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Prove that $$u=e^{-4t}\cos\omega x$$ is a solution of the one-dimensional wave equation $$\frac{\partial u}{\partial t}=c^2\frac {\partial^2 u }{\partial x^2}.$$ I found $$\frac{\partial u }{\partial t}=-4e^{-4t}\cos\omega x$$ and $$\frac{\partial^2 u}{\partial x^2}=-\omega^2e^{-4t}\cos\omega x$$ but I can't equate the two. Please help to find a solution.

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I feel like the 4 should be dependent on $\omega$ and $c$ somehow. Also, $\dfrac{\partial u}{\partial t}=-4e^{-4t}\cos(\omega x)$. – Daryl Aug 12 '12 at 9:58
so you are telling the $\frac{\partial u}{\partial t}$ i found is wrong – Biju jose Aug 12 '12 at 10:02
Yes. Note that $\cos(\omega x)$ does not depend on $t$ and when determining the partial derivative wrt $t$, $x$ is held constant. – Daryl Aug 12 '12 at 10:06
Also, this seems more a heat equation than a wave equation. – enzotib Aug 12 '12 at 10:13
@Daryl thanks for telling the mistake and you are right it was happened because of my carelessness – Biju jose Aug 12 '12 at 10:17

I guess this means you have to find $\omega$ so that $u=e^{−4t}\cos\omega x$ satisfies $u_t=c^2u_{xx}$? By the way, this is called a heat equation, not a wave equation.

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thanks timur for the info ,well how can i find $\omega$ can you please help? – Biju jose Aug 12 '12 at 15:48
@Biju: Maybe you want to compare the two sides of the equality you found? – timur Aug 12 '12 at 16:53
equating the two the value of $\omega$ is 4 so we can tell the value of c is $2$ rgt – Biju jose Aug 13 '12 at 3:56
@Biju: You have to focus more. – timur Aug 13 '12 at 14:35
i don't know man can you please help – Biju jose Aug 14 '12 at 3:43

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