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Thread: Differential Equation

  1. #1
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    Differential Equation

    Hello,

    I have the DE....

    $\displaystyle y''(t) + \lambda^2 y(t) - x(t) = 0$

    I know that

    $\displaystyle y''(t) + \lambda^2 y(t) = 0$

    gives $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t$

    The solution in the text gives

    $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t + \frac{x(t)}{\lambda^2}$

    but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by JacksonR View Post
    but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.
    Necessarily, $\displaystyle y(t)=x(t)/\lambda^2$ should be a solution for the complete equation. Equivalently:

    $\displaystyle \dfrac{x''(t)}{\lambda^2}+ \lambda^2\dfrac{x(t)}{\lambda^2}-x(t)=0 \Leftrightarrow x''(t)=0 \Leftrightarrow x(t)=C_1t^2+C_2t+C_3$

    (Of course with $\displaystyle C_1=0$ )

    Fernando Revilla
    Last edited by FernandoRevilla; Dec 11th 2010 at 07:07 AM. Reason: I don't want to disagree with chisigma.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by JacksonR View Post
    Hello,

    I have the DE....

    $\displaystyle y''(t) + \lambda^2 y(t) - x(t) = 0$

    I know that

    $\displaystyle y''(t) + \lambda^2 y(t) = 0$

    gives $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t$

    The solution in the text gives

    $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t + \frac{x(t)}{\lambda^2}$

    but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.
    The function $\displaystyle \displaystyle y^{*} (t)= \frac{x(t)}{\lambda^{2}}$ is a particular solution of the DE $\displaystyle \displaystyle y^{''}(t) + \lambda^{2}\ y(t)= x(t)$ only if is $\displaystyle x^{''}(t)=0$, i.e. $\displaystyle x(t)= c_{0} + c_{1}\ t$. For all other $\displaystyle x(t)$ this particular solution is...

    $\displaystyle \displaystyle y^{*} (t) = x(t) * \frac{\sin \lambda t}{\lambda} = \int_{0}^{t} \frac{\sin \lambda \tau}{\lambda}\ x(t-\tau)\ d\tau$ (1)



    Merry Christmas from Italy

    $\displaystyle \chi$ $\displaystyle \sigma$
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