Given the IVP $$ y'' + y = f(t) , \qquad\quad y(0) = 0 , \quad y'(0) = 0 , \tag{1}$$ where $$ f_{k} (t) = u_{0} + 2 \sum_{k=1}^{n} (-1)^{k} u_{k \pi}(t). \tag{2}$$ We want to find the solution.
My attempt is as follows : $$ f(t) = u_{0}(t) -2u_{\pi}(t) + 2u_{2\pi}(t) -2u_{3\pi}(t) + \dots $$ Therefore, we can compute the Laplace transform on both sides of Equation 1 \begin{align} \mathcal{L}[y'' +y] &= \mathcal{L}[u_{0}(t) -2u_{\pi}(t) + 2u_{2\pi}(t) -2u_{3\pi}(t) + \dots ]\\ \mathcal{L}[y''] + \mathcal{L}[y] &=\mathcal{L}[u_{0}(t)] - 2\mathcal{L}[u_{\pi}(t)] + 2\mathcal{L}[u_{2\pi}(t)] -2 \mathcal{L}[u_{3\pi}(t)] + \dots\\ Y(s)(s^{2} + 1) &= \frac{1}{s}-2\frac{e^{-\pi s}}{s}+2\frac{e^{-2\pi s}}{s}-2\frac{e^{-3\pi s}}{s}+\dots\\ \therefore Y(s) &= \frac{1}{s(s^{2}+1)} - 2\frac{e^{-\pi s}}{s(s^{2}+1)}+2\frac{e^{-2\pi s}}{s(s^{2}+1)} -2\frac{e^{-3\pi s}}{s(s^{2}+1)} + \dots \\ \implies y(t) &= \mathcal{L}^{-1} \left[\frac{1}{s(s^{2}+1)} \right] - 2\mathcal{L}^{-1} \left[\frac{e^{-\pi s}}{s(s^{2}+1)} \right] +2 \mathcal{L}^{-1} \left[\frac{e^{-2\pi s}}{s(s^{2}+1)} \right] + \dots\\ & \because \frac{1}{s(s^2 +1)} = \left(\frac{1}{s} - \frac{s}{s^2 +1} \right), \\ &= 1-\cos(t) - 2 u_{\pi }(t) \mathcal{L}^{-1}\left[\frac{1}{s} - \frac{s}{s^2 +1} \right]+2 u_{2\pi }(t) \mathcal{L}^{-1}\left[\frac{1}{s} - \frac{s}{s^2 +1}\right] + \dots \\ &= 1-\cos(t) - 2 u_{\pi }(t) (1-\cos(t-\pi)) + 2 u_{2\pi }(t)(1-\cos(t-2\pi)) + \dots \\ \therefore y(t) &= 1-\cos(t) + 2 \sum_{k=1}^{n} (-1)^{k} u_{k\pi}(t)(1-\cos(t-k\pi)) . \end{align}
I'm not quite certain about my reasoning. Particularly, I'm unsure if the given form of the solution ,which involves a summation ,is adequate. Can anyone tell me if I went wrong anywhere ?
Looks good to me. Good job. You can avoid lots of calculation by keeping the sigma notation: $$Y(s)(s^{2} + 1) = \dfrac {1}{s } + 2 \sum_{k=1}^{n} (-1)^{k} \dfrac {e^{-k\pi s}}{s }$$ $$Y(s)= \dfrac {1}{s(s^{2} + 1) } + 2 \sum_{k=1}^{n} (-1)^{k} \dfrac {e^{-k\pi s}}{s(s^{2} + 1) }$$ $$Y(s)= \left (\dfrac 1 s -\dfrac s{s^{2} + 1 } \right) + 2 \sum_{k=1}^{n} (-1)^{k}{e^{-k\pi s}} \left (\dfrac 1 s -\dfrac s{s^{2} + 1 } \right)$$ $$y(t)= 1-\cos t + 2 \sum_{k=1}^{n} (-1)^{k}{u_{k\pi} }{(1-\cos(t-k\pi ))}$$