Differential Equations
MODULE - VIII
Calculus
32
DIFFERENTIAL EQUATIONS
Notes
Having studied the concept of differentiation and integration, we are now faced with the question
where do they find an application.
In fact these are the tools which help us to determine the exact takeoff speed, angle of launch,
amount of thrust to be provided and other related technicalities in space launches.
Not only this but also in some problems in Physics and Bio-Sciences, we come across relations
which involve derivatives.
ds
ds
 4.9 t 2 where s is distance and t is time. Therefore,
dt
dt
represents velocity (rate of change of distance) at time t.
One such relation could be
Equations which involve derivatives as their terms are called differential equations. In this lesson,
we are going to learn how to find the solutions and applications of such equations.
OBJECTIVES
After studying this lesson, you will be able to :
l
define a differential equation, its order and degree;
l
determine the order and degree of a differential equation;
l
form differential equation from a given situation;
l
l
illustrate the terms "general solution" and "particular solution" of a differential equation
through examples;
solve differential equations of the following types :
dy
 f ( x)
dx
(ii)
dy
 f ( x) g( y )
dx
dy
f ( x)
(iii) dx  g  y 
(iv)
dy
 P xy  Q x
dx
(i)
l
find the particular solution of a given differential equation for given conditions.
EXPECTED BACKGROUND KNOWLEDGE
l
Integration of algebraic functions, rational functions and trigonometric functions
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32.1 DIFFERENTIAL EQUATIONS
As stated in the introduction, many important problems in Physics, Biology and Social Sciences,
when formulated in mathematical terms, lead to equations that involve derivatives. Equations
dy d 2 y
,
(or differentials) etc. and
dx dx 2
Notes
independent and dependent variables are called differential equations.
which involve one or more differential coefficients such as
For example,
dy
 cos x
dx
(i)
(ii)
d2 y
dx 2
y0
(iii)
2
3
 d2y 
2  dy 

x
(iv)  2 

 0
dx


dx


(vi)
dy
 dy 
y
 1 

dx
 dx 
xdx  ydy  0
2
32.2 ORDER AND DEGREE OF A DIFFERENTIAL EQUATION
Order : It is the order of the highest derivative occurring in the differential equation.
Degree : It is the degree of the highest order derivative in the differential equation.
Differential Equation
(i)
dy
 sin x
dx
(ii)
 dy 
2

  3y  5x
dx


(iii)
4
 d 2s 
2  ds 
 2  t 
 0
 dt 
 dt 
Order
Degree
One
One
One
Two
Two
Two
Three
One
Four
Two
2
2
(iv)
d3v
dr3

2 dv
0
r dr
2
(v)
5
3
 d4y 
3 d y 

x
 4 
 3   sin x
dx


 dx 
Example 32.1 Find the order and degree of the differential equation :
d2 y 
dy  2 

 1  
 0
 dx  
dx 2 
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MATHEMATICS
Differential Equations
MODULE - VIII
Calculus
Solution : The given differential equation is
2
d2 y 
 dy    0

1



 
 dx  
dx 2 
or
2
d2 y
 dy 

 1
2
 dx 
dx
Hence order of the diferential equation is 2 and the degree of the differential equation is 1.
Notes
Note : Degree of a differential equation is defiend if it is a palynomial equation in terms of its
derivatives.
32.3 LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONS
A differential equation in which the dependent variable and all of its derivatives occur only in the
first degree and are not multiplied together is called a linear differential equation. A differential
equation which is not linear is called non-linear differential equation . For example, the differential
equations
d2 y
dx 2
 y  0 and
cos2 x
d3 y
dx 3
 x3
dy
 y  0 are linear..
dx
2
y
dy
 dy 
The differential equation 
is two.
   log x is non-linear as degree of
x
dx
 dx 
Further the differential equation y
y and its derivative
d2 y
dx 2
d2y
dx 2
 4  x is non-linear because the dependent variable
are multiplied together..
32.4 FORMATION OF A DIFFERENTIAL EQUATION
Consider the family of all straight lines passing through the origin (see Fig. 28.1).
This family of lines can be represented by
y = mx
.....(1)
Differentiating both sides, we get
dy
m
dx
.....(2)
From (1) and (2), we get
MATHEMATICS
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Calculus
y x
dy
dx
.....(3)
dy
represent the same family..
dx
Clearly equation (3) is a differential equation.
So y  mx and y  x
Notes
Working Rule : To form the differential equation
corresponding to an equation involving two variables,
say x and y and some arbitrary constants, say, a, b,
c, etc.
(i)
Differentiate the equation as many times as the
number of arbitrary constants in the equation.
(ii) Eliminate the arbitrary constants from these
equations.
Fig. 32.1
Remark
If an equation contains n arbitrary constants then we will obtain a differential equation of nth
order.
Example 32.2 Form the differential equation representing the family of curves.
y  ax 2  bx.
......(1)
Differentiating both sides, we get
dy
 2 ax  b
dx
......(2)
Differentiating again, we get
d2 y
dx 2
2 a
......(3)
1 d2y
......(4)

