diferential equation

1. HOMOGENEOUS
DIFFERENTIAL
EQUATIONS

2. Homogeneous Function
A function f(x,y) is called Homogeneous of
degree n if
Where t is a nonzero real number. Thus
are
Homogeneous function of degree 1, 8 and 0
respectively
)
,
(
)
,
( y
x
f
t
y
x
f n











y
x
and
y
x
y
x
xy sin
...
, 2
2
10
10

3. Homogeneous Equation
A first order DE of the form
Is said to be Homogeneous if the function f does
not depend on x and y separately, but only on
ratio . Thus first order homogeneous
equation are of the form ---------(1)
A homogeneous equation
Is transformed into a separable equation (in the
variables y and x) by the substitution y = vx
)
,
( y
x
f
dx
dy

x
y







x
y
g
dx
dy







x
y
g
dx
dy

4. Put y = vx and in eq (1)
This can be separated and be solved
)
(v
g
dx
dv
x
v 









dx
dv
x
v
dx
dy
0
)
( 



dx
dv
x
v
g
v
0
)]
(
[ 


 xdv
dx
v
g
v

 



x
dx
v
g
v
dv
)
(

5.   0
2 2
2


 dy
x
xy
y
2
2
2
x
xy
y
dx
dy 


vx
y 
Put and
Solve
Soln:
dx
dv
x
v
dx
dy


-----------(1)
So eq (1) becomes
 
v
v
x
xvx
x
v
dx
dv
x
v 2
2 2
2
2
2







 
v
v
dx
dv
x 3
2


   
 



x
dx
v
v
dv
3

6.   c
x
v
v log
log
3
log
3
1
log
3
1






 










x
dx
dv
v
v 3
1
1
3
1
x
c
v
v 1
log
3
3
log 


3
1
log
3
log 














x
c
v
v
3
3
1
3
3 x
c
x
c
v
v










3
3
x
c
x
y
x
y













)
3
(
3


 y
c
y
x

7.   0
)
4
(
5
2 


 dy
y
x
dx
y
x
)
4
(
)
5
2
(
y
x
y
x
dx
dy




vx
y 
Put and
Solve
Soln:
dx
dv
x
v
dx
dy


when y(1)=4
So eq(1) becomes
)
4
(
)
5
2
(
)
4
(
)
5
2
(
v
v
vx
x
vx
x
dx
dv
x
v









v
v
v
dx
dv
x 




)
4
(
)
5
2
(
  
 




x
dx
v
v
dv
v
2
)
1
(
)
4
(
-----------(1)

8.   c
x
v
v log
log
2
log
2
)
1
log( 





  




x
dx
v
dv
v
dv
2
2
1
cx
v
v
log
)
2
(
)
1
(
log 2




2
]
2
[
]
1
[ 



x
y
cx
x
y 2
]
2
[
]
[ x
y
c
x
y 



cx
v
v



 2
)
2
(
)
1
(
2
]
2
4
[
]
1
4
[ 


 c
2
]
2
4
[
]
1
4
[ 


 c
12
1

 c 2
]
2
[
]
[
12 x
y
x
y 




9. dx
y
x
xdx
ydy 2
2



Solve
Solve
1
)
1
(
..
..
0
)
( 2
2



 y
when
dy
x
dx
y
xy

10. EQUATIONS
REDUCIBLE TO
HOMOGENEOUS
FORM

11. Equation Reducible to Homogeneous Form
The DE
Is not homogeneous. It can be reduced to
homogeneous form as explained below
Case-I If then make the
transformation x = X + h, y = Y + k
0
)
(
)
( 2
2
2
1
1
1 




 dy
c
y
b
x
a
dx
c
y
b
x
a
2
1
2
1
b
b
a
a

0
)
(
)
(
2
2
2
2
2
1
1
1
1
1










dY
c
k
b
h
a
Y
b
X
a
dX
c
k
b
h
a
Y
b
X
a

12. Let h and k be the solution of the system of
equations
Then for calculated values of h and k eq (1) will be
reduced to homogeneous form
In the variables X and Y
0
0
2
2
2
1
1
1






c
k
b
h
a
c
k
b
h
a
0
)
(
)
( 2
2
1
1 


 dY
Y
b
X
a
dX
Y
b
X
a

13. Case-II If
then put
And the given equation will reduce to a separable
equation in the variables x and z
y
b
x
a
z 1
1 

2
1
2
1
b
b
a
a


14. Solve
Soln: Let x = X+h and y = Y+k, then
Now
5h + 5 = 0
h = -1
k = 1
3
2
1
2





y
x
y
x
dx
dy
3
)
(
2
1
)
(
2









k
Y
h
X
k
Y
h
X
dx
dy
3
2
2
1
2
2










k
h
Y
X
k
h
Y
X
dx
dy











0
3
2
0
1
2
2
k
h
k
h

15. Put Y = vX
3
2
1
2
1
1
2
2










Y
X
Y
X
dx
dy
Y
X
Y
X
dx
dy
2
2














X
Y
X
Y
dx
dy
2
1
2
v
v
dX
dv
X
v
2
1
2





v
v
v
dX
dv
X 




2
1
2

16. c
X
v
v ln
ln
2
)
1
ln(
tan 2
1




 
v
v
v
v
v
v
dX
dv
X
2
1
)
1
(
2
2
1
2
2 2
2










 



X
dX
dv
v
v
2
1
)
2
1
(
2

  




X
dX
v
vdv
v
dv
2
1
2
1 2
2
c
X
v
v ln
ln
)
1
ln(
tan 2
2
1




 

17. 2
2
1
)
1
(
ln
tan X
v
c
v 

 
2
2
2
1
1
ln
tan X
X
Y
c
X
Y










 
)
(
ln
tan 2
2
1
Y
X
c
X
Y


 
]
)
1
(
)
1
[(
ln
)
1
(
)
1
(
tan
2
2
1














 
y
x
c
x
y

18. Solve
Soln: Let z = 3x – 4y then
3
4
3
2
4
3





y
x
y
x
dx
dy
dx
dy
dx
dz
4
3 

dx
dz
dx
dy
)
4
1
(
4
3



3
2
)
4
1
(
4
3





z
z
dx
dz
3
2
4
3
)
4
1
(





z
z
dx
dz

19. )
3
(
4
)
1
(
)
4
1
(





z
z
dx
dz
)
3
(
)
1
(





z
z
dx
dz

 



 dx
z
dz
z
)
1
(
)
3
(


 



 dx
z
dz
dz
)
1
(
4

20. )
1
ln(
4
1 



 z
c
x
z
)
1
4
3
ln(
4
4
3 1 





 y
x
c
x
y
x
)
1
4
3
ln(
4
1





 y
x
c
y
x
)
1
4
3
ln( 




 y
x
c
y
x
Put z = 3x – 4y

21. Solve
5
1





x
y
x
y
dx
dy
Solve
1
2
5
2





y
x
y
x
dx
dy