Solve ODE by Laplace transform y''+4y=sint

Short answer

Apply Laplace transform, use partial fractions to write Y = 1/3 [1/(s^2+1) - 1/(s^2+4)], then invert to get y(t) = (1/3) sin t - (1/6) sin 2t.

Using Laplace transform, solve: y'' + 4y = sin t; y(0) = 0 = y'(0).

y″ + 4y = sin t, y(0) = y′(0) = 0.

Step 1: Transform. With Y = L[y]:

s²Y + 4Y = 1/(s² + 1)
Y = 1/[(s² + 1)(s² + 4)]

Step 2: Partial fractions. Since (s² + 4) − (s² + 1) = 3:

Y = (1/3) [ 1/(s² + 1) − 1/(s² + 4) ]

Step 3: Invert. L⁻¹[1/(s² + 4)] = (1/2) sin 2t.

Answer: y(t) = (1/3) sin t − (1/6) sin 2t

Check: y(0) = 0; y′(0) = 1/3 − 1/3 = 0; y″ + 4y = −(1/3) sin t + (2/3) sin 2t + (4/3) sin t − (2/3) sin 2t = sin t ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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