# Homogeneous Definition by fkx75474

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```									                         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 # &
dx                " x%
! 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\$ ' .
dx      xy        y  x # 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,
( x )n M " 1, %       " y%
y
\$
# x&  '    M \$ 1, '
# x&
dy      M(x, y)                                       " y%
=!          =!                    =!           = g\$ '
# x&
( x )n N " 1, %       " y%
dx      N(x, y)                   y
\$
# x&  '    N \$ 1, '
# 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\$
Given the homogeneous O.D.E.           = g# & ,
dx     " x%
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 =
v !1
2
x
integrating
2v          dx
" v2 ! 1 dv = " x
ln v2 ! 1 = ln x + ln c

v2 ! 1 = xc
replacing v = y/x,
y2
2
! 1 = xc !or! y 2 ! x 2 = xc x 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\$ '
dx         xy         y       x     # 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
" !e dv = " x
!v

e !v = ln x + ln c = ln ( x c )
y
!
e       x   = ln ( xc )

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