I am having a difficulty to derive the solution to the ODE \begin{align*} Ay'' - By = G(x) \end{align*} where \begin{align*} G(x) = \begin{cases} C \sin (kx) & |kx| < \pi \\ 0 & \text{Otherwise} \end{cases} \end{align*} The solution roughly reads \begin{align*} y = \begin{cases} p(\sin(kx) + kqe^{-\frac{\pi}{kq}} \sinh(\frac{x}{q})) & |kx| < \pi \\ pqk \sinh(\frac{\pi}{kq}) e^{-\frac{|x|}{q}} sgn(x) & \text{Otherwise} \end{cases} \end{align*} where $p$, $q$ are some constants in terms of $A$ and $B$. I have tried to use Laplace Transform that leads to something resembling the first part of the solution ($|kx| < \pi$), but I cannot obtain anything that looks like the second part (otherwise). I think the given solution should be correct (and physically feasible), but I have no idea how to proceed. I also doubt that if my approach of Laplace Transform is correct, since the initial condition is kind of not given (in hindsight $y(0)=0$, and I try to make my solution physically bounded). I attempt the question by writing $G(x)$ with the help of Heaviside step function and do the transform (and inverse). Any help would be appreciated.
Update: From the book (c > u)
My attempt:
In order to simplify notation, I will rewrite your first two equations as $$ y''-a^2y=C\sin kx\,[1-H(x-\pi/k)], \tag{1} $$ where $H$ denotes the Heaviside step function. Taking the Laplace transform of both sides, and imposing the condition $y(0)=0$, we obtain $$ -y'(0)+(s^2-a^2)\mathcal{L}[y]=\frac{Ck}{s^2+k^2}(e^{-\pi s/k}+1), \tag{2} $$ which yields $$ \mathcal{L}[y]=\frac{y'(0)}{s^2-a^2}+\frac{Ck}{a^2+k^2} \left(\frac{1}{s^2-a^2}-\frac{1}{s^2+k^2}\right)(e^{-\pi s/k}+1). \tag{3} $$ Taking the inverse Laplace transform we obtain \begin{align*} y(x)=& \left[y'(0)\,\frac{\sinh ax}{a} + \frac{Ck}{a^2+k^2}\left( \frac{\sinh ax}{a}-\frac{\sin kx}{k}\right) \right]H(x) \\ & +\frac{Ck}{a^2+k^2}\left[\frac{\sinh a(x-\pi/k)}{a}-\frac{\sin k(x-\pi/k)}{k}\right] H(x-\pi/k) \\ =&-\frac{C}{a^2+k^2}\,\sin kx\,[H(x)-H(x-\pi/k)] \\ &+\left[y'(0)+\frac{Ck}{a^2+k^2}\right]\frac{\sinh ax}{a}\,H(x) +\frac{Ck}{a^2+k^2}\,\frac{\sinh a(x-\pi/k)}{a}\,H(x-\pi/k). \tag{4} \end{align*} For $x>\pi/k$ only the last line of $(4)$ contributes to $y(x)$. To examine its behavior for $x\to\infty$, it is convenient to rewrite it in terms of exponentials: \begin{align*} y(x)|_{x>\pi/k}=&\left[y'(0)+\frac{Ck}{a^2+k^2} \,(1+e^{-\pi a/k})\right]\frac{e^{ax}}{2a} \\ &-\left[y'(0)+\frac{Ck}{a^2+k^2}\, (1+e^{\pi a/k})\right]\frac{e^{-ax}}{2a}.\tag{5} \end{align*} For $y(x)$ to be bounded, the coefficient of $e^{ax}$ in $(5)$ must be $0$, and this condition determines the value of $y'(0)$: $$ y'(0)=-\frac{Ck}{a^2+k^2}\,(1+e^{-\pi a/k}). \tag{6} $$ Combining results $(4)-(6)$ we finally obtain the following expression for $y(x)$: $$ y(x)=-\frac{C}{a^2+k^2}\times\begin{cases} \sin kx+(k/a)\,e^{-\pi a/k}\,\sinh ax &\mathrm{if}\,0\leq x<\pi/k, \\ (k/a)\,\sinh(\pi a/k)\,e^{-ax} &\mathrm{if}\,x>\pi/k. \tag{7} \end{cases} $$ To obtain $y(x)$ for $x<0$ take the odd extension of $(7)$.