2 dx 2
(The equation (1) contains two arbitrary constants. Therefore, we differentiate this equation
two times and eliminate 'a' and 'b').
a
On putting the value of 'a' in equation (2), we get
dy
d2 y
 x
 b
dx
dx 2

410
b 
dy
d2y
 x
dx
dx 2
......(5)
MATHEMATICS
Differential Equations
Substituting the values of 'a' and 'b' given in (4) and (5) above in equation (1), we get
MODULE - VIII
Calculus
 1 d2y 
 dy
d2y 
y  x2 

x

x



2
dx 2 
 2 dx 
 dx
2
x 2 d2y
dy
2 d y

x

x
2 dx 2
dx
dx 2
or
y
or
yx
or
x 2 d2y
dy
x
y0
2
2 dx
dx
Notes
dy x 2 d 2 y

dx
2 dx 2
which is the required differential equation.
Example 32.3 Form the differential equation representing the family of curves
y = a cos (x + b).
Solution :
y = a cos (x +b)
Differentiating both sides, we get
.....(1)
dy
 a sin(x  b)
dx
....(2)
Differentiating again, we get
d2 y
dx 2
 a cos  x  b 
.....(3)
From (1) and (3), we get
d2 y
dx 2
 y
or
d2 y
dx 2
y0
which is the required differential equation.
Example 32.4 Find the differential equation of all circles which pass through the origin and
whose centres are on the x-axis.
Solution : As the centre lies on the x-axis, its coordinates will be (a, 0).
Since each circle passes through the origin, its radius is a.
Then the equation of any circle will be
 x  a 2  y 2  a 2
(1)
To find the corresponding differential equation, we differentiate equation (1) and get
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Calculus
2  x  a   2y
dy
0
dx
xay
or
dy
0
dx
dy
a y
x
or
Notes
dx
Substituting the value of 'a' in equation (1), we get
2
dy


 dy

 x   y2   y
x
xy
dx


 dx

2
2
2
dy
 dy 
 dy 
2
2
y
 y  x y
  2xy
dx
 dx 
 dx 
or
y 2  x 2  2xy
or
dy
dx
which is the required differential equation.
Remark
If an equation contains one arbitrary constant then the corresponding differential equation is
of the first order and if an equation contains two arbitrary constants then the corresponding
differential equation is of the second order and so on.
dr
k
dt
which is the required differential equation.
or
CHECK YOUR PROGRESS 32.1
1.
Find the order and degree of the differential equation
2.
dy
1
dx
Write the order and degree of each of the following differential equations.
yx
4
(a)
d 2s
 ds 

  3s 2  0
 dt 
dt
(b)
3
 d 2s 
 ds 
 2   3
 40
 dt 
 dt 
2
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MATHEMATICS
Differential Equations
3.
MODULE - VIII
Calculus
State whether the following differential equations are linear or non-linear.
 xy2  x  dx   y  x 2 y  dy  0
(a)
(b)
dx  dy  0
dy
dy
 cos x (d)
 sin 2 y  0
dx
dx
Form the differential equation corresponding to
(c)
4.
 x  a 2   y  b 2  r 2
5.
(a)
by eliminating 'a' and 'b'.
Form the differential equation corresponding to

