By Reinhard Diestel
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Example text
For integral n. Thus, this function can be viewed as an interpolation of the factorial. Now we let f (t) = ta , where a > −1. It is then clear that f has a Laplace transform, and we find, for s > 0, ∞ f (s) = ta e−st dt 0 = ∞ 1 sa+1 0 st = u dt = du/s ∞ = 0 u a −u du e s s Γ(a + 1) ua e−u du = . 6. 6 Find the transform of f (t) = (e) e2t sin 3t 1/ε for 0 < t < ε, 0 otherwise. (t − 1)2 for t > 1, 0 otherwise. (f) t3 sin 3t. 8 Compute 47 1 − e−u du. u t e−3t sin t dt. 9 Find the Laplace transform of f , if we define f (t) = t sin t for 0 ≤ t ≤ π, f (t) = 0 otherwise.
18. Another example of the same type, though more complicated. The function f (x) = |x2 − 1| can be rewritten as f (x) = (x2 − 1)H(x − 1) + (1 − x2 )(H(x + 1) − H(x − 1)) +(x2 − 1)(1 − H(x + 1)) = (x2 − 1) 2H(x − 1) − 2H(x + 1) + 1 . This formula can be differentiated, using the rule for differentiating a product: f (x) = 2x 2H(x − 1) − 2H(x + 1) + 1 + (x2 − 1) 2δ(x − 1) − 2δ(x + 1) = 2x 2H(x − 1) − 2H(x + 1) + 1 . 6). One more differentiation gives f (x) = 2 2H(x − 1) − 2H(x + 1) + 1 + 2x 2δ(x − 1) − 2δ(x + 1) = 2 2H(x − 1) − 2H(x + 1) + 1 + 4δ(x − 1) + 4δ(x + 1).
10. Find g(t), if g(s) = 2s . (s2 + 1)2 Solution. We recognize the transform as a derivative: g(s) = − 1 d . 2 and the known transform of the sine we get g(t) = t sin t. 11. Solve the initial value problem y + 4y + 13y = 13, y(0) = y (0) = 0. Solution. Transformation gives (s2 + 4s + 13)y = 13 13 ⇐⇒ y = . s s (s + 2)2 + 9 50 3. Laplace and Z transforms Expand into partial fractions: y= s+4 1 s+2 1 − = − − s (s + 2)2 + 9 s (s + 2)2 + 9 2 3 · 3 . (s + 2)2 + 9 The solution is found to be y(t) = 1 − e−2t (cos 3t + 2 3 sin 3t) H(t).