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 (1.6)
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 . (1.7)

If f and g are continuous functions, the integrals in eq. (1.7) exist, and by evaluating them we obtain a
general solution of eq. (1.6). This method of solving ODEs is called the method of separating
variables, and eq. (1.6) 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

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 HOMOGENEOUS DIFFERENTIAL EQUATIONS

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

f tx,ty   t n f x, y  (1)

for every t>0 such that (tx, ty) is in the domain of f.

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: Let
1 x
f  x, y   2 2
e y
x y

Now
1 tx
f tx,ty   2 2 2 2
e ty

t x t y
 t  2 f  x, y 
Therefore f is homogeneous of degree -2.

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 (2)

where P and Q are homogeneous functions of the same degree.

Another form of homogeneous differential equation is expressed as:

dy  y  y
 f   or y'  f  (3)
dx x x
Solution Method of Equation (2)

Equations (2) is solved in two steps

Step-1:
The aim of this step is to transform the given homogeneous ordinary differential equation into
separable form of ODE by means of substitutions.

A homogeneous differential equation is an equation which can be written in the form

P  x, y dx  Q  x, y dy  0 (4)

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where P and Q are homogeneous functions of the same degree. Equations of this type can be
transformed into separable equations by means of substitutions

y  xv, where v  v x  (5)

and

dy  vdx  xdv
Thus, substitution of xv for y in (4) yields

P  x, xv dx  Q  x, xv vdx  xdv   0

If P and Q are homogeneous functions of degree n, then

Px, xv   x n P1,v  and Qx, xv   x n Q1,v 

Substituting in the preceding differential equation and dividing both sides by xn we obtain

P1,v dx  Q1,v vdx  xdv  0 (6)

This equation can be written in the separable form

1 Q1,v  (7)
dx  dv  0
x P1,v   vQ1,v 

provided non-zero denominators occur.

Equation (7) is separable ordinary differential equation.

Step-2:
The aim of this step is to solve the transformed ODE by direct integration.
Now solve equation (7).

1 Q(1, v)
 x dx   P(1, v)  v Q(1,v) dv C (8)

The solution will be of the form:

x  h (v )  c
y
 y (Since v  x
 x  h   c ) (9)
x
Equation (9) is the solution to equation (2).

Solution Method of Equation (3)

Equation (3) is solved in two steps

Step-1:

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The aim of this step is to transform the given homogeneous ordinary differential equation into
separable form of ODE by means of substitution.

Let y  ux, where u  u  x 

and

dy du
ux
dx dx
dy  y
but dx
 f  
x
dy
  f u 
dx
Hence we have
du
f (u )  u  x
dx
du
 f (u )  u  x
dx

dx du
  (10)
x f (u )  u

Equation (4) is separable ordinary differential equation. (A first order differential equation is
separable if it can be written with one variable on the left and other variable on the right.)

Step-2:
The aim of this step is to solve the transformed ODE by directly integrating both the sides.
Now solve equation (10).

dx du
  c (11)
x f (u )  u
The solution will be of the form:
x  g (u )  c
y
 y (Since u  x
 x  g   c ) (12)
x
Equation (12) is the solution to equation (3).

Note:

(1). Sometimes, substitution reduces an ODE to the homogeneous form. If a e  bd , then h and k
can be chosen so that x  u  h and y  v  k reduces the following ODE
 ax  by  c 
y '  F   (13)
 dx  ey  f 
to a homogeneous ODE.

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(2). If a e  bd , then also, an ODE of the form
y  y
y '  g ( x ) h 
x  x
can be reduced to the separable form by substituting y  ux, where u  u  x  .
(3). Let the ODE be y '  F ( ax  by  c ), b  0 .

Now substituting v  ax  by  c reduces the equation to a separable form.

If b  0 , then it is already in separable form.
Example
Solve the differential equation

y 2
 xy dx  x 2 dy  0.
Solution
If P(x, y) =y2-xy and Q(x,y)=x2, then the functions P and Q are both homogeneous of degree 2. Hence
the differential equation is homogeneous and we substitute

y  ux, where u  u  x 
 dy  udx  xdu
This leads to the following chain of equations

x 2

u 2  x 2 u dx  x 2  udx  xdu  0

 x 2 u 2  u dx  x 2  udx   xdu   0
 
 u 2  u dx  udx  xdu  0
 u 2 dx  xdu  0
1 1
 dx  2 du  0
x u

Integrating each term gives us

1 x x
ln x   C1  ln | x |   C1  y 
u y ln | x | C1
Hence the solution.

Example: Find the general solution of the differential equation: 2 xydx  ( x 2  y 2 )dy  0 .

Solution: A simple check reveals that the equation is homogeneous

Let y  vx and dy  vdx  xdv , and substitute into the differential equation to obtain
2 x(vx)dx  ( x 2  v 2 x 2 )(vdx  xdv)  0
 x 2 (v  v 3 )dx  x 3 (v 2  1)dv  0 .

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 3 3
Dividing by x (v  v ) separates the variables. That is
dx (v 2  1)dv
+ 0
x v(1  v 2 )

dx dv 2v dv
   0
x v v2 1

Integration yields
dx dv 2v dv
 x

v
 2
v 1
 ln c

 ln x  ln v  ln(v 2  1)  ln c or
 x(v 2  1)  cv
y
Rewriting the original variables by substituting v  ,
x
we obtain y 2  x 2  cy as the general solution.

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

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 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
 y  x cot 1 (c  x 2 ) Ans.

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)
 y  2 tan(2 x  c)  4 x Ans.

2
Q3. xy '  y  y

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 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
x
y Ans.
cx

Q4. y '  ( x  y  2) 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

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 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)
(1)
 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  3x 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
 2
  3 x 2 dx
cos u
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 ( x 3  c)
x
 y  x tan 1 ( x 3  c). (1)
All we need to do is to determine c from IVP

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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
Hence the particular solution is: y  x tan 1 ( x3  1) Ans.

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