y2  m a 2  x 2
(b)
Notes

Form the differential equation corresponding to
y 2  2ay  x 2  a 2 , where a is an arbitrary constant.
(c)
(d)
(e)
Find the differential equation of the family of curves y  Ae 2x  Be3x where A
and B are arbitrary constants.
Find the differential equation of all straight lines passing through the point (3,2).
Find the differential equation of all the circles which pass through origin and whose
centres lie on y-axis.
32.5 GENERAL AND PARTICULAR SOLUTIONS
Finding solution of a differential equation is a reverse process. Here we try to find an
equation which gives rise to the given differential equation through the process of
differentiations and elimination of constants. The equation so found is called the primitive or
the solution of the differential equation.
Remarks
(1) If we differentiate the primitive, we get the differential equation and if we integrate the
differential equation, we get the primitive.
(2) Solution of a differential equation is one which satisfies the differential equation.
Example 32.5 Show that y  C1 sin x  C2 cos x, where C1 and C2 are arbitrary
constants, is a solution of the differential equation :
d2 y
dx 2
y0
Solution : We are given that
y  C1 sin x  C2 cos x
Differentiating both sides of (1), we get
dy
 C1 cos x  C2 sin x
dx
MATHEMATICS
.....(1)
.....(2)
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Differential Equations
MODULE - VIII
Calculus
Differentiating again, we get
d2 y
dx 2
Substituting the values of
Notes
d2 y
dx 2
or
d2 y
dx 2
d2 y
dx 2
 C1 sin x  C2 cos x
and y in the given differential equation, we get
 y  C1 sin x  C2 cos x  (C1 sin x  C2 cos x)
 y 0
In integration, the arbitrary constants play important role. For different values of the constants
we get the different solutions of the differential equation.
A solution which contains as many as arbitrary constants as the order of the differential equation
is called the General Solution or complete primitive.
If we give the particular values to the arbitrary constants in the general solution of differential
equation, the resulting solution is called a Particular Solution.
Remark
General Solution contains as many arbitrary constants as is the order of the differential equation.
Example 32.6 Show that y  cx 
a
(where c is a constant) is a solution of the differential
c
equation.
yx
Solution : We have y  cx 
dy
dx
a
dx
dy
a
c
.....(1)
Differentiating (1), we get
dy
c
dx
On substituting the values of

414

dx 1

dy c
dx
dy
and
in R.H.S of the differential equation, we have
dy
dx
a
1
x  c   a    cx   y
c
c
R.H.S. = L.H.S.
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MODULE - VIII
Calculus
a
Hence y  cx  is a solution of the given differential equation.
c
Example 32.7
If y  3x 2  C is the general solution of the differential equation
dy
 6x  0 , then find the particular solution when y = 3, x = 2.
dx
Notes
Solution : The general solution of the given differential equation is given as
y  3x 2  C
....(1)
Now on substituting y = 3, x = 2 in the above equation , we get
3 = 12 + C
or
C  9
By substituting the value of C in the general solution (1), we get
y  3x 2  9
which is the required particular solution.
32.6 TECHNIQUES OF SOLVING IN G. A DIFFERENTIAL EQUATION
32.6.1 When Variables are Separable
(i) Differential equation of the type
dy
 f (x)
dx
Consider the differential equation of the type
dy
 f (x)
dx
or
dy = f (x) dx
On integrating both sides, we get
 dy   f  x  dx
y   f  x  dx  c
where c is an arbitrary constant. This is the general solution.
Note : It is necessary to write c in the general solution, otherwise it will become a particular
solution.
Example 32.8 Solve
x  2
dy
 x 2  4x  5
dx
Solution : The given differential equation is  x  2 
MATHEMATICS
dy
 x 2  4x  5
dx
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Differential Equations
MODULE - VIII
Calculus
Notes
or
dy x 2  4x  5

dx
x2
or
dy x 2  4x  4  4  5

dx
x2
or
2
dy  x  2 
9


dx
x2
x2
or
dy
9
 x2
dx
x2
9 

dy   x  2 
 dx
x2

On integrating both sides of (1), we have
or
.....(1)
x2
9 

y

 2x  9 log | x  2 |  c ,
 dy    x  2  x  2  dx or
2
where c is an arbitrary constant, is the required general solution.
Example 32.9 Solve
dy
 2x 3  x
dx
given that y = 1 when x = 0
Solution : The given differential equation is
or

dy
 2x 3  x
dx

dy  2x 3  x dx
...(1)
On integrating both sides of (1), we get
 dy 
or


2x 3  x dx
y
or
y  2.
x4 x 2

C
2
2
x 4 x2

C
4
2
.....(2)
where C is an arbitrary constant.
Since y = 1 when x = 0, therefore, if we substitute these values in (2) we will get
C=1

