Solve the differential equation xy′ - 2y = x5 for given initial values

Solve the following differential equation:

$xy' - 2y = x^5 \qquad \text{on} \qquad (0, +\infty)$

with $y = 1$ when $x = 1$ .

From Theorem 8.3 (page 310 of Apostol) we know that a differential equation of the form

$y' + P(x)y = Q(x), \qquad \text{with} \quad f(a) = b,$

on an interval $I$ has solution given by

$f(x) = be^{-A(x)} + e^{-A(x)} \int_a^x Q(t)e^{A(t)} \, dt$

where

$A(x) = \int_a^x P(t) \, dt.$

In this particular case first divide through by $x$ (which is never 0 on the given interval),

$xy' - 2y = x^5 \qquad \implies \qquad y' - \frac{2}{x} y = x^4.$

Then, we apply the theorem with

$P(x) = -\frac{2}{x}, \qquad Q(x) = x^4, \qquad a = b = 1.$

This gives us

$A(x) = \int_1^x -\frac{2}{t} \, dt = -2 \log x.$

Therefore,

$\begin{align*} f(x) &= be^{-A(x)} + e^{-A(x)} \int_a^x Q(t)e^{A(t)} \, dt \\[9pt] &= x^2 + x^2 \int_1^x (t^4)(t^{-2}) \, dt \\[9pt] &= x^2 + x^2 \left( \frac{1}{3} t^3 \right) \Bigr \rvert_1^x \\[9pt] &= x^2 + \frac{1}{3}x^5 - \frac{1}{3} x^2 \\[9pt] &= \frac{1}{3}x^5 + \frac{2}{3} x^2. \end{align*}$

Stumbling Robot

Solve the differential equation xy′ – 2y = x⁵ for given initial values

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