Solve ODE by Laplace transform y''+4y'+3y=t

y″ + 4y′ + 3y = t, y(0) = 0, y′(0) = 1.

Step 1: Transform (Y = L[y]).

(s²Y − 0 − 1) + 4(sY − 0) + 3Y = 1/s²
(s² + 4s + 3) Y = 1 + 1/s² = (s² + 1)/s²
Y = (s² + 1) / [ s² (s + 1)(s + 3) ]

Step 2: Partial fractions.

Y = A/s + B/s² + C/(s+1) + D/(s+3)
B = 1/(1·3) = 1/3
C (s = −1) = (1 + 1)/(1·2) = 1
D (s = −3) = (9 + 1)/(9·(−2)) = −5/9
A: A + C + D = 0 (large s) → A = −4/9

Step 3: Invert.

Answer: y(t) = t/3 − 4/9 + e^(−t) − (5/9) e^(−3t)

Check: y(0) = −4/9 + 1 − 5/9 = 0 ✓; y′(0) = 1/3 − 1 + 5/3 = 1 ✓.