Solve the differential equation y′ – 3y = e^2x for given initial values

by RoRi

January 13, 2016

Solve the following differential equation:

$y' - 3y = e^{2x} \qquad \text{on} \quad (-\infty, +\infty)$

with $y =0$ when $x =0$ .

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 with

$P(x) = -3, \qquad Q(x) = e^{2x}, \qquad a = b= 0.$

This gives us

$A(x) = \int_0^x -3 \, dt = -3x.$

Therefore,

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

Stumbling Robot

Solve the differential equation y′ – 3y = e^2x for given initial values

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