Find the general solution of y′′ + y′ - 2y = e2x

Find the general solution of the second-order differential equation

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

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

The general solution of the homogeneous equation

$y'' + y' - 2y = 0$

is given by Theorem 8.7 with $a = 1$ and $b = -2$ . This gives us $d = a^2 - 4b = 9$ ; hence, $k = \frac{1}{2} \sqrt{d} = \frac{3}{2}$ . Thus,

$\begin{align*} y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx}) \\ &= e^{-\frac{x}{2}} ( c_1 e^{\frac{3}{2}x} + c_2 e^{-\frac{3}{2}x}) \\ &= c_1 e^x + c_2 e^{-2x}. \end{align*}$

To find a particular solution of $y'' + y' - 2y = e^{2x}$ 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,

$\begin{align*} &&p''(x)e^{2x} + 4p'(x) e^{2x} + 4p(x) e^{2x} + 2p(x) e^{2x} + p'(x)e^{2x} - 2p(x)e^{2x} &= e^{2x} \\ \implies && 4p(x) + 5p'(x) + p''(x) &= 1. \end{align*}$

Then, $p(x) = \frac{1}{4}$ is as solution to this so we have

$y_1 = \frac{1}{4} e^{2x}.$

Hence, the general solution is given by

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

Stumbling Robot

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

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