1st order homogeneous diff. equation

**vonflex1** · January 31st 2010, 08:27 AM

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:

$z`x + z = 1 - z + z^2$

where do i go from here?

thanks

**Danny** · January 31st 2010, 08:33 AM

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:

$z`x + z = 1 - z + z^2$

where do i go from here?

thanks

Move the z to the right and separate.

**vonflex1** · January 31st 2010, 09:31 AM

how do i separate it?

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

**Danny** · January 31st 2010, 09:49 AM

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}. $

**vonflex1** · January 31st 2010, 10:05 AM

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?

**Danny** · January 31st 2010, 10:34 AM

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.

**vonflex1** · January 31st 2010, 10:36 AM

thanks, i get it now.

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