1.

3 Separable Ordinary Differential Equations, Modelling

Generally, it is difficult to solve first-order ordinary differential equations y   f ( x, y ) in the sense

that no formulae exist for obtaining its solution in all cases. However, there are certain standard types
of first-order differential equations of the first degree for which routine methods of solution are
available. In this chapter, we shall discuss a few of these types.

Separable Differential Equation:

Differential equations of the form
g ( y ) dy  f ( x) dx (1)
are called equations with separated variables, the solutions of which are obtained
by direct integration.
Thus, its solution is given by

 g ( y)dy   f ( x)dx  c . (2)

If f and g are continuous functions, the integrals in Eq. (2) exist, and by evaluating them we obtain a
general solution of Eq. (1). This method of solving ODEs is called the method of separating
variables, and Eq. (1) is called a separable equation.
Example: 1 Solve y  1  y 2 .
Solution: The given equation can be written as
dy
 dx .
1  y2
Integrating both the sides, we have
tan 1 y  x  c
or y  tan( x  c) .
dy
Example: 2 Solve  e 2 x  y  x 3e  y .
dx
Solution: The given equation can be written as
dy
 (e 2 x  x 3 )e  y
dx
Separating the variables, we have
e y dy  (e2 x  x3 )dx
Integrating both the sides
e2 x x 4
ey   c.
2 4

1
 HOMOGENEOUS DIFFERENTIAL EQUATIONS

A function �(�, �) of two variables is said to be homogeneous of degree n if

� ��, �� = �� � �, � (3)
for every t>0 such that (tx, ty) is in the domain of �.
Example: Let f  x, y   2 x 4  x 2 y 2  5 xy 3
Now
f tx,ty   2tx 4  tx 2 ty 2  5tx ty 3

 t 4 2 x 4  x 2 y 2  5 xy 3 
 t 4 f  x, y 
Therefore, f is homogeneous of degree 4.
Example:
x
f ( x, y )  sin  
 y
 tx  x
f (tx, ty )  sin    sin    t 0 f ( x, y )
 ty   y
x
Thus f ( x, y )  sin   is a homogeneous function of degree 0.
 y

Homogeneous Differential Equation

A homogeneous differential equation is a differential equation which can be written in the form:
P x, y dx  Q x, y dy  0 (4)
where P and Q are homogeneous functions of the same degree.
Example: The differential equation ( x 2  xy )dx  y 2dy  0 is a homogeneous differential
equation because ( x 2  xy ) and y 2 are homogeneous functions of degree 2.
Note:
 y
1. A first order differential equation y  f ( x, y )  g   is known as homogeneous equation
x
 y
in which f ( x, y )  g   is a homogeneous function of degree 0.
x
2. Homogeneous differential equation does not involve constant terms.
Reduction to Separable form:
Type-I: If the given homogeneous differential equation of 1st order can be written
f ( x, y )
as y  then the solution of this type equation can be determined as follows:
g ( x, y )
Procedure:
I. Put y  ux
dy du
II. Calculate ux
dx dx
III. 
Substitute y and y in the given ODE. The given ODE reduces to a separable differential
equation.

2
 IV. Separate the variables and integrate both sides to get the general solution.

y
V. Replace u  and find the general solution of the given ODE.
x
Example: Solve the ODE 2 xyy  y 2  x 2 .
Solution: Given ODE is
y 2  x2

y  (5)
2 xy
du
Let y  ux,  y  u  x
dx

Substituting y and y in the given ODE (5) we find
du x 2 (u 2  1) u 2  1
ux  
dx 2ux 2 2u
2
du u  1 u  1  2u 2  (1  u 2 )
2
x  u  
dx 2u 2u 2u
This is a separable differential equation.
Separating variables and integrating we get
2u dx
 1  u 2 du    x
 ln 1  u 2   ln x  ln C
C
 1  u2 
x
y
Putting u  we obtain
x
2
y x2  y 2 C
 1 2  
x x2 x
 x 2  y 2  Cx , This is the required general solution of the given ODE.
Type-II: An equation of the form
y  f (ax  by  c)
Can be reduced to a separable equation by substituting ax  by  c  v.
Example: Solve y  ( y  4 x) 2
Solution: Given ODE is
y  ( y  4 x ) 2 (6)
Let y  4 x  v
Differentiating both sides w.r.to x we get
dv dv
y  4  ,  y  4
dx dx
The given ODE (6) becomes
dv dv
 4  v2 ,   v2  4
dx dx
This is a separable differential equation.
Separating variables and integrating both sides we get

3
 dv
   dx
v2  4
1 v
 tan 1   x  C1
2 2
v
 tan 1   2 x  2C1  2 x  C
2
 v  2 tan( 2 x  C )
Putting y  4 x  v we find
 y  4 x  2 tan( 2 x  C )
This is the required solution of the given ODE.
Example: Solve the ODE y  cos( x  y  1)
Solution: Let x  y  1  v
Differentiating w.r.to x we get
dv
y  1
dx
The given ODE becomes
dv
 1  cos v  2 cos 2 (v / 2) , This is a separable differential equation.
dx
Separating variables and integrating both sides we find
1
 sec 2 (v / 2)dv   dx
2
 tan(v / 2)  x  C
 x  y 1
 tan    xC
 2 
This is the general solution of the given differential equation.