1 00C
Now, on putting the value of C in (2), we get
y
1 4
x  x 2  1 or
2


y
1 2 2
x x 1 1
2


which is the required particular solution.
(ii) Differential equations of the type
dy
 f (x)  g(y)
dx
Consider the differential equation of the type
dy
 f (x)  g(y)
dx
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Calculus
.....(1)
dy
 f (x) dx
g(y)
or
In equation (1), x's and y's have been separated from one another. Therefore, this equation is
also known differential equation with variables separable.
To solve such differential equations, we integrate both sides and add an arbitrary constant on
one side.
Notes
To illustrate this method, let us take few examples.
Example 32.10 Solve
 1  x 2  dy   1  y2  dx
Solution : The given differential equation
 1  x 2  dy   1  y2  dx
dy
can be written as
1
y2

dx
1  x2
(Here variables have been seperated)
On integrating both sides of (1), we get
dy
dx
 1  y2   1  x 2
tan 1 y  tan 1 x  C
or
where C is an arbitrary constant.
This is the required solution.
Example 32.11 Find the particular solution of
dy
2x
 2
dx 3y  1
when y (0) = 3 (i.e. when x = 0, y = 3).
Solution : The given differential equation is
dy
2x
 2
dx 3y  1 or
 3y2  1  dy  2x dx (Variables separated)
....(1)
If we integrate both sides of (1), we get
  3y2  1  dy  2x dx ,
where C is an arbitrary constant.
y3  y  x 2  C
MATHEMATICS
....(2)
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It is given that, y  0   3.
 on substituting y = 3 and x = 0 in (2), we get
Notes
27  3  C
C = 30

Thus, the required particular solution is
y3  y  x 2  30
32.6.2 Homogeneous Differential Equations
Consider the following differential equations :
dy
dy
 xy
dx
dx
(i)
y2  x 2
(iii)
dy x 3  xy 2

dx
y2 x
(ii)
 x 3  y3  dx  3xy2dy  0
In equation (i) above, we see that each term except
dy
is of degree 2
dx
[as degree of y 2 is 2, degree of x 2 is 2 and degree of xy is 1 + 1 =2]
In equation (ii) each term except
dy
is of degree 3.
dx
In equation (iii) each term except
dy
is of degree 3, as it can be rewritten as
dx
dy
 x3  x y2
dx
Such equations are called homogeneous equations.
y2 x
Remarks
Homogeneous equations do not have constant terms.
For example, differential equation
 x 2  3yx  dx   x3  x  dy  0
is not a homogeneous equation as the degree of the function except
dy
in each term is not the
dx
same. [degree of x 2 is 2, that of 3yx is 2, of x 3 is 3, and of x is 1]
32.6.3 Solution of Homogeneous Differential Equation :
To solve such equations, we proceed in the following manner :
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MATHEMATICS
Differential Equations
(i)
(ii)
(iii)
MODULE - VIII
Calculus
write one variable = v. (the other variable).
( i.e. either y = vx or x = vy )
reduce the equation to separable form
solve the equation as we had done earlier.
Example 32.12 Solve
Notes
 x 2  3xy  y2  dx  x 2dy  0
Solution : The given differential equation is
 x 2  3xy  y2  dx  x 2dy  0
dy x 2  3xy  y 2

dx
x2
or
.....(1)
It is a homogeneous equation of degree two. (Why?)
Let y  vx . Then
dy
dv
vx
dx
dx
 From (1), we have
dv x 2  3x.vx   vx 
vx

