Find all solutions of xy′ + y = y2x2 log x for given initial values

Find all solutions of the nonlinear differential equation

$xy' + y = y^2 x^2 \log x$

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

$y = \frac{1}{2} \qquad \text{when} \qquad x = 1.$

From a previous exercise (Section 8.5, Exercise #13) we know that a function $y = f(x)$ which is never zero on an interval $I$ is a solution of the initial value problem

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

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

$y' + \frac{1}{x}y = y^2 x \log x \qquad \text{on } (0, +\infty) \qquad \text{with } y = \frac{1}{2} \text{ when } x = 1.$

Therefore, to apply the previous exercise we have

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

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^{-1} = g(x)$ where $g(x)$ is the unique solution to

$v' - \frac{1}{x} v = -x \log x \qquad \text{with } g(1) = 2.$

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

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

giving us

$\begin{align*} g(x) &= 2e^{\log x} + e^{\log x} \int_1^x (-t \log t) \cdot e^{-\log t} \, dt \\ &= 2x - x \int_1^x \log t \, dt \\ &= 2x - x (x \log x - x + 1) \\ &= x^2 + x - x^2 \log x. \end{align*}$

Therefore,

$(f(x))^{-1} = g(x) \quad \implies \quad f(x) = \frac{1}{x^2+x-x^2 \log x}.$

Stumbling Robot

Find all solutions of xy′ + y = y²x² log x for given initial values

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