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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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.