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Find the general solution of y′′ + y′ – 2y = ex + e2x

Find the general solution of the second-order differential equation

    \[ y'' + y' - 2y = e^x + e^{2x}. \]

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^x is a solution. Then,

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

Therefore,

    \begin{align*}  &&p''(x) + 2p'(x) e^x + p(x) e^x + p'(x) e^x + p(x)e^x - 2p(x)e^x &= e^x (1+e^x) \\  \implies && 3p'(x) + p''(x) &= 1 + e^x.  \end{align*}

Then, let p(x) = Ax + B + Ce^x. This implies

    \begin{align*}  p'(x) &= A + Ce^x \\  p''(x) &= Ce^x. \end{align*}

Therefore,

    \[ 3A + 3Ce^x + Ce^x = 1 + e^x \quad \implies \quad A = \frac{1}{3}, \quad C = \frac{1}{4}. \]

Hence, p(x) = \frac{1}{3} + \frac{1}{4}e^x which gives us the particular solution

    \[ y_1 = \left( \frac{1}{3}x + \frac{1}{4} e^x \right) e^x  = \frac{1}{3}xe^x + \frac{1}{4}e^{2x}. \]

Finally, the general solution is then

    \[ y = c_1 e^x + c_2 e^{-2x} + \frac{1}{3} xe^x + \frac{1}{4}e^{2x} = \frac{1}{4}e^{2x} + \left( \frac{1}{3}x + c_1 \right) e^x + c_2 e^{-2x}. \]

2 comments

  1. S says:

    A particular solution to this excercise can also be found as the sum of the solutions of the previous two excercises (9 and 10 in section 8.17). The general solution follows from the theorem 8.8.

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