# Homogeneous differential equations

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• February 25th 2009, 01:57 PM
chella182
Homogeneous differential equations
I have never understood this. The question I have is...

Find the general solution of the equation $(x-y)\frac{dy}{dx}=x+y$.

So that'd get...

$\frac{dy}{dx}=\frac{x+y}{x-y}$

In my lecture notes I have written...

How do we solve this type of equation?

Given $y'=F(\frac{y}{x})$, set $\frac{y(x)}{x}=v(x)$

i.e. $y(x)=x v(x)$, so $v+xv'=F(v)$

or $xv'=F(v)-v$ which is a separable equation.

$\int\frac{dv}{F(v)-v}=\int\frac{dx}{x}$ where $x\neq 0; F(v)-v\neq 0$

I understand that that equation is separable and stuff, but the stuff before I just don't get at all. Like where F(v) suddenly comes from and that. Can anyone help? (Crying)

Note: $y'$ and $v'$ are derivatives with respect to x.
• February 25th 2009, 01:59 PM
Krizalid
$\frac{x+y}{x-y}=\frac{1+\dfrac{y}{x}}{1-\dfrac{y}{x}}.$

Put $y=xz.$
• February 25th 2009, 01:59 PM
chella182
See that's the kinda thing I don't get (Worried) I think that process is what I need explaining to me.
• February 25th 2009, 02:02 PM
Krizalid
$y'=z+xz'$ and the ODE becomes $z+xz'=\frac{1+z}{1-z}$ and after some algebra this becomes a separable one!
• February 25th 2009, 02:03 PM
chella182
Nah, still not getting it. Sorry . Where did the y' = whatever come from? (Crying)
• February 25th 2009, 02:08 PM
Krizalid
That's the derivative. We're making a change of variables. We have $y'=\frac{dy}{dx}$ and $z'$ is $\frac{dz}{dx}.$
• February 25th 2009, 02:09 PM
chella182
Nah, none of this is going in at all, sorry. I'm gonna have to ask tomorrow after his lecture.
• February 25th 2009, 02:09 PM
Krizalid
Tell me exactly what you don't get.
• February 25th 2009, 02:18 PM
chella182
I don't know. I don't like the whole $\frac{y}{x}$ thing you've gotta do. I just don't understand any of it. What I've written in my notes just doesn't make sense to me. I also hate the notation $y'$ and what not, but I doubt thats anything to do with it haha.

Okay, I've gotten this far with it.

$\frac{dy}{dx}=\frac{x+y}{x-y}$

$\frac{dy}{dx}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}$

$y=xv$ so $\frac{dy}{dx}=v+x\frac{dv}{dx}$

And now I'm lost...
• February 25th 2009, 02:20 PM
Krizalid
And that becomes $v+x\,\frac{dv}{dx}=\frac{1+v}{1-v}.$
• February 25th 2009, 02:22 PM
chella182
And what do I do with that? I just need walking through it all from this point...
• February 25th 2009, 02:23 PM
Krizalid
$x\,\frac{dv}{dx}=\frac{1+v}{1-v}-v\implies x\,\frac{dv}{dx}=\frac{1+v-v+v^{2}}{1-v},$ thus $\frac{1-v}{1+v^{2}}\,dv=\frac{dx}x.$
• February 25th 2009, 02:25 PM
chella182
Wouldn't it be $\int\frac{dx}{x}$ on the right hand side? I dislike that left-hand side also haha. But I know how to do it (roughly).
• February 25th 2009, 02:26 PM
Krizalid
Oh yes, I've edited.
• February 25th 2009, 02:29 PM
chella182
Okay right, on integrating both sides would you get...

$-\frac{1}{2}\ln{(1+v^2)}=\ln{(x)}+A$ where A = constant of integration?

The left-hand side is likely to by wrong somehow; I never could do those tricky $\int\frac{f'(x)}{f(x)}dx$ problems.
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