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Solve ODE by Laplace transform y''+3y'+2y=e^(−t)

Short answer

Take Laplace transform, solve Y = 1/[(s+1)^2(s+2)], use partial fractions and invert. The solution is y(t) = (t-1)e^(-t) + e^(-2t).

Using Laplace transform, solve the initial value problem y'' + 3y' + 2y = e^(−t), y(0) = 0, y'(0) = 0.

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

s²Y + 3sY + 2Y = 1/(s + 1)
(s + 1)(s + 2) Y = 1/(s + 1)
Y = 1/[(s + 1)²(s + 2)]

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

B = 1/(s + 2) at s = −1  = 1
C = 1/(s + 1)² at s = −2 = 1
Compare s² terms: A + C = 0 → A = −1
Y = −1/(s + 1) + 1/(s + 1)² + 1/(s + 2)

Inverse transform (L⁻¹{1/(s + a)²} = t e^(−at)):

y(t) = −e^(−t) + t e^(−t) + e^(−2t)

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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