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Solve ODE by Laplace transform y''+2y=r(t)

Short answer

Write r(t) = 1 - u(t-1). Transform gives Y = (1 - e^(-s)) / (s(s²+2)). Partial fractions yield g(t) = ½(1 - cos√2 t). By second shift, y(t) = g(t) - g(t-1)u(t-1). For 0<t<1: y = ½(1 - cos√2 t); for t>1: y = ½[cos√2(t-1) - cos√2 t].

By using Laplace transform, solve the initial value problem y'' + 2y = r(t), y(0) = y'(0) = 0, where r(t) = 1 for 0 < t < 1 and r(t) = 0 otherwise.

The right side is a square pulse: r(t) = 1 − u(t − 1), where u is the unit step function.

Step 1: Transform both sides (y(0) = y′(0) = 0).

s²Y + 2Y = 1/s − e^(−s)/s = (1 − e^(−s))/s
Y = (1 − e^(−s)) / [ s(s² + 2) ]

Step 2: Partial fractions.

1/[s(s² + 2)] = (1/2) [ 1/s − s/(s² + 2) ]
L⁻¹ = g(t) = (1/2)(1 − cos √2 t)

Step 3: Second shifting theorem. L⁻¹[e^(−s) G(s)] = g(t − 1) u(t − 1). So

y(t) = g(t) − g(t − 1) u(t − 1)

Answer:

y(t) = ½ (1 − cos √2 t)                     for 0 < t < 1
y(t) = ½ [cos √2 (t − 1) − cos √2 t]        for t > 1

The solution and its first derivative are continuous at t = 1 (since y″ has only a jump there).

✓ Answer checked by Claude Opus 5.5 (high effort)

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