Let [y] = y(s)

$\therefore$ Taking Laplace transform of above given equation consider L.H.S.

$\therefore L\bigg[\frac{d^2y}{dt^2}\bigg] = L[y"] = s^2y(s) - sy(0) - y'(0)$

$L\bigg[\frac{dy}{dt}\bigg] = L[y'] = sy(0) - y(0)$

Now consider R.H.S.

$L[sint] = \frac{1}{s^2 + 1}$

$\therefore$ The equation becomes

$[s^2 y(s) - sy(0) - y'(0)] + 2[sy(s) - y(0)] - 3y(s)$

Substituting y(0) = 0, y'(0)

$\therefore s^2y(s) + 2sy(s) - 3y(s) = \frac{1}{s^2 + 1}$

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

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

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

By using partial fraction method

$\frac{1}{(s + 3) (s - 1) (s^2 + 1)} = \frac{A}{(s + 3)} + \frac{B}{(s - 1)} + \frac{Cs + D}{s^2 + 1}$

$\therefore 1 = A(s-1)(s^2 + 1) + B(s + 3)(s^2 + 1) + (Cs + D)(s + 3)(s - 1)$

Now put s = -3, 1 = A(-4)(9 + 1) $\therefore A = \frac{-1}{40}$

Now put s = 1, 1 = B(4)(2) $\therefore B = \frac{1}{8}$

Now equating 'degree 3' coefficients,

$\therefore A+B+C = 0$

$\therefore \frac{-1}{40} + \frac{1}{8} + C = 0$

$\therefore C = \frac{-4}{40} = \frac{-1}{10}$

$\therefore C = \frac{-1}{10}$

Now equating 'degree 0' coefficients

-A + 3B - 3D = 1

$\therefore \frac{1}{40} + \frac{3}{8} - 3D = 1$

$\therefore \frac{16}{40} - 1 = 3D$

$\therefore \frac{-24}{40} * \frac{1}{3} = D$

$\therefore D = \frac{-1}{5}$

$\therefore A = \frac{-1}{40} , B = \frac{1}{8}, C = \frac{-1}{10}, D = \frac{-1}{5}$

$\therefore y(s) = \frac{-1}{40(s + 3)} + \frac{1}{8(s - 1)} - \frac{1}{5} \frac{s}{s^2 + 1} - \frac{1}{5(s^2 + 1)}$

Taking inverse Laplace Transform

$y(t) = \frac{-1}{40} e^{-3t} + \frac{1}{8} e^t - \frac{1}{5}cost - \frac{1}{5}sint$

$y(t) = \frac{-1}{40} e^{-3t} + \frac{1}{8} e^t - \frac{1}{5}(cost + sint)$