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For the equation

$\frac{\partial^2}{{\partial}x^2}(IE\frac{{\partial^2}u}{{\partial}x^2}) = \mu(\frac{{\partial^2}u}{{\partial}t^2})$

with $E$ a function of x, derive two ODE's by separation of variables.


I greatly appreciate your help/advice. I'm pretty unsure with how to handle derivatives like this. This is my best attempt, I've been working on this one for quite a while.

http://mathbin.net/51398

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    oh of course..... how silly of me :( thank you Hans Lundermark.2010-09-04

1 Answers 1

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Let $u(x,t)=X(x)T(t)$ ,

Then $\dfrac{\partial^2}{\partial x^2}\biggl(IE(x)\dfrac{\partial^2(X(x)T(t))}{\partial x^2}\biggr)=\mu\dfrac{\partial^2(X(x)T(t))}{\partial t^2}$

$T(t)\dfrac{d^2}{dx^2}\biggl(IE(x)\dfrac{d^2X(x)}{dx^2}\biggr)=\mu X(x)\dfrac{d^2T(t)}{dt^2}$

$\dfrac{1}{\mu X(x)}\dfrac{d^2}{dx^2}\biggl(IE(x)\dfrac{d^2X(x)}{dx^2}\biggr)=\dfrac{1}{T(t)}\dfrac{d^2T(t)}{dt^2}=-(f(s))^2$

$\begin{cases}\dfrac{d^2}{dx^2}\biggl(IE(x)\dfrac{d^2X(x)}{dx^2}\biggr)+\mu(f(s))^2X(x)=0\\\dfrac{d^2T(t)}{dt^2}+(f(s))^2T(t)=0\end{cases}$