Find solutions of xy′ - 2y = 4x3y1/2 for given initial values

Find all solutions of the nonlinear differential equation

$xy' - 2y = 4x^3y^{\frac{1}{2}}$

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

$y = 0 \qquad \text{when} \qquad x = 1.$

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

$xy' - 2y = 4x^3 y^{\frac{1}{2}} \qquad \text{on } (-\infty, +\infty) \qquad \text{with } y = 0 \text{ when } x = 1.$

Therefore, to apply the previous exercise we have

$P(x) = -\frac{2}{x}, \quad Q(x) = 4x^2, n = \frac{1}{2}, k = 1 - n = \frac{1}{2}.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^{\frac{1}{2}} = g(x)$ where $g(x)$ is the unique solution to

$v' - \frac{1}{x}v = 2x^2 \qquad \text{with } g(1) = 0.$

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

$\begin{align*} g(x) &= e^{\log x} \int_1^x 2t^2 \cdot e^{-\log t} \,dt \\ &= x \int_1^x 2t \, dt \\ &= x \left( t^2 \Bigr \rvert_1^x \right) \\ &= x (x^2 - 1) \\ &= x^3 - x. \end{align*}$

Therefore,

$(f(x))^{\frac{1}{2}} = g(x) \quad \implies \quad f(x) = (x^3-x)^2.$

Stumbling Robot

Find solutions of xy′ – 2y = 4x³y^1/2 for given initial values

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