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Math Help - Please give me a hand!

  1. #1
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    Please give me a hand!

    this Q

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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..

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

    \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 u\left(x,t\right)=\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right) satisfies the heat equation.

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  3. #3
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    u=e^{-\alpha^2k^2t}\sin(kx)

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

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

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

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

    = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)=u_t\quad \blacksquare
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  4. #4
    Moo
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    A Cute Angle Moo's Avatar
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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 :

    <br /> <br />
u=e^{-\alpha^2k^2t\sin(kx)}<br />
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