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Guest2
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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?
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?
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