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Find the general solution of y′′ + 2y′ + y = e-x/x2

Find the general solution of the second-order differential equation

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

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


The general solution of the homogeneous equation

    \[ y'' + 2y' + y = 0 \]

is given by Theorem 8.7 with a = 2 and b = 1. This gives us d = a^2 - 4b = 0. Hence,

    \begin{align*}  y &= e^{-\frac{ax}{2}} (c_1 + c_2x) \\  &= e^{-x} (c_1 + c_2 x). \end{align*}

We then find a particular solution of the non-homogeneous equation by assuming y_1 = p(x) e^{-x} is a solution. This has derivatives

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

Therefore,

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

Integrating twice we can solve for p(x),

    \begin{align*}  p''(x) = \frac{1}{x^2} && \implies && p'(x) &= \int \frac{1}{x^2} = -\frac{1}{x} \\  && \implies && p(x) &= -\int \frac{1}{x} = - \log |x|. \end{align*}

(We can ignore the constants of integration here since we only need p(x) to satisfy the equation p''(x) = x^{-2}. Any constants of integration are not going to matter when we take derivatives of p(x).)
Therefore the particular solution is

    \[ y_1(x) = -e^{-x} \log |x| \]

and the general solution (by Theorem 8.8) is

    \[ y = e^{-x}(c_1 + c_2 x - \log |x|). \]

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