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Homogeneous Definition

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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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