2nd order IVP via Laplace transform

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Find the solution $y(t)$ through the Laplace transform for an input $\sin(t)$. $$\ddot y(t) + \dot y(t) = \sin(t), \quad \dot y (0) = 2, \quad y (0) = 1$$

This is how far I got:

$$(s^2 y(s) - a - 2) + s y(s) - 1 + y(s) = 0$$

$$y(s) = \frac{a+3}{s(s^2 + 1 + s)}$$

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$$\ddot y(t) + \dot y(t) = \sin(t), \quad \dot y (0) = 2, \quad y (0) = 1$$ The Laplace transform of $\sin t $ is $\dfrac 1 {s^2+1}$: $$s^2Y(s)-s-2+sY(s)-1=\dfrac 1 {s^2+1}$$ $$Y(s)(s^2+s)-s-3=\dfrac 1 {s^2+1}$$ $$Y(s)=\dfrac 1 {(s^2+1)(s^2+s)}+ \dfrac {s+3}{s^2+s}$$ Decompose into simple fractions and apply inverse Laplace transform.