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Math Help - Method of undetermined coefficients

  1. #1
    MHF Contributor alexmahone's Avatar
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    Method of undetermined coefficients

    Apply the method of undetermined coefficients to find a particular solution of each of the systems. Primes denote derivatives with respect to t.

    1) x'=x+y+2t,\ y'=x+y-2t

    2) x'=2x+y+1,\ y'=3x+2y+e^{4t}
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    Re: Method of undetermined coefficients

    Quote Originally Posted by alexmahone View Post
    Apply the method of undetermined coefficients to find a particular solution of each of the systems. Primes denote derivatives with respect to t.

    1) x'=x+y+2t,\ y'=x+y-2t

    2) x'=2x+y+1,\ y'=3x+2y+e^{4t}
    1) From the first equation

    \displaystyle \begin{align*} \frac{dx}{dt} &= x + y + 2t \\ \frac{d^2x}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 \end{align*}

    and from the second

    \displaystyle \begin{align*} \frac{dy}{dt} &= x + y - 2t \\ \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} - 2 \end{align*}

    Therefore

    \displaystyle \begin{align*} \frac{d^2x}{dt^2} - \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 - \left( \frac{dx}{dt} + \frac{dy}{dt} - 2  \right) \\ \frac{d^2x}{dt^2} + \frac{d^2y}{dt^2} &= 4 \\ \frac{d^2y}{dt^2} &= 4 - \frac{d^2x}{dt^2} \\ \frac{dy}{dt}  &= 4t - \frac{dx}{dt} + C_1 \\ y &= 2t^2 - x + C_1t + C_2 \\ x + y &= 2t^2 + C_1t + C_2 \\ \frac{dx}{dt} - 2t &= 2t^2 + C_1t + C_2 \\ \frac{dx}{dt} &= 2t^2 + (C_1 + 2)t + C_2 \\ x &= \frac{2}{3}t^3 + \left(\frac{C_1 + 2}{2}\right)t^2 + C_2t + C_3 \end{align*}

    and from a similar process you can find \displaystyle \begin{align*} y \end{align*}
    Last edited by Prove It; November 29th 2011 at 03:41 PM.
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    MHF Contributor alexmahone's Avatar
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    Re: Method of undetermined coefficients

    Quote Originally Posted by Prove It View Post
    Am I correct in assuming that you're using \displaystyle \begin{align*} x' \end{align*} to represent \displaystyle \begin{align*} \frac{dx}{dt} \end{align*} and \displaystyle \begin{align*} y' \end{align*} to represent \displaystyle \begin{align*} \frac{dy}{dt} \end{align*} ...
    Yes. (I said that in my second sentence. )
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    Re: Method of undetermined coefficients

    Quote Originally Posted by alexmahone View Post
    Yes. (I said that in my second sentence. )
    I edited my post.
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    Re: Method of undetermined coefficients

    Quote Originally Posted by Prove It View Post
    1) From the first equation

    \displaystyle \begin{align*} \frac{dx}{dt} &= x + y + 2t \\ \frac{d^2x}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 \end{align*}

    and from the second

    \displaystyle \begin{align*} \frac{dy}{dt} &= x + y - 2t \\ \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} - 2 \end{align*}

    Therefore

    \displaystyle \begin{align*} \frac{d^2x}{dt^2} - \frac{d^2y}{dt^2} &= \frac{dx}{dt} + \frac{dy}{dt} + 2 - \left( \frac{dx}{dt} + \frac{dy}{dt} - 2  \right) \\ \frac{d^2x}{dt^2} + \frac{d^2y}{dt^2} &= 4\end{align*}
    Shouldn't that last equation be \frac{d^2x}{dt^2}-\frac{d^2y}{dt^2}=4?
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    Re: Method of undetermined coefficients

    Quote Originally Posted by alexmahone View Post
    Shouldn't that last equation be \frac{d^2x}{dt^2}-\frac{d^2y}{dt^2}=4?
    Yes it should be, sorry, so there'll be sign errors through the rest of the working, but the process still works
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