Exercise 4.4, problem 24, page 144 of the book on differential equations by KKOP [here][1].
Solve the initial value problem $(D^2+2aD+b^2)y=\sin \omega t$, $y(0)=y'(0)=0$, where $a,b,\omega$ are real constants, $a<b$. Consider separately the cases $\omega \ne \sqrt{b^2-a^2}$ and $\omega = \sqrt{b^2-a^2}$ and sketch the solution curve in each case.
I am getting not so nice looking terms. So, I would like to ask if Im proceeding in the right direction, if somebody could verify my steps or help me with the solution, that would be great. My solution is as below.
Solution(My attempt).
The associated homogeneous differential equation is –
$(D^2+2aD+b^2)y=0$
The characteristic equation is:
$\begin{align}
D^2+2aD+b^2&=0\\
(D+a)^2 + (b^2-a^2)&=0\\
(D+a)^2 &= -(b^2-a^2)\\
D &= -a \pm \sqrt{b^2-a^2}i
\end{align}$
The general solution of the homogeneous linear differential equation is :
$\bbox[5px, border: 2px solid blue]{y_h = e^{-ax}(c_1 \cos (\sqrt{b^2-a^2}x) + c_2 \sin (\sqrt{b^2-a^2}x))}$
A particular solution of the nonhomogeneous linear differential equation can be constructed of the form:
$\bbox[5px, border: 2px solid blue]{y_p = c_1(x) e^{-ax} \cos (\sqrt{b^2-a^2}x) + c_2(x) e^{-ax} \sin (\sqrt{b^2-a^2}x)}$
where $c_1(x)$ and $c_2(x)$ are given by the equations:
$\begin{align}
y_1(x)c_1'(x)+y_2(x)c_2'(x)&=0\\
y_1'(x)c_1'(x)+y_2'(x)c_2'(x)&=h(x)\\
\end{align}$
That is,
$\begin{align}
e^{-ax} \cos (\sqrt{b^2-a^2}x) c_1'(x)+e^{-ax} \sin (\sqrt{b^2-a^2}x)c_2'(x)&=0\\
\{-ae^{-ax}\cos (\sqrt{b^2-a^2}x)-\sqrt{b^2-a^2}e^{-ax}\sin (\sqrt{b^2-a^2}x)\}c_1'(x)\\+
\{-ae^{-ax}\sin (\sqrt{b^2-a^2}x)+\sqrt{b^2-a^2}e^{-ax}\cos (\sqrt{b^2-a^2}x)\}c_2'(x)&=\sin (\omega x)\\
\end{align}$
The Wronskian of the functions $e^{-ax}\cos (\sqrt{b^2-a^2}x), e^{-ax}\sin (\sqrt{b^2-a^2}x)$ is:
$\begin{align}
W(x) &=
\scriptsize
e^{-2ax}
\begin{vmatrix}
\cos (\sqrt{b^2-a^2}x) & \sin (\sqrt{b^2-a^2}x)\\
-a\cos (\sqrt{b^2-a^2}x)-\sqrt{b^2-a^2}\sin (\sqrt{b^2-a^2}x) & -a\sin (\sqrt{b^2-a^2}x)+\sqrt{b^2-a^2}\cos (\sqrt{b^2-a^2}x)
\end{vmatrix}\\
&= \sqrt{b^2-a^2}e^{-2ax}
\end{align}$
Consequently,
$\begin{align}
\scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{\sqrt{b^2-a^2}}[\cos (\omega – \sqrt{b^2-a^2}x) – \cos (\omega + \sqrt{b^2-a^2}x)]\\
\scriptsize c_2'(x)&=\scriptsize e^{ax}\frac{\cos \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =\frac{e^{ax}}{\sqrt{b^2-a^2}}[\sin (\omega + \sqrt{b^2-a^2}x) + \sin (\omega – \sqrt{b^2-a^2}x)]
\end{align}$
Integrating these yields –
$\begin{align}
\scriptsize c_1(x)&=\scriptsize \left[\frac{e^{ax}}{a^2+(\omega + \sqrt{b^2-a^2})^2}\{a \sin (\omega + \sqrt{b^2-a^2}) – (\omega + \sqrt{b^2-a^2}) \cos (\omega + \sqrt{b^2-a^2})\}\\
-\frac{e^{ax}}{a^2+(\omega – \sqrt{b^2-a^2})^2}\{a \sin (\omega – \sqrt{b^2-a^2}) – (\omega – \sqrt{b^2-a^2}) \cos (\omega – \sqrt{b^2-a^2})\}\right]\\
\scriptsize c_2(x)&=\scriptsize \left[\frac{e^{ax}}{a^2+(\omega + \sqrt{b^2-a^2})^2}\{a \cos (\omega + \sqrt{b^2-a^2}) + (\omega + \sqrt{b^2-a^2}) \sin (\omega + \sqrt{b^2-a^2})\}\\
+\frac{e^{ax}}{a^2+(\omega – \sqrt{b^2-a^2})^2}\{a \cos (\omega – \sqrt{b^2-a^2}) + (\omega – \sqrt{b^2-a^2}) \sin (\omega – \sqrt{b^2-a^2})\}\right]\\
\end{align}$
Best Answer
