HomoEquations.pdf

8/12/2019 HomoEquations.pdf

1/4

Homogeneous Equations

A homogeneous equation can be transformed into a separable equation by a change ofvariables.

Definition : An equation in differential form M(x,y) dx + N(x,y) dy = 0 is said to behomogeneous , if when written in derivative form

dydx

= f(x,y) = g yx

! " #

$ % &

there exists a function g such that f(x,y) = g y

x

! " #

$ % & .

Example :(x2 3y 2) dx xy dy = 0 is homogeneous since

dydx

=

x2 ! 3y 2xy

=

xy ! 3

yx

=

yx

" # $

% & '

!1

! 3yx

= g yx

" # $

% & ' .

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) = t n F(x,y)

Example :Given F(x,y) = x 3 4x 2y + y 3, it is a homogeneous function of degree 3 since

F(tx,ty) = (tx)3

4(tx)2

(ty) + (ty)3

= t3

(x3

4x2

y + y3

).

Theorem : The O.D.E. in differential form M(x,y) dx + N(x,y) = 0 is a homogeneousO.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) = t nM(x,y) and N(x,y) = t n N(x,y)Assume that the parameter t = 1/x, then

M 1,yx

! " #

$ % &

= M 1x

x,1x

y! " # $

% & =

1x

! " #

$ % &

n

M(x,y) and N 1,yx

! " #

$ % &

= N 1x

x,1x

y! " # $

% & =

1x

! " #

$ % &

n

N(x,y)

then,

dydx

= ! M(x,y)N(x,y)

= !x( )n M 1, y

x" # $

% & '

x( )n N 1, yx

" # $

% & '

= !M 1,

yx

" # $

% & '

N 1,yx

" # $

% & '

= g yx

" # $

% & '

this shows that the equation is homogeneous in the sense of the first definition.


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Theorem : Given a homogeneous O.D.E., the change of variable y = vx transforms theequation into a separable equation in the variables v and x.Proof:

Given the homogeneous O.D.E.dydx

= g yx

! " #

$ % & ,

Let y = vx or v =y

x (notice that v depends on x)

thendydx

= v + x dvdx

and the equation is transformed into

v + x dvdx

= g(v) or v ! g(v)[ ]dx + xdv = 0 , it is a separable equation.

If we separate the variables, we getdv

v ! g(v)+

dxx

= 0

integrating, we getdv

v ! g(v)+

dxx

= c" " or G(v) + ln x = c = ln m since we can name the arbitrary constant any way we want.

Example : Solve the equations1) (x 2 3y 2) dx + 2xy dy = 0M(x,y) = x 2 3y 2 and N(x,y) = 2xy are homogeneous functions of degree 2.Lets express the equation in derivative form:

dy

dx

= ! x

2y

+3y

2x

Take the transformation y = vx anddydx

= v + x dvdx

then,

v + x dv

dx= !

1

2v+

3v

2

or

x dv

dx= ! v !

1

2v+

3v

2=

! 2v 2 ! 1 + 3v 2

2v=

v 2 ! 12v

separating variables

2vv 2 ! 1

dv = dxx

integrating2v

v 2 ! 1dv =

dx

x" "


3/4

ln v 2 ! 1 = ln x + ln c

v 2 ! 1 = xc

replacing v = y/x,

y2

x2 ! 1 = xc or y2 ! x2 = xc x 2

2) Solve the I.V.P.(x2 xy + y 2)dx xy dy = 0

y(1) = 0

the equation in derivative form isdydx

=x2 ! xy ! y2

xy=

xy

! 1 + yx

= g yx

" # $

% & '

it is homogeneous.Take the transformation y = vx, and replace

v + x dv

dx=

1

v! 1 + v or x

dv

dx=

1

v! 1 =

1 ! vv

separate the variables and integratev

1 ! vdv =

dxx

dxx

" + vv ! 1

dv = c" dx

x" + v ! 1

+ 1

v ! 1dv = c"

dxx

" + v ! 1v ! 1

dv +dv

v ! 1" = c" ln x + v + ln v ! 1 = ln m

v = ! ln v ! 1 x m( )e ! v = v ! 1 x m

s =yx

! 1 x eyx or s = y ! x e

yx

using the initial conditions y = 0 when x = 1,|s|= |0 1| e 0 = 1

then s = 1, but |y x| > 0 and e y/x > 0, then s = 1The solution I.V.P. is: 1 = |y x| e y/x

3) (x e y/x y)dx + x dy = 0the equation in derivative form is:


4/4

dydx

= ! eyx +

yx

take the transformation y = vx

v + x dv

dx= ! e v + v or x

dv

dx= ! e v

separate variables

! evdv =dxx

! e ! vdv = dxx

" " e! v = ln x + ln c = ln x c( )

e! y

x = ln xc( )

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HomoEquations.pdf