Sugestions to solve this equation

pedro_ani

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


[itex]y'' +\frac{b'}{b} y' - \frac{a^2}{b^2}y=0[/itex]

where [itex]a[/itex] 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, [itex]b[/itex] is a function, if it were a constant its derivative would be zero and the term with [itex]y'[/itex] 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

[itex]by''+b'y'-\frac{a^2}{b}y=0[/itex]

to the standard form:

[itex]z''=-c^2(t)z[/itex],

with

[itex]c(t)=-\frac{2a + b'(t)}{2b}[/itex]

the general solution of the transformed equation is then

[itex]z=Asin(cx)+Bcos(cx)[/itex]

Then get the solution of y by transforming back:
[itex]y=z e^{\int{\frac{b'}{2b}dx}} = z\frac{b'}{2b}[/itex]

[itex]y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}[/itex]

That's of course, assuming a,b are such that all steps are valid

haruspex

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:

[itex]y=A\sinh(\int -\frac{a}{b(x)}dx) + B\cosh(\int -\frac{a}{b(x)}dx)[/itex]

hope this helps a bit...

Vargo

That suggests the substitution

[itex] t =∫(a/b(x)) dx [/itex]

which works wonders :).