Sugestions to solve this equation

[pedro_ani]

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

 

[Number Nine]

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?).

 

[pedro_ani]

No, $b$ is a function, if it were a constant its derivative would be zero and the term with $y'$ would disappear

 

[haruspex]

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.

 

[bigfooted]

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

 

[haruspex]

$y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}$

Looks to me that if you substitute that in the original equation there'll be an unbalanced b''' term.

 

[bigfooted]

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...

 

[Vargo]

That suggests the substitution

$t =∫(a/b(x)) dx$

which works wonders :).