Solve the differential equation x′ + x = e2t for given initial values

Solve the following differential equation:

$\frac{dx}{dt} + x = e^{2t} \qquad \text{on} \qquad (-\infty, +\infty)$

with $x = 1$ when $t = 0$ .

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 we apply the theorem (noting that in this problem we have $x$ is a function of $t$ , rather than $y$ a function of $x$ as in the theorem… of course, the names of the variables doesn’t matter, but we take care to apply the theorem properly) with

$P(t) = 1, \qquad Q(t) = e^{2t}, \qquad a = 0, \ b = 1.$

This gives us

$A(t) = \int_0^t P(s) \, dt = \int_0^t \,ds = t.$

Therefore,

$\begin{align*} f(t) &= be^{-A(t)} + e^{-A(t)} \int_a^s Q(s) e^{A(s)} \, ds \\[9pt] &= e^{-t} + e^{-t} \int_0^s e^{2s} e^{s} \, ds \\[9pt] &= e^{-t} + e^{-t} \int_0^s e^{3s} \, ds \\[9pt] &= e^{-t} + e^{-t} \left( \frac{1}{3}e^{3t} - \frac{1}{3} \right) \\[9pt] &= \frac{1}{3}e^{2t} + \frac{2}{3}e^{-t}. \end{align*}$

Stumbling Robot

Solve the differential equation x′ + x = e^2t for given initial values

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