You first mistake is at the step $$\begin{align} \scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{\sqrt{b^2-a^2}}\left[\cos \left(\omega - \sqrt{b^2-a^2}x\right) - \cos \left(\omega + \sqrt{b^2-a^2}x\right)\right]\\ \scriptsize c_2'(x)&=\scriptsize e^{ax}\frac{\cos \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =\frac{e^{ax}}{\sqrt{b^2-a^2}}\left[\sin \left(\omega + \sqrt{b^2-a^2}x\right) + \sin \left(\omega - \sqrt{b^2-a^2}x\right)\right] \end{align}$$ It should have been $$\begin{align} \scriptsize c_1'(x)&=\scriptsize -e^{ax}\frac{\sin \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =-\frac{e^{ax}}{2\sqrt{b^2-a^2}}\left[\cos \left(\omega - \sqrt{b^2-a^2}\right)x - \cos \left(\omega + \sqrt{b^2-a^2}\right)x\right]\\ \scriptsize c_2'(x)&=\scriptsize e^{ax}\frac{\cos \sqrt{b^2-a^2}x \cdot sin \omega x}{\sqrt{b^2-a^2}} =\frac{e^{ax}}{2\sqrt{b^2-a^2}}\left[\sin \left(\omega + \sqrt{b^2-a^2}\right)x + \sin \left(\omega - \sqrt{b^2-a^2}\right)x\right] \end{align}$$ Part of the error was due to translating your manuscript to MathJax, but you are missing the factors of $2$ in $$2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta)$$ and $$2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$$ But you could have corrected the factor of $2$ if you differentiated your solution and substituted into the original differential equation. More serious is the fact that $$\int e^{ax}\cos\beta x\,dx=\frac{e^{ax}\left(a\cos\beta x+\beta\sin\beta x\right)}{a^2+\beta^2}+C$$ and $$\int e^{ax}\sin\beta x\,dx=\frac{e^{ax}\left(-\beta\cos\beta x+a\sin\beta x\right)}{a^2+\beta^2}+C$$ which may be checked by differentiation but you seem to have applied switched these formulas around, not to mention the mistake of dropping the factor of $1/\sqrt{b^2-a^2}$ and forgetting about the factor of $x$ in the arguments to the trig functions in transcription to MathJax.
There is a much easier way to solve this kind of problem with a sinusoidally driven damped harmonic oscillator: since $\Im\left(e^{i\omega x}\right)=\sin\omega $ just solve the complex problem $$z^{\prime\prime}+2az^{\prime}+b^2z=e^{i\omega x}$$ and then take the imaginary part of the solution. Assume a solution $z=Be^{i\omega x}$ for some complex number $B$. Then on differentiation and substitution we get $$B\left(-\omega^2+2ia\omega+b^2\right)e^{i\omega x}=e^{i\omega x}$$ So $$\begin{align}z&=Be^{i\omega x}=\frac{e^{i\omega x}}{b^2-\omega^2+2ia\omega}=\frac{\left(b^2-\omega^2-2ia\omega\right)\left(\cos\omega x+i\sin\omega x\right)}{\left(b^2-\omega^2\right)^2+4a^2\omega^2}\\ &=\frac{\left(b^2-\omega^2\right)\cos\omega x+2a\omega\sin\omega x+i\left(\left(b^2-a^2\right)\sin\omega x-2a\omega\cos\omega x\right)}{\left(b^2-\omega^2\right)^2+4a^2\omega^2}\end{align}$$ And our solution is $$y=\Im z=\frac{\left(b^2-a^2\right)\sin\omega x-2a\omega\cos\omega x}{\left(b^2-\omega^2\right)^2+4a^2\omega^2}$$ When you eventually get your method to work you can expand trig functions like $$\cos\left(\omega+\sqrt{b^2-a^2}\right)x=\cos\omega x\cos\left(\sqrt{b^2-a^2}x\right)-\sin\omega x\sin\left(\sqrt{b^2-a^2}x\right)$$ and then group terms with like powers of trig functions of $\sqrt{b^2-a^2}x$ and $\omega x$ and simplify to compare with the easy answer.