# Help with 1st order ODE of degree 4

• Apr 17th 2010, 02:41 AM
rebghb
Help with 1st order ODE of degree 4
Hello everyone,

I have this ODE and I am afraid I've got the wrong approach...
Here's the DE:
$y'=\frac{y^{4}+x^{2}\,y^{2}-9\,x^{4}} {x\,y^{3}}$

I'm thinking that we may replace $y'$ by $dy/dx$ so that the DE becomes separable. Then we set $x=vy$ so that we get rid of x and the only 2 variable remaining would be v and y...

but I got to the point where $-dy/dy=\frac{dv}{v^{3}\,(9v-1)}$

Have I got the right approach...
Thanks for any help...
• Apr 17th 2010, 05:16 AM
mr fantastic
Quote:

Originally Posted by rebghb
Hello everyone,

I have this ODE and I am afraid I've got the wrong approach...
Here's the DE:
$y'=\frac{y^{4}+x^{2}\,y^{2}-9\,x^{4}} {x\,y^{3}}$

I'm thinking that we may replace $y'$ by $dy/dx$ so that the DE becomes separable. Then we set $x=vy$ so that we get rid of x and the only 2 variable remaining would be v and y...

but I got to the point where $-dy/dy=\frac{dv}{v^{3}\,(9v-1)}$

Have I got the right approach...
Thanks for any help...

Divide the numerator and denominator by y^4 and you get $\frac{dy}{dx} = \frac{1 + \left( \frac{x}{y}\right)^2 - 9 \left( \frac{x}{y}\right)^4}{ \left( \frac{x}{y}\right)}$.

The usual technique is to now substitute $\frac{y}{x} = v \Rightarrow y = x v \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$.

Therefore $v + x \frac{dv}{dx} = \frac{v^4 + v^2 - 9}{v^3}$, which is seperable.