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- 2080 Bhadra · 5 marks
Solve ODE by Laplace transform y''+y=sin3t
Short answer
y(t) = (3/8) sin t - (1/8) sin 3t.
Solve the initial value problem by the Laplace transform method: y'' + y = sin 3t; y(0) = 0, y'(0) = 0.
Transform y'' + y = sin 3t with y(0) = y'(0) = 0. Let Y = L{y}.
s²Y + Y = 3/(s² + 9)
Y = 3/[(s² + 1)(s² + 9)]
Since 1/[(s² + 1)(s² + 9)] = (1/8)[1/(s² + 1) − 1/(s² + 9)]:
Y = (3/8) · 1/(s² + 1) − (3/8) · 1/(s² + 9)
y(t) = (3/8) sin t − (3/8)(sin 3t)/3
Answer: y(t) = (3/8) sin t − (1/8) sin 3t. Check: y(0) = 0; y'(0) = 3/8 − 3/8 = 0 ✓.
✓ Answer checked by Claude Opus 5.5 (high effort)
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