i) = ex − s i n x (j) = cos3x − sin3x
dx 2 dx 3
dv
8. If = gcos α − kv,g, α, k being constants. Find v in term of t if v = 0, when t = 0.
dt
9. Solve the differential equation
d2 y dy
= logx , given that y = 1 and = −1 , when x = 1.
2 dx
dx
10. If the temperature of air is 290 K and a substance cools from 370 K to 330 K in 10
minutes, find when the temperature will be 295 K.

MATHEMATICS 483
 Differential Equations
MODULE - V
Calculus ANSWERS
CHECK YOUR PROGRESS 28.1
1. Order is 1 and degree is 2.
Notes 2. (a) Order 2, degree 1 (b) Order 1, degree 2
(c) Order 1, degree 1 (d) Order 2, degree 2
3. (a) Non- linear (b) Linear
(c) Linear (d) Non-linear
3 2
  dy  2   d 2y 
1 +    =r  2 
2
4.
  dx    dx 

dy 2 2
5. (a) xy
d2 y
dx 2
+ x   −y
 dx 
dy
dx
= 0 (b) ( x 2 − 2y2 )  dy 

dx 
− 4xy
dy
dx
− x2 = 0

d 2y dy dy
(c) + − 6y = 0 (d) y = ( x − 3) +2
2 dx dx
dx

(e) ( x 2 − y 2 ) dy − 2xy = 0
dx

CHECK YOUR PROGRESS 28.2

1. (i) Yes (ii) No
2. (i), (ii) and (iv) are particular solutions (iii) is the general solution
3. (ii), (iv) are homogeneous
5. (a) y = tan x (b) y = tanx + 3

6. (a) y =
1 6
6
1
( )
1
x tan −1 x3 − x3 + tan −1 x3 + C
6 6
( )
1 1
(b) y = cos5 x − cos3 x + ( x − 1 ) e x + C
5 3
1 1 1
(c) y = log x 2 + 1 + C (d) y = x 3 − cos3x + C
2 3 3
7. y = −4e− x + 7
x 1 1 x2
8. (a) log y = C + x +y (b) log y + 1 = x + +C
2
x3
(c) tan x t a n y = C (d) ey = ex + +C
3
484 MATHEMATICS
 Differential Equations
MODULE - V
9. 3e −4y + 4e3x = 7
Calculus
(
10. (a) x = C x 2 − y2 ) (b) x + cy = ylog | x |

−1  y y
(c) sin   = log | x | +C (d) tan = Cx
x 2x
11. y ( sec x + tan x ) = sec x + tan x − x + C
Notes

12. (a) y = tan −1 x −1 +Ce − t a n x (b) y = tanx − 1 + Ce − t a n x

C
(c) y = log x +
log x
13. (a) x = Ce y − ( y + 2 ) (b) x = y 2 + Cy

CHECK YOUR PROGRESS 28.3

1 2 π 1 1 x4 3 2
1. y= x − x + cos2x − 2. y= + x − x + cosx − 1
4 4 8 8 24 2
1 2
3. (a) y = − x s i n x − 2cosx + C1x + C2 (b) y = C1x + C2 − logsinx − x
2
1 2 1
(c) y = x + cos2x + C1 x + C2 (d) y = − cosx + sinx + C1x + C2
4 8
1 2
4. (a) y = x ( log x − 1 ) + 2 (b) y= x − sinx + x
2
(c) y = x 2 + x + 1

x5
5. (a) y = e x + + C1x 2 + C2 x + C3 (b) y = e x − sin x + C1x 2 + C2 x + C3
60
(c) y = C1x 2 + C2 x + C3

CHECK YOUR PROGRESS 28.4

1. t = 10log2 years 2. 13.55 Minutes
2
3. 12 minutes 4. 243 Times
3
CHECK YOUR PROGRESS 28.5
2. 4e 3x + 3e −4y = 7
1 2
3. (a) log ( y + 1 ) = x +x −4 (b) y sin x + 5ecosx = 7
2

4 5x
4. y= e +C
5
MATHEMATICS 485
 Differential Equations
MODULE - V
− ( cot −1 x ) +C
3 2
Calculus 5. (a) x = e − y ( C + tan y ) (b) y = 2log 1 + x 2
2
x2 y2
(c) log x + 2log 1 − y = − −2y +C
2 2

Notes TERMINAL EXERCISE

1. (a) Order 2, degree 3 (e) Order 2, degree 1
(b) Order 1, degree 1 (f) Order 2, degree 1
(c) Order 4, degree 1 (g) Order 1, degree 2
(d) Order 1, degree 1 (h) Order 2, degree 1
2. (a), (d), (e) are linear; (b), (c) are non-linear
dy 2
3. ( )
x 2 − 2y2   dy  − x 2 = 0
 − 4xy 
 dx 

 dx 

d3 y
4. = 0 , Order 3, degree 1.
dx3
5. (a) Two (b) Three (c) Five
x3
7. (a) y + 3 cosec x + 4 cot x = C (b) ey = ex + +C
3
x2
(c) siny = Ce− s i n x (d) log ( 1 − y ) + =C
2
x m +1 −1 y
(e) y = cosx + C c o s x (f) x = tan −1 y − 1 + Ce− tan
m +1
x 2 cos2x
(g) y= − + C1 x +C2
4 8
 x 2 1 3
(h) y= +  sin −1 x + x 1 −x 2 +C1x +2
C
 2 4 4
sin3x cos3x C1x 2
(i) y = e x +sinx +C1x +C2 (j) y = − − + +C2 x +C3
27 27 2
gcos α
8. v=
k
(
1 − e − kt )
x 2 log x 3 2 7
9. y= − x +
2 4 4
10. 40 minutes.

486 MATHEMATICS