Solving Homogeneous ODE: $\displaystyle x(y-3x)\frac{dy}{dx}=2y^2-9xy+8x^2$

[Guest2]

I'm trying to solve $\displaystyle x(y-3x)\frac{dy}{dx} = 2y^2-9xy+8x^2$

Let $y = vx$ then $\displaystyle \frac{dy}{dx}= v+x\frac{dv}{dx}$ and I end up with

$\displaystyle \log(cx) = \frac{1}{2}\log(y^2/x^2-6y/x+8)$

Is this correct and what am I supposed to do after this?

 

[Ackbach]

Re: Differential equation

It looks good so far! I would multiply both sides by 2, exponentiate, and then solve for $y$. Don't forget to check by plugging back into the original.