Lowest frequency non-uniform string

[Lengalicious]

Homework Statement

See attachment (stuck with part b at the moment)

Homework Equations

The Attempt at a Solution

\phi=D(x)T(t)
so
(1+bx)D''(x)T(t)-D(x)T''(t)=0
(1+bx)\frac{D''(x)}{D(x)}=\frac{T''(t)}{T(t)}

let
\frac{T''}{T}=\sigma (1)
use trial solution T=be^{rt}

subbing into (1) and solve for r.
r=\pm\sqrt{\sigma}

use same trial solution and repeat steps for
(1+bx)\frac{D''}{D}=\sigma

r=\pm\sqrt{\frac{\sigma}{1+bx}}

from principle of superposition

D(x)=a_1e^{\sqrt{\frac{\sigma}{1+bx}}x}+a_2e^{-\sqrt{\frac{\sigma}{1+bx}}x}

T(t)=b_1e^{\sqrt{\sigma}t}+b_2e^{-\sqrt{\sigma}t}

Then I get confused with boundary conditions can someone let me know if I am on the right lines so far and give me any advice for proceeding?

Thanks

 

[dauto]

Your solution for D is not correct. In particular, since r turned out to be a function of x, your calculation of D'' must take that into consideration. Also, σ turns out to be negative. You must be aware of that when looking for the ansatz solutions.

 

[Lengalicious]

not really sure I understand, am I correct in using D=ae^{rx} and subbing that into(1+bx)\frac{D''}{D}=\sigma to find r? Or is my r incorrect?

EDIT: I see what you mean nevermind..

EDIT: I don't suppose you could give me a hint on how to find the trial solution? I have not come across an ODE in the form of f(x)y''-ay=0, what is this type of differential equation called?