Find solutions of 2xyy′ + (1+x)y2 = ex for given initial values

Find all solutions of the nonlinear differential equation

$2xyy' + (1+x)y^2 = e^x$

on the interval $(0, +\infty)$ satisfying the initial conditions

$y = \sqrt{e}$ when $x = 1$ ;
$y = -\sqrt{e}$ when $x = 1$ ;
a finite limit as $x \to 0$ .

previous exercise

$y = f(x)$

$I$

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b$

if and only if $(f(x))^k = g(x)$ on $I$ where $g(x) = v$ is the unique solution of the initial-value problem

$v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b.$

In the present problem we have the equation
$y' + \frac{1+x}{2x} y = \frac{e^x}{2x} y^{-1} \qquad \text{on } (0, +\infty) \qquad \text{with } y = \sqrt{e} \text{ when } x = 1.$

Therefore, to apply the previous exercise we have

$P(x) = \frac{1+x}{2x}, \quad Q(x) = \frac{e^x}{2x}, n = -1, k = 1 - n = 2.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^2 = g(x)$ where $g(x)$ is the unique solution to

$v' + \frac{1+x}{x}v = \frac{e^x}{x} \qquad \text{with } g(1) = e.$

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

$A(x) = \int_1^x \frac{1+t}{t} \,dt = \log x + x - 1,$

giving us

$\begin{align*} g(x) &= e \cdot e^{-\log x - x +1} + e^{-\log x - x +1} \int_1^x \frac{e^t}{t} e^{\log t + t - 1} \, dt \\[9pt] &= \frac{e^2}{xe^x} + \frac{e}{xe^x} \int_1^x e^{2t-1} \, dt \\[9pt] &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \int_1^x e^{2t} \, dt\\[9pt] &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \left( \frac{1}{2} e^{2t} \Bigr \rvert_1^x \right) \\[9pt] &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \left( \frac{e^{2x}}{2} - \frac{e^2}{2} \right) \\[9pt] &= \frac{e^x + e^2 - x}{2x} \end{align*}$

Therefore,

$(f(x))^2 = g(x) \quad \implies \quad f(x) = \left( \frac{e^x+e^{2-x}}{2x} \right)^{\frac{1}{2}}.$
From part (a) we have $g(x) = (f(x))^2$ which implies $f(x) = \pm (g(x))^{\frac{1}{2}}$ . Since $f(1) = -\sqrt{e}$ we then have
$f(x) = - \left( \frac{e^x + e^{2-x}}{2x} \right)^{\frac{1}{2}}.$
Let $\lim_{x \to 0} f(x) = b$ where $b$ is some finite real number. Then, (using the calculations we already completed in part (a) and just changing the initial-values $a$ and $b$ )
$\begin{align*} v &= be^{-x-\log x + a + \log a} + e^{-x - \log x + a + \log a} \int_a^x \let( \frac{e^t}{t} \right) \left( e^{t + \log t - a - \log a} \right) \, dt \\[9pt] &=\frac{bae^a}{xe^x} + \left( \frac{1}{xe^x} \right) \int_a^x e^{2t} \, dt \\[9pt] &= \frac{bae^a}{xe^x} + \frac{e^{2x} - e^{2a}}{2xe^x} \\[9pt] &= \frac{e^{2x} - c}{2xe^x}. \end{align*}$

Then,

$\lim_{x \to 0} \frac{e^{2x} - c}{2xe^x} = L \quad \implies \quad \lim_{x \to 0} (e^{2x} - c) = 0 \quad \implies \quadd c = 1.$

Therefore,

$y^2 = \frac{e^{2x}-1}{2xe^x} = \frac{e^x - e^{-x}}{2x} = \frac{\sinh x}{x}.$