University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2020
Views:

234

Pages:

2
Author:

ilhisry3

SOLUTIONS to the exercises in the complex Laplace handout 1. Use Euler’s Formula to show that, if a, b are constants, and s is a parameters, then lim e−(s−(a+ib))t = 0 t→∞ SOLUTION: First write out the expression. e−(s−(a+ib))t = e−(s−a)t e−i(bt) Since e−ibt is on the unit circle, its magnitude is 1. Therefore, the magnitude of the exponential function is simply: |e−(s−(a+ib))t | = e−(s−a)t and as long as s − a > 0, the magnitude will go to zero as t → ∞, which means the original function must also go to zero. 2. Use complex exponentials to compute R −2t e sin(3t) dt. R (−2+3i)t SOLUTION: We’ll find the imaginary part of Z e(−2+3i)t dt = e dt: 1 −2 − 3i e(−2+3i)t = e−2t (cos(3t) + i sin(3t)) · = −2 + 3i 4+9 e−2t ((−2 cos(3t) + 3 sin(3t)) + i(−3 cos(3t) − 2 sin(3t)) 13 The imaginary part gives us the answer: Z e −2t e−2t (−3 cos(2t) − 2 sin(3t)) sin(3t) dt = 13 3. Use complex exponentials to compute the Laplace transform of sin(at). SOLUTION: Use cos(at) + i sin(at) = eiat . Then we’ll compute the transform of eiat using the table and pull off the imaginary part. L(eiat ) = Therefore, the imaginary part is: s + ia 1 = 2 s − ai s + a2 a s2 +a2 4. Use complex exponentials to compute the Laplace transform of eat sin(bt) and eat cos(bt). SOLUTION: Use the fact that e(a+ib)t gives us both expressions, and L(e(a+ib)t = 1 1 (s − a) + ib = = s − (a + ib) (s − a) − ib (s − a)2 + b2 Therefore, we get both transforms: L(eat cos(bt)) = s−a (s − a)2 + b2 1 L(eat sin(bt)) = b (s − a)2 + b2 5. Prove that et goes to infinity faster than any polynomial by considering the limit below. SOLUTION: Use l’Hospital’s rule n times to get: tn ntn−1 n(n − 1)tn−2 n! lim = lim = lim = · · · = lim t = 0 t t t→∞ et t→∞ t→∞ t→∞ e e e 6. Use the Racetrack Principle to show that, for t ≥ 1, ln(t) < t. SOLUTION: Let f (t) = ln(t) and g(t) = t. Then: • f (1) = 0 and g(1) = 1, so f (1) < g(1). • f 0 (x) = 1/x and g 0 (x) = 1, so for x > 1, f 0 (x) < g 0 (x). Therefore, by the Racetrack Principle, ln(t) < t for all t ≥ 1. 7. If f (t) is bounded by a number A, then |f (t)| ≤ A = Keat , t ≥ M. So f is of exponential order with K = A, a = 0 and M anything. 8. Exponential order practice: (a) sin(t): Since | sin(t)| ≤ 1, we can do the same thing as in the previous exercise with A = 1. (b) tan(t) has a vertical asymptote at t = π/2 (and multiples of π thereafter), so it is NOT of exponential order. 3 (c) t3 = eln(t ) = e3 ln(t) ≤ e3t , so this is of exponential order. 2 (d) et is not of exponential order, since the exponent is of order 2. t (e) 5t = eln(5 ) = et ln(5) (f) tt is not of exponential order: t tt = eln(t ) = et ln(t) 2

Solutions to the exercises in the complex Laplace handout

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