Find the general solution of y′′ + 4y = e-2x

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,

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

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

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

Therefore,

$\begin{align*} &&4p(x) e^{-2x} - 4p'(x)e^{-2x} + p''(x)e^{-2x} + 4p(x) e^{-2x} &= e^{-2x} \\ \implies && p''(x) - 4p'(x) + 8p(x) &= 1 \\ \implies && p(x) &= \frac{1}{8}. \end{align*}$

Thus, $y_1 = \frac{1}{8}e^{-2x}$ . Hence, the general solution is given by

$y = c_1 \cos (2x) + c_2 \sin (2x) + \frac{1}{8} e^{-2x}.$

Stumbling Robot

Find the general solution of y′′ + 4y = e^-2x

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