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Find all solutions of the differential equation x(x+1)y′ + y = x(x+1)2 e-x2

Find all solutions of the differential equation

    \[ x(x+1)y' + y = x(x+1)^2 e^{-x^2} \]

on the interval (-1,0). Prove that all of these solutions go to 0 as x \to -1, but that there is exactly one solution which has a finite limit as x \to \pi.


Proof. Since x(x+1) \neq 0 for any x \in (-1,0) we divide both sides by x(x+1) to get

    \[ x(x+1)y' + y = x(x+1)^2 e^{-x^2} \qquad \implies \qquad y' + \frac{1}{x(x+1)} y = (x+1)e^{-x^2}. \]

Then, we apply Theorem 8.3 (page 310 of Apostol) with

    \[ P(x) = \frac{1}{x(x+1)}, \qquad Q(x) = (x+1)e^{-x^2}. \]

We choose a = -\frac{1}{2} and compute solutions in terms of b. Using partial fractions we evaluate A(x),

    \begin{align*}   A(x) &= \int_{-\frac{1}{2}}^x \frac{1}{t(t+1)} \, dt \\  &= \int_{-\frac{1}{2}}^x \frac{1}{t} \, dt - \int_{-\frac{1}{2}}^x \frac{1}{t+1} \, dt  &= \log |x| - \log |x+1|. \end{align*}

Therefore,

    \begin{align*}  f(x) &= be^{-\log|x| + \log|x+1|} + e^{-\log |x| + \log|x+1|} \int_{-\frac{1}{2}}^x (t+1)e^{-t^2}e^{\log|t|-\log|t+1|} \, dt  \\[9pt]  &= b \left( \frac{1+x}{x} \right) + \frac{1+x}{x} \int_{-\frac{1}{2}}^x te^{-t^2} \, dt \\[9pt]  &= b \left( \frac{1+x}{x} \right) + \frac{1+x}{x} \left( -\frac{1}{2} e^{-t^2} \right) \Bigr \rvert_{-\frac{1}{2}}^x \\[9pt]  &= b \left( \frac{1+x}{x} \right) + \frac{1+x}{x} \left( -\frac{e^{-x^2}}{2} + \frac{e^{-\frac{1}{4}}}{2} \right) \\[9pt]  &= b \left( \frac{1+x}{x} \right) - \frac{1}{2} \left( \frac{x+1}{x} \right) (e^{-x^2} - e^{-\frac{1}{4}} ) \\[9pt]  &= \frac{1}{2} \left( 1 + \frac{1}{x} \right) (2b - e^{-x^2} + e^{-\frac{1}{4}} ) \\[9pt]  &= \frac{1}{2} \left( 1 + \frac{1}{x} \right)(C - e^{-x^2}). \end{align*}

This gives us all solutions to the differential equation in terms of the constant C = 2b + e^{-\frac{1}{4}}.

Next, we take the limit

    \begin{align*}  \lim_{x \to -1} f(x) &= \lim_{x \to -1} \left( \frac{1}{2} \left( 1 + \frac{1}{x} \right) \left( C - e^{-x^2} \right) \right) \\[9pt]  &= 0 \end{align*}

for all C (since 1 + \frac{1}{x} \to 0 as x \to -1 and all of the other terms in the product are tending to finite limits as well). Next, if

    \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{2} \left( 1 + \frac{1}{x} \right) \left( C - e^{-x^2} \right) \right) \]

is finite then we must have \lim_{x \to 0} (C - e^{-x^2} ) = 0; thus, C = 1. \qquad \blacksquare

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