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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,

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

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