Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.9 The Convolution Integral - Problems - Page 716: 30

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Given: $y''-a^2y=f(t)$ Since $y(0)=\alpha\\ y'(0)=\beta$ obtain: $L[y]=\frac{F(s)}{s^2-a^2}+\frac{s\alpha}{s^2-a^2}+\frac{\beta}{s^2-a^2}$ Using the Convolution Theorem $y(t)=L^{-1}[\frac{F(s)}{s^2-a^2}]+L^{-1}[\frac{s\alpha}{s^2-a^2}]+L^{-1}[\frac{\beta}{s^2-a^2}]\\ =\frac{1}{A}\beta\sin(at)+\alpha \cos (at)+\int^t_0 \sin(a\tau)f(t-\tau)d \tau$