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$(x^2+xy)\displaystyle\frac{dy}{dx}-(3xy+y^2)=0$

Here is my idea , $\displaystyle\frac{dy}{dx}=\displaystyle\frac{y}{x}-\displaystyle\frac{2x}{x+y}+2$

Let $t=\displaystyle\frac{y}{x} $ , then

$RHS=t-\displaystyle\frac{2}{1+t}+2$

But I don't know how to do the LHS

Is that a right way to solve this question??

If the idea is wrong , please teach me.

Thanks a lot

2 Answers 2

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Let $t=\frac{y}{x}$. This implies that $y=tx$ and $\frac{dy}{dx}=t+x\frac{dt}{dx}$. Substitute these into the differential equation, we have $(x^2+tx^2)\left(t+x\frac{dt}{dx}\right)-(3tx^2+t^2x^2)=0.$ Divide the whole equation by $x^2$, we have $(1+t)\left(t+x\frac{dt}{dx}\right)-(3t+t^2)=0$ or $x\frac{dt}{dx}=\frac{3t+t^2}{1+t}-t$ which can be solved by separation of variables.

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    yes , I apply the Lambert W function^^2012-12-21
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A related problem. Your approach is correct. Here is how to find LHS

$ y = tx \implies \frac{dy}{dx} = x \frac{dt}{dx}+t. $

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    @cwk709394: You are welcome.2012-12-21