# Lowest frequency non-uniform string

1. May 2, 2014

### Lengalicious

1. The problem statement, all variables and given/known data
See attachment (stuck with part b at the moment)

2. Relevant equations

3. 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

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Last edited: May 2, 2014
2. May 2, 2014

### 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.

3. May 2, 2014

### 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?

Last edited: May 2, 2014