dx
x2
or
vx
dv
 1  3v  v 2
dx
x
or
dv
 v 2  2v  1
dx
dv
or
2
 v  1
2

dx
x
or
or
or
vx
x
 1  3v  v 2 
dv
 x2 

dx
x2


dv
 1  3v  v 2  v
dx
dv
v 2  2v  1

dx
x
.....(2)
Further on integrating both sides of (2), we get
1
 C  log x ,
v 1
where C is an arbitrary constant.
On substituting the value of v, we get
x
 log x  C
yx
which is the required solution.
dx P  x, y 
Note: If the Homoqeneous differential equation is written in the form qy  Q x, y then
 
x=vy is substituted to find solution.
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32.6.4 Differential Equation of the type
dy
 py  Q ,where P and Q are functions of x
qx
only.
Consider the equation
dy
 Py  Q
.....(1)
dx
Notes
where P and Q are functions of x. This is linear equation of order one.
To solve equation (1), we first multiply both sides of equation (1) by e Pdx
(called integrating factor) and get
e
Pdx
dy
Pdx
Pdx
 Py e 
 Qe 
dx
d
Pdx
ye 
dx

  Qe
 d

  e
 dx  ye
or
Pdx
Pdx
Pdx
.....(2)
dy
Pdx 
 Py.e 

dx
On integrating, we get
ye 
Pdx

 Qe
Pdx
dx  C
.....(3)
where C is an arbitrary constant,
 Pdx 
Pdx
ye 
Qe  dx  C 
 

or
Note : e  Pdx is called the integrating factor of the equation and is written as I.F in short.
Remarks
(i)
We observe that the left hand side of the linear differential equation (1) has become
d
Pdx
ye 
after the equation has been multiplied by the factor e  Pdx .
dx
(ii) The solution of the linear differential equation


dy
 Py  Q
dx
P and Q being functions of x only is given by
ye 
Pdx
(iii) The coefficient of

 Q  e
Pdx
 dx  C
dy
, if not unity, must be made unity by dividing the equation by it
dx
throughout.
(iv) Some differential equations become linear differential equations if y is treated as the
independent variable and x is treated as the dependent variable.
420
MATHEMATICS
Differential Equations
For example, dx  Px  Q , where P and Q are functions of y only, is also a linear
dy
differential equation of the first order.
Pdy
I.F.  e 
and the solution is given by
MODULE - VIII
Calculus
In this case
x  I.F.  
Notes
 Q.  I.F. dy  C
Example 32.13 Solve
dy y
  e x
dx x
Solution : Here P 
1
, Q  e x (Note that both P an Q are functions of x)
x
1
I.F. (Integrating Factor) e  Pdx  e  x
 Solution of the given equation is:
yx 
 xe
 x dx
dx
 elog x  x
x  0
C
where C is an arbitrary constant
or
xy  xe  x   e  x dx  C
or
xy   xe  x  e  x  C
or
xy  e  x  x  1   C
or
 x  1  x C
y  
e 
x
 x 
Note: In the solution x > 0.
Example 32.14 Solve :
sin x
dy
 y cos x  2 sin 2 x cos x
dx
Solution : The given differential equation is
sin x
dy
 y cos x  2 sin 2 x cos x
dx
or
dy
 y cot x  2 sin x cos x
dx
Here
P  cot x, Q  2 sin x cos x
.....(1)
Pdx
cot xdx
I.F.  e 
 e
 e log sin x  sin x
MATHEMATICS
421
Differential Equations
MODULE - VIII
Calculus
 Solution of the given equation is:
y sin x   2 sin 2 x cos x dx  C
where C is an arbitrary constant (sin x >0)
y sin x 
or
Notes
2 3
sin x  C ,
3
which is the required solution.
 tan 1 y  x
 1  y2  dx
dy
Example 32.15 Solve
Solution : The given differential equation is
 tan 1 y  x
 1  y2  dx
dy
or
dx tan 1 y
x


dy
1  y2 1  y2
or
dx
x
tan 1 y


dy 1  y 2
1  y2
which is of the form
dx
 Px  Q , where P and Q are the functions of y only..
dy
I.F.  e

....(1)
 Pdy

e
1
1 y2
dy
 e tan
1 y
Solution of the given equation is:
 e tan 1 y
x  e tan 1 y  
2
 1 y
 tan 1 y
dy  c.
e

