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13.5: Differential equations

  • Page ID
    6555
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    Coverup Method

    We are going to use partial fractions and the coverup method. We will assume you have seen partial fractions. If you don’t remember them well or have never seen the coverup method.

    Example \(\PageIndex{1}\)

    Solve \(y'' - y = e^{2t}\), \(y(0) = 1\), \(y'(0) = 1\) using Laplace transform.

    Solution

    Call \(\mathcal{L} (y) = Y\). Apply the Laplace transform to the equation gives

    \[(s^2 Y - sy(0) - y'(0)) - Y = \dfrac{1}{s - 2}\nonumber \]

    A little bit of algebra now gives

    \[(s^2 - 1) Y = \dfrac{1}{s - 2} + s + 1.\nonumber \]

    So

    \[Y = \dfrac{1}{(s - 2)(s^2 - 1)} + \dfrac{s + 1}{s^2 - 1} = \dfrac{1}{(s - 2)(s^2 - 1)} + \dfrac{1}{s - 1}\nonumber \]

    Use partial fractions to write

    \[Y = \dfrac{A}{s - 2} + \dfrac{B}{s - 1} + \dfrac{C}{s + 1} + \dfrac{1}{s - 1}.\nonumber \]

    The coverup method gives \(A = 1/3, B = -1/2, C = 1/6.\)

    We recognize

    \[\dfrac{1}{s - a}\nonumber \]

    as the Laplace transform of \(e^{at}\), so

    \[y(t) = Ae^{2t} + Be^t + Ce^{-t} + e^t = \dfrac{1}{3} e^{2t} - \dfrac{1}{2} e^t + \dfrac{1}{6} e^{-t} + e^t.\nonumber \]

    Example \(\PageIndex{2}\)

    Solve \(y'' - y = 1\), \(y(0) = 0\), \(y'(0) = 0\).

    Solution

    The rest (zero) initial conditions are nice because they will not add any terms to the algebra. As in the previous example we apply the Laplace transform to the entire equation.

    \[s^2 Y - Y = \dfrac{1}{s}, \text{ so } Y = \dfrac{1}{s(s^2 - 1)} = \dfrac{1}{s(s - 1)(s+1)} = \dfrac{A}{s} + \dfrac{B}{s -1} + \dfrac{C}{s + 1}\nonumber \]

    The coverup method gives \(A = -1, B = 1/2, C = 1/2\). So,

    \[y = A + Be^t + Ce^{-t} = -1 + \dfrac{1}{2} e^t + \dfrac{1}{2} e^{-t}.\nonumber \]


    This page titled 13.5: Differential equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.