# 1st order homogeneous diff. equation

• Jan 31st 2010, 08:27 AM
vonflex1
1st order homogeneous diff. equation
hi all,

i have this differential equation to solve...

i know it is homogeneous, so i can put $z = \frac{y}{x}$

but i still can't solve this...

$y = \frac{x^2-xy+y^2}{x^2}$

i get to:

$zx + z = 1 - z + z^2$

where do i go from here?

thanks
• Jan 31st 2010, 08:33 AM
Jester
Quote:

Originally Posted by vonflex1
hi all,

i have this differential equation to solve...

i know it is homogeneous, so i can put $z = \frac{y}{x}$

but i still can't solve this...

$y = \frac{x^2-xy+y^2}{x^2}$

i get to:

$zx + z = 1 - z + z^2$

where do i go from here?

thanks

Move the z to the right and separate.
• Jan 31st 2010, 09:31 AM
vonflex1
how do i separate it?

there's that 1, so i can't divide by z...
• Jan 31st 2010, 09:49 AM
Jester
Quote:

Originally Posted by vonflex1
how do i separate it?

there's that 1, so i can't divide by z...

Here's what I mean

$
x \frac{dz}{dx} = z^2-2z+1 = (z-1)^2
$

so

$
\frac{dz}{(z-1)^2} = \frac{dx}{x}.
$
• Jan 31st 2010, 10:05 AM
vonflex1
i understand what you did there,

and i will get $ln(z-1)^2 = lnx$

but what next?

how do i get back afterward back to y?
• Jan 31st 2010, 10:34 AM
Jester
Quote:

Originally Posted by Danny
Here's what I mean

$
x \frac{dz}{dx} = z^2-2z+1 = (z-1)^2
$

so

$
\frac{dz}{(z-1)^2} = \frac{dx}{x}.
$

Integrating gives

$
\frac{-1}{z-1} = \ln x + c
$
.

Solve for $z$ and then use your substitution.
• Jan 31st 2010, 10:36 AM
vonflex1
thanks, i get it now.