1
where C is an arbitrary constant let t  tan y therefore dt 
or
 etan 1 y  x   et .tdt  C ,
or
 etan1 y  x  te t   et  C
or
or
or
422
 etan 1 y  x  tet  et  C
 etan1 y  x  tan 1 y etan1 y  etan1 y  C
x  tan 1 y  1  Ce tan
1
1  y2
dy
(on putting t  tan 1 y )
1 y
MATHEMATICS
Differential Equations
MODULE - VIII
Calculus
CHECK YOUR PROGRESS 32.2
1.
2.
(i)
Is y  sin x, a solution of
(ii)
3
Is y  x , a solution of x
d2 y
dx 2
 y  0?
dy
 4y  0 ?
dx
Notes
Given below are some solutions of the differential equation
dy
 3x .
dx
State which are particular solutions and which are general solutions.
(i)
3.
4.
5.
2y  3x 2
y
(ii)
3 2
x 2
2
3 2
x 3
2
State whether the following differential equations are homogeneous or not ?
y
(iii)
2y  3x 2  C
(iv)
(i)
dy
x2

dx 1  y 2
(ii)
(iii)
 x  2
(a)
Show that y = a sin 2x is a solution of
(b)
3
2
Verify that y  x  ax  c is a solution of
dy
 x 2  4x  9
dx
 3xy  y2  dx   x 2  xy  dy  0
(iv)
 x 3  yx 2  dy   y3  x3  dx  0
d2y
dx 2
 4y  0
d3 y
dx 3
6
The general solution of the differential equation
dy
 sec 2 x is y  tan x  C .
dx
Find the particular solution when
6.
7.

2
,y 1
,y  0
(b) x 
4
3
Solve the following differential equations :
(a)
x 
(a)
dy
 x 5 tan 1 x 3
dx
(b)
dy
 sin 3 x cos 2 x  xe x
dx
(c)
x
 1  x 2  dy
dx
(d)
dy
 x 2  sin 3x
dx
 
x
Find the particular solution of the equation e
MATHEMATICS
dy
 4 , given that y = 3 , when x = 0
dx
423
Differential Equations
MODULE - VIII
Calculus
Notes
8.
9.
10.
11.
Solve the following differential equations :
(a)
dy
 y2  xy 2  0
 x 2  yx 2  dx
(b)
dy
 xy  x  y  1
dx
(c)
sec 2 x tan y dx  sec 2 y tan x dy  0
(d)
dy
 ex  y  e y x 2
dx
Solve the following differential equations :
 x 2  y2  dx  2xy dy  0
(b)
x
(c)
dy

dx
x 2  y2  y
x
(d)
dy y
y
  sin  
x
dx x
dy
 y sec x  tan x
dx
Solve the following differential equations :
Solve :
(a)
12.
dy y2

y
dx
x
(a)
dy
 y  tan 1 x
 1  x 2  dx
2
(b) cos x
dy
 y  tan x
dx
dy
 y  2 log x , x > 1
dx
Solve the following differential equations:
(c)
x log x
(a)
 x  y  1
[Hint:
(b)
dy
1
dx
dx
dx
dx
 x  y  1 or
 x  y  1 which is of the form
 Px  Q ]
dy
dy
dy
dy
 y, y > 0
 x  2y2  dx
[Hint: y
dx
dx x
 x  2y 2 or
  2y ]
dy
dy y
Example 32.16 Verify if y  e m sin 1x is a solution of

1  x2
d2 y
 m2 y  0
 dx 2  dy
dx
Solution : We have,
y  e m sin
1 x
.....(1)
Differentiating (1) w.r.t. x , we get
1
dy me m sin x


dx
1  x2
or
424
1  x2
my
1  x2
dy
 my
dx
MATHEMATICS
Differential Equations
MODULE - VIII
Calculus
Squaring both sides, we get

