Step 1: Solve the initial-value problem y'' + 3y' + 2y = 0 with y(0) = 1, y'(0) = 0 using Laplace transforms. Step 2: Take the Laplace transform of both sides. Let Y(s) = L{y}. Then L{y''} = s²Y − s·y(0) − y'(0) = s²Y − s, and L{y'} = sY − y(0) = sY − 1. Step 3: The transformed equation is (s²Y − s) + 3(sY − 1) + 2Y = 0. Step 4: Collect Y: Y(s² + 3s + 2) = s + 3, so Y = (s+3)/(s²+3s+2) = (s+3)/[(s+1)(s+2)]. Step 5: Partial fractions: (s+3)/[(s+1)(s+2)] = 2/(s+1) − 1/(s+2). Step 6: Invert term by term: L⁻¹{2/(s+1)} = 2e^(−t), L⁻¹{1/(s+2)} = e^(−2t). Result: y(t) = 2e^(−t) − e^(−2t). Check: y(0) = 2 − 1 = 1 ✓, and substitution confirms the ODE.