Solve ODE by Laplace transform y''+3y'+2y=eᵗ

Take the Laplace transform of y'' + 3y' + 2y = eᵗ with y(0) = y'(0) = 0, Y = L{y}.

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

Partial fractions: Y = A/(s − 1) + B/(s + 1) + C/(s + 2).

A = 1/[(1 + 1)(1 + 2)] = 1/6
B = 1/[(−1 − 1)(−1 + 2)] = −1/2
C = 1/[(−2 − 1)(−2 + 1)] = 1/3

Inverse transform:

y(t) = (1/6) eᵗ − (1/2) e^(−t) + (1/3) e^(−2t)

Answer: y(t) = eᵗ/6 − e^(−t)/2 + e^(−2t)/3. Check: y(0) = 1/6 − 1/2 + 1/3 = 0 ✓; y'(0) = 1/6 + 1/2 − 2/3 = 0 ✓.