Home » Blog » Prove existence of a complex valued solution to a given differential equation

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

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,

    \[ L(y) = Ae^{i \omega x}. \]

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