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

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