Skip to main content

Solve ODE by Laplace transform

Short answer

Transform, solve for Y(s), partial fractions yield Y(s) = (7/4)/(s+1) + (1/2)/(s+1)² − (3/4)/(s+3). Inverse gives y(t) = (7/4 + t/2)e^(−t) − (3/4)e^(−3t).

Use the Laplace transform to solve d²y/dt² + 4 dy/dt + 3y = e^(−t), y(0) = y'(0) = 1.

d²y/dt² + 4 dy/dt + 3y = e^(−t), y(0) = y′(0) = 1.

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

(s²Y − s·1 − 1) + 4(sY − 1) + 3Y = 1/(s + 1)
(s² + 4s + 3) Y = 1/(s + 1) + s + 5
Y = 1/[(s+1)²(s+3)] + (s + 5)/[(s+1)(s+3)]

Step 2: Partial fractions.

1/[(s+1)²(s+3)] = −(1/4)/(s+1) + (1/2)/(s+1)² + (1/4)/(s+3)
(s+5)/[(s+1)(s+3)] = 2/(s+1) − 1/(s+3)
Y = (7/4)/(s+1) + (1/2)/(s+1)² − (3/4)/(s+3)

Step 3: Invert (L⁻¹[1/(s+1)²] = t e^(−t)).

Answer: y(t) = (7/4) e^(−t) + (t/2) e^(−t) − (3/4) e^(−3t)

Check: y(0) = 1 ✓ and y′(0) = −7/4 + 1/2 + 9/4 = 1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →