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Thread: Please give me a hand!

  1. #1
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    Please give me a hand!

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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Zachary View Post
    this Q

    Just substitute the function into the heat equation..

    $\displaystyle \frac{\partial u}{\partial t}=-\alpha^2k^2\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right)$

    $\displaystyle \frac{\partial^2 u}{\partial x^2}=-k^2\exp\left(-\alpha^2k^2t\right)\sin\left(kt\right)$

    It should be evident now that $\displaystyle u\left(x,t\right)=\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right)$ satisfies the heat equation.

    Does this help?
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  3. #3
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    $\displaystyle u=e^{-\alpha^2k^2t}\sin(kx)$

    $\displaystyle u_t=-\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$

    $\displaystyle u_x=e^{-\alpha^2k^2t}\cos(kx)\cdot k$

    $\displaystyle u_{xx}=-e^{-\alpha^2k^2t}\sin(kx)\cdot k^2$

    $\displaystyle \alpha^2u_{xx}=\alpha^2(-e^{-\alpha^2k^2t}\sin(kx)\cdot k^2)$

    $\displaystyle = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)=u_t\quad \blacksquare$
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  4. #4
    Moo
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    Hi,



    And now what ?
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  5. #5
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    No wonder i got wrong even though i have tried N-times. Because i thought the equation is like this :

    $\displaystyle

    u=e^{-\alpha^2k^2t\sin(kx)}
    $
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