## Sugestions to solve this equation

Any sugestions on how to find the solutions to this equation?

$y'' +\frac{b'}{b} y' - \frac{a^2}{b^2}y=0$

where $a$ is a constant
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 I assume that b is a constant as well. That's a pretty standard homogeneous linear equation; just use the standard methods (can you write out its characteristic equation?).
 No, $b$ is a function, if it were a constant its derivative would be zero and the term with $y'$ would disappear

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## Sugestions to solve this equation

You can obtain a lot of solutions by setting b(x) = Axn and solving for y. That won't give you a general solution though.
 rewrite $by''+b'y'-\frac{a^2}{b}y=0$ to the standard form: $z''=-c^2(t)z$, with $c(t)=-\frac{2a + b'(t)}{2b}$ the general solution of the transformed equation is then $z=Asin(cx)+Bcos(cx)$ Then get the solution of y by transforming back: $y=z e^{\int{\frac{b'}{2b}dx}} = z\frac{b'}{2b}$ $y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}$ That's of course, assuming a,b are such that all steps are valid

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 Quote by bigfooted $y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}$
 Oops, you're right, I was thinking too simple! The transformation to standard form does not lead to an expression that can be easily solved by a ricatti equation. Actually, maple gave the following: $y=A\sinh(\int -\frac{a}{b(x)}dx) + B\cosh(\int -\frac{a}{b(x)}dx)$ hope this helps a bit...
 That suggests the substitution $t =∫(a/b(x)) dx$ which works wonders :).