13.5: Differential equations

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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

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

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

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,

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