2
 dy 
 2
1  x2 
 m y
 dx 

Differentiating both sides, we get
2
2
dy
 dy 
2 dy d y
2x 
. 2  2m 2 y
  2 1 x
dx dx
dx
 dx 
x
or

or
1  x2

Notes

dy
d2 y
 1  x2
 m2 y
2
dx
dx


d2y
dy
 m2 y  0
 dx 2  x dx
1
Hence y  em sin x is the solution of
2
dy
 m2 y  0
 1  x 2  ddxy2  x dx
Example 32.17 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
dx
y
x
Integrating both sides of equation (1), we get
or
dy 
.....(1)
1 y
y

 x 1
 dy    x  dx



1

 

1
or
  y  1  dy    1  x  dx
or
log y  y  x  log x  C
.....(2)
Since the curve is passing through the point (1,1), therefore,
substituting x = 1, y = 1 in equation (2), we get
MATHEMATICS
425
Differential Equations
MODULE - VIII
Calculus
1=1+C
or
C=0
Thus, the equation of the required curve is
log y  y  x  log x
log  xy   x  y
Notes or
dy 3e 2x  3e 4x

Example 32.18 Solve
dx
e x  e x
Solution : We have
dy 3e 2x  3e 4x

dx
e x  e x


3x
ex  e x
dy 3e

dx
ex  ex
or
dy
 3e3x
dx
or
dy  3e3x dx
Integrating both sides of (1), we get
or
.....(1)
y   3e3x dx  C
where C is an arbitrary constant.
e3x
C
or
3
which is required solution.
y3

y  e3x  C
or


y  1  ax  1  a 2  x  1  ay  1  a 2

which is the required solution.
CHECK YOUR PROGRESS 32.3
1.
(a)

1  x2
d2 y
d2 y
dy
0
 dx 2  2x dx
(a)
dy
 2y  0
dx
dx 2
Find the equation of the curve represented by
(b)
dy
 xy  x  y  1 and passing through the point (2, 0)
dx
Find the equation of the curve represented by
(b)
2.
If y 
tan 1 x , prove that
y
ex
sin x , prove that
2
dy
 
 y cot x  5ecos x and passing through the point  , 2 
2 
dx
426
MATHEMATICS
Differential Equations
3.
1A
%
l
l
l
l
l
l
l
l
(a)
dx  xdy  e  y sec 2 y dy
(b)
 4x  3 cot 1 x
 1  x 2  dy
dx
(c)
 1  y  xy dy   1  x 2   1  y  dx
Notes
C
4.
l
MODULE - VIII
Calculus
dy 4e3x  4e5x

Solve :
dx
ex  e x
Solve the following differential equations :
+
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.
Degree of a differential equation exists, if it is a polynomial equation in terms of its derivatives.
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.
The solution of the differential equation of the type
dy
 f  x  is obtained by integrating
dx
both sides.
l
l
l
l
The solution of the differential equation of the type
dy
 f  x  g (y) is obtained after
dx
separating the variables and integrating both sides.
The differential equation M (x, y) dx + N ( x , y ) dy = 0 is called homogeneous if
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.
The solution of the first order linear equation
MATHEMATICS
dy
 Py  Q is
dx
427
Differential Equations
MODULE - VIII
Calculus
ye 
Pdx

 Q  e
Pdx
 dx  C ,
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.
Notes
SUPPORTIVE WEB SITES
http://www.youtube.com/watch?v=9Wfn-WWV1aY
http://www.youtube.com/watch?v=6YRGEsQWZzY
TERMINAL EXERCISE
1.
Find the order and degree of the differential equation :
2
(a)
(c)
(e)
2.
4
 d2y 
2  dy 
 2  x 
 0
 dx 
 dx 
d4 y
dx 4
x2
4
d2 y
dx 2
dy
 4y  5 cos 3x
dx
 xy
dy
y
dx
(b)
xdx  ydy  0
(d)
dy
 cos x
dx
(f)
d2 y
dx 2
y0
Find which of the following equations are linear and which are non-linear
(a)
dy
 cos x
dx
(c)
2
 d2y 
2  dy 

x
 2 

 0
 dx 
 dx 
(e)
dx  dy  0
(b)
dy y
  y 2 log x
dx x
(d)
x
3
dy
4 x
dx
3.
Form the differential equation corresponding to y 2  2ay  x 2  a 2 by eliminating a.
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)
428
Second order differential equation.
MATHEMATICS
Differential Equations
6.
(b)
Differential equation of order three.
(c)
Differential equation of order five.
Show that y  a cos  log x   b sin  log x  is a solution of the differential equation
x2
7.
MODULE - VIII
Calculus
d2y
dx 2
x
dy
y0
dx
Notes
Solve the following differential equations:
(b)
dy
 e x  y  x 2e  y
dx
dy cos x sin y

