Homogeneous Equations
A homogeneous equation can be transformed into a separable equation by a change of
variables.
Definition: An equation in differential form M(x,y) dx + N(x,y) dy = 0 is said to be
homogeneous, if when written in derivative form
dy
! y$
= f(x, y) = g # &
" x%
dx
! y$
there exists a function g such that f(x,y) = g # & .
" x%
Example:
(x2 – 3y2) dx – xy dy = 0 is homogeneous since
!1
dy x 2 ! 3y 2 x
y " y%
y
" y%
=
= ! 3 = $ ' ! 3 = g$ ' .
# x&
dx
xy
y
x # x&
x
We can use another approach to define a homogeneous equation.
Definition: A function F(x,y) of the variables x and y is called homogeneous of degree n
if for any parameter t
F(tx, ty) = tn F(x,y)
Example:
Given F(x,y) = x3 – 4x2y + y3, it is a homogeneous function of degree 3 since
F(tx,ty) = (tx)3 – 4(tx)2(ty) + (ty)3 = t3(x3 – 4x2y + y3).
Theorem: The O.D.E. in differential form M(x,y) dx + N(x,y) = 0 is a homogeneous
O.D.E. if M(x,y) and N(x,y) are homogeneous functions of the same degree.
Proof:
Assume M(x,y) and N(x,y) are homogeneous functions of degree n, then
M(tx,ty) = tnM(x,y) and N(x,y) = tnN(x,y)
Assume that the parameter t = 1/x, then
n
n
! y$
! 1 1 $ ! 1$
! y$
! 1 1 $ ! 1$
M # 1, & = M # x, y & = # & M(x, y)!and!N # 1, & = N # x, y & = # & N(x, y)
" x%
" x x % " x%
" x%
" x x % " x%
then,
" y%
" y%
n
x ) M $ 1, '
M $ 1, '
(
# x&
# x&
dy
M(x, y)
" y%
=!
=!
=!
= g$ '
# x&
y
dx
N(x, y)
" y%
N $ 1, '
( x )n N "$# 1, %'&
# x&
x
this shows that the equation is homogeneous in the sense of the first definition.
Theorem: Given a homogeneous O.D.E., the change of variable y = vx transforms the
equation into a separable equation in the variables v and x.
Proof:
dy
! y$
= g# & ,
Given the homogeneous O.D.E.
" x%
dx
y
Let y = vx or v = (notice that v depends on x)
x
dy
dv
then
= v+ x
dx
dx
and the equation is transformed into
dv
v+ x
= g(v)!or![ v ! g(v)] dx + xdv = 0 , it is a separable equation.
dx
If we separate the variables, we get
dv
dx
+
=0
v ! g(v) x
integrating, we get
dv
dx
" v ! g(v) + " x = c !or!!G(v) + ln x = c = ln m
since we can name the arbitrary constant any way we want.
Example: Solve the equations
1) (x2 – 3y2) dx + 2xy dy = 0
M(x,y) = x2 – 3y2 and N(x,y) = 2xy are homogeneous functions of degree 2.
Let’s express the equation in derivative form:
dy
x 3y
=!
+
dx
2y 2x
dy
dv
Take the transformation y = vx and
= v+ x
dx
dx
then,
dv
1 3v
v+ x
=! +
dx
2v 2
or
dv
1 3v !2v2 ! 1 + 3v2 v2 ! 1
x
= !v !
+
=
=
dx
2v 2
2v
2v
separating variables
2v
dx
dv =
2
x
v !1
integrating
2v
dx
" v2 ! 1 dv = " x
ln v2 ! 1 = ln x + ln c
v2 ! 1 = xc
replacing v = y/x,
y2
! 1 = xc !or! y 2 ! x 2 = xc x 2
2
x
2) Solve the I.V.P.
(x2 –xy + y2)dx – xy dy = 0
y(1) = 0
the equation in derivative form is
dy x 2 ! xy ! y 2 x
y
" y%
=
= ! 1+ = g$ '
# x&
dx
xy
y
x
it is homogeneous.
Take the transformation y = vx, and replace
dv 1
dv 1
1! v
v+ x
= ! 1 + v!!or!x
= !1=
dx v
dx v
v
separate the variables and integrate
v
dx
dv =
1! v
x
dx
v
" x + " v ! 1 dv = c
dx
v !1+1
" x + " v ! 1 dv = c
dx
v!1
dv
" x + " v ! 1 dv!+ " v ! 1 = c
ln x + v + ln v ! 1 = ln m
v = ! ln ( v ! 1 x m )
e !v = v ! 1 x m
y
y
y
s = ! 1 x e x !or! s = y ! x e x
x
using the initial conditions y = 0 when x = 1,
|s|= |0 – 1| e0 = 1
then s = ±1, but |y – x| > 0 and ey/x > 0, then s = 1
The solution I.V.P. is: 1 = |y – x| ey/x
3) (x ey/x – y)dx + x dy = 0
the equation in derivative form is:
y
dy
y
= !e x +
dx
x
take the transformation y = vx
dv
dv
v+ x
= !e v + v!!or!!x
= !e v
dx
dx
separate variables
dx
!e vdv =
x
dx
!v
" !e dv = " x
e !v = ln x + ln c = ln ( x c )
e
!
y
x
= ln ( xc )