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[tex] u_{xx} - u_{tt} - au_{t} - bu = 0; 0<x<L, t>0 [/tex]

says if you multiply it by

[tex]2u_{t} [/tex]

you can get

[tex]\left( 2u_{t}u_{x}\right)_{x} - \left( u^{2}_{x} + u^{2}_{t} + bu^{2}\right)_{t} -2au^{2}_{t} = 0 [/tex]

or

[tex]\frac{\partial}{\partial x} \left( 2 \frac{\partial u}{\partial t}\frac{\partial u}{\partial x} \right) - \frac{\partial}{\partial t} \left[ \left( \frac{\partial u}{\partial x} \right) ^{2} + \left( \frac{\partial u}{\partial t} \right) ^{2} + bu^{2} \right] - 2a\left(\frac{\partial u}{\partial t} \right)^{2} =0; [/tex]

So far I have:

[tex] 2 \frac{\partial u}{\partial t} \left( \frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2}u}{\partial t^{2}} - a\frac{\partial u}{\partial t} - b\frac{\partial u}{\partial t} \right) = 0; [/tex]

[tex] 2\frac{\partial u}{\partial t} \ \frac{\partial^{2}u}{\partial x^{2}} \ - \ 2 \frac{\partial u}{\partial t} \ \frac{\partial^{2}u}{\partial t^{2}} \ - \ 2a\left( \frac{\partial u}{\partial t} \right)^{2} - 2b\left(\frac{\partial u}{\partial t} \right)^2 = 0; [/tex]

I can pull a d/dx out of the first term, to get [tex] \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial t} \ \frac{\partial u }{\partial x}\right) [/tex], and the [tex]- \ 2a\left( \frac{\partial u}{\partial t} \right)^{2} [/tex] is already there. How can I get the rest of it?

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# PDE derivation

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