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posted by Anonymous
posted 2y 261d ago
\frac{\partial^2}{{\partial}x^2}(IE\frac{{\partial^2}u}{{\partial}x^2}) = \mu(\frac{{\partial^2}u}{{\partial}t^2})



Let E be a function of x, derive two ODE's by seperation of variables.
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Untitled
posted by Anonymous
posted 2y 261d ago
OP Here, this is my best attempt. Is there anything wrong with this? I am unsure and would very much appreciate if someone could look this over.

\frac{\partial ^{2}}{\partial x^{2}}\left( I\mbox{E}\frac{\partial ^{2}u}{\partial x^{2}} \right)=\mu \frac{\partial ^{2}u}{\partial t^{2}}


u\left( x,t \right)=Q\left( x \right)h\left( t \right)


Taking second derivative of u(x,t) with respect to x:

\frac{d^{2}}{dx^{2}}u\left( x,t \right)=\frac{d^{2}Q\left( x \right)}{dx^{2}}h\left( t \right)


Taking second derivative of u(x,t) with respect to t:

\frac{d^{2}}{dt^{2}}u\left( x,t \right)=\frac{d^{2}h\left( t \right)}{dt^{2}}Q\left( x \right)



Subbing back into original equation:

\frac{d^{2}}{dx^{2}}\left( I\mbox{E}\left( x \right)\frac{d^{2}Q\left( x \right)}{dx^{2}}h\left( t \right) \right)=\mu \frac{d^{2}h\left( t \right)}{dt^{2}}Q\left( x \right)


Separating functions of x and t:

\frac{1}{Q\left( x \right)}\frac{d^{2}}{dx^{2}}\left( \mbox{E}\left( x \right)\frac{d^{2}Q\left( x \right)}{dx^{2}} \right)=\mu \frac{1}{h\left( t \right)I}\frac{d^{2}h\left( t \right)}{dt^{2}}

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Untitled
posted by Anonymous
posted 2y 261d ago
Accidentally posted before I finished...

Next step:

\frac{1}{Q\left( x \right)}\frac{d^{2}}{dx^{2}}\left( \mbox{E}\left( x \right)\frac{d^{2}Q\left( x \right)}{dx^{2}} \right)\; =\; \mu \frac{1}{h\left( t \right)I}\frac{d^{2}h\left( t \right)}{dt^{2}}\; =\; -\lambda


Two ODE's:

\frac{d^{2}}{dx^{2}}\left( \mbox{E}\left( x \right)\frac{d^{2}Q\left( x \right)}{dx^{2}} \right)\; =\; -\lambda Q\left( x \right)


\frac{d^{2}h\left( t \right)}{dt^{2}}\; =\; \frac{-\lambda h\left( t \right)I}{\mu }


Would greatly appreciate if someone could tell me if I'm right.
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