Solve ODE by Laplace transform y''+2y'−3y=sint

Take the Laplace transform of y'' + 2y' − 3y = sin t, with Y(s) = L{y}, y(0) = y'(0) = 0.

L{y''} = s²Y − s y(0) − y'(0) = s²Y
L{y'}  = sY − y(0) = sY
L{sin t} = 1/(s² + 1)

(s² + 2s − 3) Y = 1/(s² + 1)
Y = 1/[(s − 1)(s + 3)(s² + 1)]

Partial fractions: Y = A/(s − 1) + B/(s + 3) + (Cs + D)/(s² + 1).

A = 1/[(1 + 3)(1 + 1)] = 1/8           (put s = 1)
B = 1/[(−3 − 1)(9 + 1)] = −1/40        (put s = −3)
Coefficient of s³: A + B + C = 0  → C = −1/10
Put s = 0: −1/3 = −A + B/3 + D        → D = −1/5
Y = 1/(8(s − 1)) − 1/(40(s + 3)) − (s + 2)/(10(s² + 1))

Inverse transform:

y(t) = (1/8) eᵗ − (1/40) e^(−3t) − (1/10) cos t − (1/5) sin t

Answer: y(t) = eᵗ/8 − e^(−3t)/40 − (cos t)/10 − (sin t)/5. Check: y(0) = 1/8 − 1/40 − 1/10 = 0 ✓.