0
dx
cos y
(d)
dy  xydx  xdx
dy
 y tan x  x m cos mx
dx
(f)
 tan 1 y  x
 1  y2  dx
dy
(a)
sin 2 x
(c)
(e)
MATHEMATICS
dy
 3 cos x  4
dx
429
Differential Equations
MODULE - VIII
Calculus
ANSWERS
CHECK YOUR PROGRESS 32.1
1.
Notes 2.
Order is 1 and degree is 1.
(a) Order 2, degree 1
(b) Order 2, degree 2
3.
(a) Non- linear
(b)
Linear
(c) Linear
(d)
Non-linear
3
4.
2
2

 dy  
2 d y 
1  
  r  2 
 dx  
 dx 

5.
(a) xy
(b)
(c)

2
d2y
dy
 dy 

x

y
0


dx
 dx 
dx 2
2
dy
 dy 
x 2  2y 2 
 x2  0
  4xy
dx
dx



d2 y
dx 2

dy
 6y  0
dx
(d) y   x  3 
(e)
2
dy
2
dx
 x 2  y 2  dy  2xy  0
dx
CHECK YOUR PROGRESS 32.2
1.
(i)
Yes
(ii)
2.
(i), (ii) and (iv) are particular solutions
3.
5.
(ii), (iv) are homogeneous
(a) y  tan x
No
(iii) is the general solution
(b) y  tan x  3
6.
430
1 6
1
1
x tan 1 x 3  x 3  tan 1 x 3  C
6
6
6
1
1
5
3
x
(b) y  cos x  cos x   x  1  e  C
5
3
(a)
y
 
 
MATHEMATICS
Differential Equations
1
2
(c) y  log x  1  C
2
(d)
1
1
y  x 3  cos 3x  C
3
3
7.
y  4e  x  7
8.
x
1 1
(a) log y  C  x  y
(b)
log y  1  x 
(c) tan x tan y  C
(d)
e y  ex 
(b)
x  cy  y log | x |
(d)
tan
9.

(a) x  C x 2  y 2

1  y 
(c) sin    log | x |  C
x
x2
C
2
MODULE - VIII
Calculus
Notes
x3
C
3
y
 Cx
2x
10. y  sec x  tan x   sec x  tan x  x  C
11. (a) y  tan 1 x  1  Ce  tan x
(c) y  log x 
(b)
y  tan x  1  Ce  tan x
(b)
x  y 2  Cy
C
log x
12. (a) x  Ce y   y  2 
CHECK YOUR PROGRESS 32.3
2.
3.
4.
(a) log  y  1  
y
1 2
x x4
2
(b)
y sin x  5ecos x  7
(b)
y  2 log 1  x 2 
4 5x
e C
5
(a) x  e  y  C  tan y 
(c) log x  2 log 1  y 
3  1  2
cot x  C
2
x 2 y2

 2y  C
2
2
TERMINAL EXERCISE
1.
2.
3.
(a) Order 2, degree 3
(b)
Order 1, degree 1
(c) Order 4, degree 1
(d)
Order 1, degree 1
(e) Order 2, degree 1
(f)
Order 2, degree 1
(a), (d), (e) are linear; (b), (c) are non-linear
x
2
 2y
MATHEMATICS
2

2
 dy 
 dy 
2

  4xy 
x 0
 dx 
 dx 
431
Differential Equations
MODULE - VIII
Calculus
d3 y
4.
5.
dx 3
 0 , Order 3, degree 1.
(a) Two
(b) Three
(c) Five
Notes
7.
(b)
e y  ex 
(c) sin y  Ce sin x
(d)
log  1  y  
(f)
x  tan 1 y  1  Ce tan
(e)
432
x3
C
3
(a) y + 3 cosec x + 4 cot x = C
y
x m 1
cos x  C cos x
m 1
x2
C
2
1 y
MATHEMATICS