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I've come to the following ODE, and I'm not even sure where to start on finding the general solution: $$4\alpha\xi y''(\xi)+(\xi+2\alpha)y'(\xi)=0$$ Here, $\alpha$ is some positive constant. Can anyone help me out?

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2 Answers 2

up vote 4 down vote accepted

$y'(\xi) = f(\xi)$. We have $$4 \alpha \xi f'(\xi) = (\xi + 2 \alpha) f(\xi) \implies \dfrac{f'(\xi)}{f(\xi)} = \dfrac{\xi + 2\alpha}{4 \alpha \xi} = \dfrac1{4 \alpha} + \dfrac1{2 \xi}$$ Can you finish it off from here by finding $f$ and hence $y$?

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Indeed I can! Thanks! –  Cameron Buie Sep 25 '12 at 18:15

Generally, if we have $F(x,y',y'')=0$ (i.e. we don't have $y$ in our equation) then by putting $y'=u$ and so, $$y''=\frac{d}{dx}\bigg(\frac{dy}{dx}\bigg)=\frac{du}{dx}$$ the original equation $F(x,y',y'')=0$ will be the first order diff equation $F(x,u,\frac{du}{dx})=0$

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Much appreciated! –  Cameron Buie Sep 25 '12 at 18:15

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