Prove that a given differential equation has a given particular solution

by RoRi

February 27, 2016

Let $L$ be defined by

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

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

Assume $c \in \mathbb{R}$ and $b \neq \omega^2$ . Prove that the differential equation

$L(y) = c \sin (\omega x)$

has a particular solution of the form

$y = A \sin (\omega x + \alpha)$

and find expressions for $A$ and $\alpha$ in terms of $a,b,c,$ and $\omega$ .

Proof. Incomplete.

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2 comments

June 1, 2019 at 1:12 am

Evangelos says:

\textit{Proof}. Let

$\begin{align*} A &= \frac{c}{i \sqrt{(a \omega)^{2} + (b - \omega^{2})^{2}}} \end{align*}$

$\begin{align*} \sin(\alpha) &= \frac{a\omega}{i \sqrt{(a\omega)^2 + (b - \omega^{2})}} \\ \cos(\alpha) &= \frac{\omega^{2} - b}{i \sqrt{(a\omega)^2 + (b - \omega^{2})}} \end{align*}$

Then, for

$\begin{align*} y &= A\sin(\omega x + \alpha) \\ &= \frac{A}{2i}(e^{i(\omega x + \alpha)} - e^{-i(\omega x + \alpha)}) \end{align*}$

We have

$\begin{align*} L(y) &= \frac{A}{2i}(e^{i(\omega x + \alpha)}(b - \omega^{2} + ai\omega) + e^{-i(\omega x + \alpha)}(\omega^{2} - b + ai\omega)) \\ &= \frac{A}{2i}(e^{i\omega x}(b - \omega^{2} + ai\omega)(\cos(\alpha) + i\sin(\alpha)) + e^{-i\omega x}(\omega^{2} - b + ai\omega)(\cos(\alpha) - i\sin(\alpha)) \\ &= \frac{c}{2i}e^{i\omega x}\frac{(ai\omega + \omega^{2} - b)}{(ai\omega + \omega^{2} - b)} - \frac{c}{2i}e^{-i\omega x}\frac{(ai\omega + b - \omega^{2})}{(ai\omega + b - \omega^{2})} \\ &= c\sin(\omega x) \quad \blacksquare \end{align*}$

Reply
- July 13, 2020 at 9:53 am
  
  Anonymous says:
  
  Greetings!
  
  Very nice approach!
  
  I have minor issue with the second to last line. It seems to me $(ai\omega + \omega^2 - b)$ should be $(-ai\omega +\omega^2- b)$ . Why was there no sign change for the imaginary part $ai\omega$ ?
  
  Cheers.
  
  Reply

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