Prove existence of a complex valued solution to a given differential equation

by RoRi

February 27, 2016

Let $L$ be defined by

$L(y) = y'' + ay' + by,$

for $a, b \in \mathbb{R}$ constants.

Let $\omega \in \mathbb{R}$ and $A \in \mathbb{C}$ . If either $b \neq \omega^2$ or $a \omega \neq 0$ prove that the differential equation

$L(y) = Ae^{i \omega x}$

has a solution

$y = Be^{i \omega x}.$

Express the value of $B$ in terms of $a$ , $b$ , $A$ , and $\omega$ .

Proof. Let $y = Be^{i \omega x}$ with either $b \neq \omega^2$ or $a \omega \neq 0$ . Then we have

$\begin{align*} L(y) &= -B \omega^2 e^{i \omega x} + a B i \omega e^{i \omega x} + bB e^{i \omega x} \\ &= e^{i \omega x} (bB - B\omega^2 + \omega a B i ). \end{align*}$

So,

$B = \frac{A}{b - \omega^2 + i a \omega} \qquad \text{if } b - \omega^2 + i a \omega \neq 0.$

But since $b - \omega^2 \neq 0$ or $a \omega \neq 0$ we have $b - \omega^2 + i a \omega \neq 0$ . Hence,

$L(y) = Ae^{i \omega x}.$

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply