Find the general solution of y′′ - 4y = e2x

Find the general solution of the second-order differential equation

$y'' - 4y = e^{2x}.$

If the solution is not valid everywhere, describe the interval on which it is valid.

The general solution of the homogeneous equation

$y'' - 4y = 0$

is given by Theorem 8.7 with $a = 0$ and $b = -4$ . This gives us $d = a^2 - 4b = 16$ ; hence, $k = \frac{1}{2} \sqrt{d} = 2$ . Thus,

$\begin{align*} y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx}) \\ &= c_1 e^{2x} + c_2 e^{-2x}. \end{align*}$

To find a particular solution of $y'' - 4y = e^{2x}$ assume $y_1 = p(x)e^{2x}$ is a solution. Then,

$\begin{align*} y_1' &= 2p(x) e^{2x} + p'(x)e^{2x} \\ y_1'' &= p''(x) e^{2x} + 4p'(x) e^{2x} + 4p(x) e^{2x}. \end{align*}$

Therefore,

$\begin{align*} &&p''(x)e^{2x} + 4p'(x) e^{2x} + 4p(x) e^{2x} - 4p(x) e^{2x} &= e^{2x} \\ \implies && p''(x) + 4p'(x) = 1. \end{align*}$

Now, let $p(x) = Ax + B$ , then we have

$4A = 1 \quad \implies \quad A = \frac{1}{4}\quad \implies \quad p(x) = \frac{1}{4}x.$

Thus, $y_1 = \frac{1}{4} x e^{2x}. \] Hence, the general solution is given by

$y = c_1 e^{2x} + c_2 e^{-2x} + \frac{1}{4} x e^{2x} = e^{2x} \left( \frac{1}{4} x + c_1 \right) + c_2 e^{-2x}.$

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