1st order homogeneous ODE

**MattWT** · December 31st 2010, 03:16 AM

Hello,

Could you please have a look at this and see if i have done it correctly? thanks!

$(x^2 + y^2) \dfrac{dy}{dx} = 2xy$

$\dfrac{dy}{dx} = \dfrac{2xy}{x^2+y^2}$

Would you multiply the numerator and denominator by

$\dfrac{1}{x^2}$

And then given the substitution that

$v = y/x$

$dy/dx = x \dfrac{dv}{dx} + v$

So then, putting both equations together,

$\dfrac{2v}{1+v^2} = v + x \dfrac{dv}{dx}$

Rearranging to give:

$\int \dfrac{(1+v^2)}{v(1-v^2)} \, dv = \int \dfrac{1}{x} \, dx$

Then use partial fractions on the first integral:

$\int \dfrac{(1+v^2)}{-v(v+1)(v-1)} \, dv = \int \dfrac{1}{x} \, dx$

$\int \dfrac{1}{v} - \dfrac{1}{v+1}- \dfrac{1}{v-1} \, dv = ln|x| + c$

then simply integrate, sub v in, then try to rearrange for y?
havan't gone homogeneous ODE's before, just trying one out with the method.

thanks!

**Miss** · December 31st 2010, 03:36 AM

Edit: Looks good to me.
But I did not check your partial fraction.

**Sudharaka** · December 31st 2010, 04:28 AM

Originally Posted by Miss

Edit: Looks good to me.
But I did not check your partial fraction.

The partial fraction is also correct.

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