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.
Please help to solve the pde problem
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pde
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0I feel like the 4 should be dependent on $\omega$ and $c$ somehow. Also, $\dfrac{\partial u}{\partial t}=-4e^{-4t}\cos(\omega x)$. – 2012-08-12
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0so you are telling the $\frac{\partial u}{\partial t}$ i found is wrong – 2012-08-12
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0Yes. Note that $\cos(\omega x)$ does not depend on $t$ and when determining the partial derivative wrt $t$, $x$ is held constant. – 2012-08-12
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0Also, this seems more a heat equation than a wave equation. – 2012-08-12
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0@Daryl thanks for telling the mistake and you are right it was happened because of my carelessness – 2012-08-12
1 Answers
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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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0thanks timur for the info ,well how can i find $\omega$ can you please help? – 2012-08-12
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0@Biju: Maybe you want to compare the two sides of the equality you found? – 2012-08-12
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0equating the two the value of $\omega$ is 4 so we can tell the value of c is $2$ rgt – 2012-08-13
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0@Biju: You have to focus more. – 2012-08-13
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0i don't know man can you please help – 2012-08-14
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0w is +2 or -2 what's next – 2013-03-08