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Marcus Right's question at Yahoo! Answers regarding a first order homogeneus ODE

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

Homogenous Differential Help with equation?

dy/dx = (-2*x^2-y^2)/(x*y)

I can see it is of homogenous form dy/dx = F(y/x)

how do I rewrite the d.e in terms of v and x
then
how do I separate it to the form x(dx) = v (dv)
and finally how do I solve it if it is indeed solvable.

I'm confused at the rewriting and separation steps :/

Any help is appreciated :)
Here is a link to the question:

Homogenous Differential Help with equation? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
13,775
Hello Marcus Right,

We are given the first order ODE:

\(\displaystyle \frac{dy}{dx}=-\frac{2x^2+y^2}{xy}=-\frac{2+\left(\frac{y}{x} \right)^2}{\frac{y}{x}}\)

So, we see, it is indeed a homogeneous equation. If we make the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\)

Now, differentiating with respect to $x$, we find:

\(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\)

and so, making the substitutions, we have:

\(\displaystyle v+x\frac{dv}{dx}=-\frac{2+v^2}{v}\)

Subtract through by $v$:

\(\displaystyle x\frac{dv}{dx}=-\frac{2+v^2}{v}-v=-\frac{2(1+v^2)}{v}\)

Separate variables and integrate:

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

\(\displaystyle -\frac{1}{4}\ln|1+v^2|=\ln|Cx|\)

\(\displaystyle 1+v^2=\frac{C}{x^4}\)

Subtract through by 1 and back substitute for $v$:

\(\displaystyle \frac{y^2}{x^2}=\frac{C-x^4}{x^4}\)

The solution is the given implicitly by:

\(\displaystyle y^2=\frac{C-x^4}{x^2}\)

To Marcus Right and any other guests viewing this topic, I invite and encourage you to post other differential equation questions here in our Differential Equations forum.

Best Regards,

Mark.