Type-III: Nonhomogeneous ODEs Reducible to Homogeneous Form

If the differential equation is of the form
a x  b1 y  c1
y  1 (7)
a2 x  b2 y  c2
a1 b1 a b1
Suppose  or 1 0
a2 b2 a2 b2
Equation (7) can be solved as follows:
a b
Let 1  1  k (constant ),  a1  a2k and b1  b2k
a2 b2
Equation (7) becomes
k (a2 x  b2 y )  c1
y  (8)
a2 x  b2 y  c2
Putting a2 x  b2 y  u , the differential equation (8) reduces to a separable equation in terms
of u and x .
x y4
Example: Solve y 
x y6
Solution: Given differential equation is

4
 x y4
y  (9)
x y6
a b
In the equation (9), 1  1  1
a2 b2
du
Putting x  y  u and y   1 in equation (9) we find
dx
du u4 du u  4 2(u  1)
1  ,   1  , This is a separable equation in terms
dx u6 dx u  6 u6
of u and x .
Separating variables and integrating both sides we find
u6  5 
 u  1 du   2dx,   1  u  1 du   2dx
 u  5 ln u  1  2 x  C
 x  y  5 ln x  y  1  2 x  C
 y  x  5 ln x  y  1  C.
Solve the following ODEs.
1. (2� − 4� + 5)�' + � − 2� + 3 = 0
2. (2� + � + 1)�� + (4� + 2� − 1)�� = 0
3. (2� − 3� + 2)�� + 3(4� − 6� − 1)�� = 0

Problems Set 1.3

Question: Find a general solution. Show the steps of derivation. Check your answer by substitution

 y
Q1: xy '  y  2 x 3 sin 2  
 x
y
Answer: Let u   y  ux
x
Differentiating we get y '  u  xu '

Now we can substitute y and y ' in the equation:

x(u  xu ' )  ux  2 x 3 sin 2 u
 xu  x 2 u '  ux  2 x 3 sin 2 u
 x 2 u '  2 x 3 sin 2 u
u' x3
 2
sin 2 u x2
u'
  2x
sin 2 u
Integrating both the sides we get

du

sin 2 u 
 2 x dx

5
 x2
  cot u  2 c
2
  cot u  x 2  c
  cot u  x 2  c
 cot u  c  x 2
 u  cot 1 (c  x 2 )
y
Using back substitution method u  , we get
x
y
 cot 1 (c  x 2 )
x Ans.
1 2
 y  x cot (c  x )

2
Q2: y '  ( y  4 x)
Answer: Let v  y  4 x  y  v  4 x
Differentiating we get y '  v '4

Now we can substitute y and y ' in the given equation:

v '4  (v  4 x  4 x) 2
 v ' v2  4
dv
  v2  4
dx
dv
 2  dx
v 4

Integrating both the sides we get

dv

v 4 
2
 dx

1 v c
 tan 1 ( )  x 
2 2 2
v
 tan 1 ( )  2 x  c
2
v
  tan( 2 x  c)
2
 v  2 tan( 2 x  c)
Using back substitution method, we get v  y  4 x
 y  4 x  2 tan( 2 x  c)
Ans.
 y  2 tan( 2 x  c)  4 x

2
Q3. xy '  y  y

6
 y
Answer: Let u   y  ux
x
Differentiating we get y '  u  xu '

Now we can substitute y and y ' in the equation:

x(u  xu ' )  (ux ) 2  ux
 xu  x 2 u '  (ux ) 2  ux
 u ' u2
du
  u2
dx
du
 2  dx
u
Integrating both the sides we get

du

u2 
 dx  c

1
  xc
u
1
 cx
u
1
u 
cx
y
Using back substitution method u  , we get
x
y 1

x cx
Ans.
x
y
cx
2
Q4. y '  ( x  y  2) , y(0)  2

Answer: Let v  x  y  2  y  v  x  2
Differentiating we get y '  v '1

Now we can substitute y and y ' in the given equation:

v '1  (v) 2
 v ' v2 1
dv
  v2 1
dx
dv
 2  dx
v 1

Integrating both the sides we get

7
 dv

v 1 
2
 dx

 tan 1 (v)  x  c
 v  tan( x  c)
Using back substitution method, we get v  x  y  2

 x  y  2  tan( x  c)
 y  tan( x  c)  x  2
All we need to do is to determine c from IVP
Given y (0)  2 i.e. when x  0 then y  2
2  tan(0  c)  0  2
 tan c  0
c0

Hence the particular solution is: y  tan x  x  2 Ans.

 y
Q5. xy '  y  3 x 4 cos 2  , y (1)  0
 x
y
Answer: Let u   y  ux
x
Differentiating we get y '  u  xu '

Now we can substitute y and y ' in the equation:

x(u  xu ' )  ux  3 x 4 cos 2 u
 xu  x 2u '  ux  3 x 4 cos 2 u
 x 2u '  3 x 4 cos 2 u
u' x4
 3
cos 2 u x2
u'
  3x 2
2
cos u
Integrating both the sides we get

du

cos 2 u 
 3 x 2 dx

x3
 tan u  3  c
3
3
 tan u  x  c
 u  tan 1 ( x 3  c)
y
Using back substitution method u  , we get
x
y
 tan 1 ( x3  c)
x
 y  x tan 1 ( x3  c).
All we need to do is to determine c from IVP

8
Given y (1)  0 i.e. when x  1 then y  0
0  1. tan 1 (1  c)
 (c  1)  tan 0
 c 1  0
 c  1
1 3
Hence the particular solution is: y  x tan ( x  1) Ans.

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