Escrevemos \[ g(t)=1-u_{\pi/2}(t). \] Aplicando a transformada de Laplace na equação diferencial, temos \[ \mathcal L\{y''\}+\mathcal L\{y\}=G(s). \] Como \(y(0)=0\) e \(y'(0)=1\), \[ \mathcal L\{y''\}=s^2Y(s)-1. \] Além disso, \[ G(s)=\frac{1}{s}-\frac{\textrm{e}^{-\pi s/2}}{s}. \] Logo \[ (s^2+1)Y(s)-1 = \frac{1}{s}-\frac{\textrm{e}^{-\pi s/2}}{s}. \]

Portanto \[ Y(s)= \frac{1}{s^2+1} + \left(1-\textrm{e}^{-\pi s/2}\right) \frac{1}{s(s^2+1)}. \] Agora \[ \mathcal L^{-1}\left\{\frac{1}{s^2+1}\right\} = \operatorname{sen}(t), \] e \[ \mathcal L^{-1}\left\{\frac{1}{s(s^2+1)}\right\} = 1-\cos(t). \]

Chamando \(h(t)=1-\cos(t)\), obtemos \[ \colorbox{green!20}{\displaystyle y(t)=\operatorname{sen}(t)+h(t)-u_{\pi/2}(t)h\left(t-\frac{\pi}{2}\right)}. \] Isto é, \[ \colorbox{green!20}{\displaystyle y(t)=\operatorname{sen}(t)+1-\cos(t) -u_{\pi/2}(t)\left(1-\cos\left(t-\frac{\pi}{2}\right)\right)}. \]