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

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The discussion focuses on solving the homogeneous ordinary differential equation given by x(y-3x)dy/dx = 2y^2 - 9xy + 8x^2. The substitution y = vx leads to the expression log(cx) = (1/2)log(y^2/x^2 - 6y/x + 8). The next steps involve multiplying both sides by 2, exponentiating, and solving for y. It's emphasized to verify the solution by substituting it back into the original equation. The approach and calculations presented are deemed correct so far.
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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?
 
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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.
 

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