$\dfrac {d^2 y}{dt^2} + y = t$ Taking Laplace Transform, $\therefore L \left [\dfrac {d^2 y}{dt^2} \right ] + L [y] = L [t]$ $\therefore s^2 \bar{y} - sy (0)- y'(0)+ \bar{y} = \dfrac {1}{s^2} $ $\therefore s^2 \bar{y}-s(1) - 0 + \bar{y}= \dfrac {1}{s^2} $ $\therefore \bar{y}(s^2 +1) = \dfrac {1}{s^2}+s$ $\therefore \bar{y} = \dfrac {1}{s^2 (s^2+1)}+ \dfrac {s}{s^2+1}$ $\therefore y=L^{-1} \left [ \dfrac {1}{s^2 (s^2+1)} + \dfrac {s}{s^2+1} \right ] \ \ \to (1)$ Let $\dfrac {1}{s^2 (s^2 +1)} = \dfrac {A}{s^2} + \dfrac {B}{s^2 +1}\ \ \ \to (2)$ $\therefore A=\dfrac {1}{s^2 +1} \Bigg \vert_{s^2 = 0} = \dfrac {1}{0^2 +1}=-1 $ $\therefore B=\dfrac {1}{s^2} \Bigg \vert_{s^2 = -1} = \dfrac {1}{-1} = -1$ Here, from (1) & (2)  $\therefore y=L^{-1} \left [ \dfrac {1}{s^2} - \dfrac {1}{s^2+1} + \dfrac{s}{s^2+1} \right ]$ $\therefore y=t - \sin t + \cos t$