Posted: January 10th, 2017

Show that if f(t) is piecewise continuous on [0, ∞) and of exponential order α, i.e. |f(t)| ≤ Meαt , for t ≥ T for some M, T > 0, then its Laplcace transform L{f(t)}(s) exists for s > α

Show that if f(t) is piecewise continuous on [0, ∞) and of exponential order α, i.e. |f(t)| ≤ Meαt , for t ≥ T for some M, T > 0, then its Laplcace transform L{f(t)}(s) exists for s > α.

[3]

(ii) Solve the following initial value problem of the form y ′′(t) + 3y ′ (t) + 2y(t) = h(t), y(0) = y ′ (0) = 0, where h(t) = { 1, if 0 ≤ t ≤ π, 0, if t > π. [5] (iii) Let f(t) and g(t) be piecewise continuous and of exponential order α functions on [0, ∞). Then the convolution of f(t) and g(t) denoted by f ∗ g is defined by (f ∗ g)(t) := ∫ f 0 (t − s) g(s) ds. Let also F(s) = L{f(t)}(s) and G(s) = L{g(t)}(s) be the Laplace transforms of f and g respectively. Then it holds that L {(f ∗ g)(t)} (s) = F(s) G(s), (1) L −1 {F(s)G(s)} (t) = (f ∗ g)(t). (2) (a) Use property (1) to solve the integro-differential equation y ′ (t) = 1 − ∫ t 0 y(t − s) e −2s ds, y(0) = 1. [6] (b) Next use property (2) to solve the initial value problem y ′′(t) − y(t) = g(t), y(0) = y ′ (0) = 1, where g(t) is piecewise continuous on [0, ∞) and of exponential order.