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

2024
Views:

301

Pages:

3
Author:

Kailyn Pham

Quiz 6 Solutions 1. Use the definition of the Laplace Transform to compute L(f (t)), where f (t) is given as: ( f (t) = t2 2+t if 0 ≤ t < 1 if t > 1 Be sure to include the details about the convergence of the improper integral (L’Hospital’s rule might come in handy). L(f (t)) = Z ∞ Z 1 e−st f (t) dt = 0 t2 e−st dt + Z ∞ 0 (2 + t)e−st dt 1 To integrate each of these, we need to integrate by parts: + t2 e−st − 2t − 1s e−st + 2 s12 e−st − 0 − s13 e−st + 2 + t e−st − 1 − 1s e−st 1 −st e + 0 s2 For the first integral, −e −st 2 t2 2t + 2+ 3 s s s !1 = −e−s 0 2 2 2 1 + 2+ 3 + 3 s s s s For the second integral, −e −st ∞ 2+t 1 + 2 s s = lim −e 1 −st t→∞ 2+t 1 + 2 s s −s +e 3 1 + 2 s s For this limit, we see that, by L’Hospital’s rule: lim −e t→∞ −st 2+t 1 + 2 s s = − lim t→∞ 2+t s + est 1 s2 = 1 lim s t→∞ sest = 0, s > 0 Overall, the solution is: −e −s 1 2 2 2 3 1 + 2 + 3 + 3 + e−s + 2 s s s s s s Note: You could check your answer by rewriting f (t) in terms of the Heaviside function, then use the table (not necessary, but shown here for completeness). If t ≥ 0, f (t) = (1 − u(t − 1))t2 + u(t − 1)(2 + t) = t2 − u1 (t)t2 + u1 (t)(2 + t) For the term: u1 (t)t2 ⇒ f (t − 1) = t2 ⇒ f (t) = (t + 1)2 = t2 + 2t + 1 So the Laplace transform is: 2 L(u1 (t)t ) = e 1 −s 2 2 1 + 2+ 3 s s s For the third term, ⇒ u1 (t)(2 + t) f (t − 1) = 2 + t ⇒ f (t) = 2 + (t + 1) = 3 + t so the Laplace transform is: L(u1 (t)(2 + t) = e −s 3 1 + 2 s s 2. Problem 14, page 313. You may either use the formula given in the text, or the formula: e(a+ib)t = eat cos(bt) + ieat sin(bt) and you may use Table Entry 2. Of course, you may NOT use Table Entry 10. Given Euler’s formula, e(a+ib)t = eat cos(bt) + ieat sin(bt) The desired Laplace transform can be found via: L(e(a+ib)t ) = L(eat cos(bt)) + iL(eat sin(bt)) Using Table Entry 2, 1 = L(eat cos(bt)) + iL(eat sin(bt)) s − (a + ib) 1 = L(eat cos(bt)) + iL(eat sin(bt)) (s − a) − ib (s − a) + i b = L(eat cos(bt)) + iL(eat sin(bt)) (s − a)2 + b2 b (s − a) +i = L(eat cos(bt)) + iL(eat sin(bt)) 2 2 (s − a) + b (s − a)2 + b2 Therefore, the Laplace transform of eat cos(bt) is (s − a) (s − a)2 + b2 s s2 + 2s − 3 We first perform Partial Fraction Decomposition: 3. Find, using the table: L−1 s2 A B 3 1 1 1 s + = + = + 2s − 3 s+3 s−1 4 s+3 4 s−1 so that the Laplace transform is: 3 −3t 1 t e + e 4 4 2 4. Find an expression for Y (s), do not solve for y(t): y 00 − 4y 0 + 4y = et cos(t), y(0) = 0, y 0 (0) = 1 Taking the Laplace transform of both sides, (s2 Y − 0 − 1) − 4(sY − 0) + 4Y = Y (s) = ((s − 1)2 s−1 (s − 1)2 + 1 1 s−1 + 2 2 + 1)(s − 4s + 4) s − 4s + 4 5. Solve the following IVP using the method of Laplace Transforms: y 00 + 3y 0 + 2y = e−t s2 Y − s + 3(sY − 1) + 2Y = 1 s+1 y(0) = 1 y 0 (0) = 0 ⇒ (s2 + 3s + 2)Y = 1 +s+3 s+1 Therefore, Y = (s + 1 s+3 + 2 + 3s + 2) s + 3s + 2 1)(s2 We could keep these separate, or combine them (the following solution will combine them). Factor the denominator: Y = 1 s+3 s2 + 4s + 4 s+2 + = = 2 2 (s + 1) (s + 2) (s + 1)(s + 2) (s + 1) (s + 2) (s + 1)2 Use partial fractions, or see that: Y = 1 1 s+1+1 = + 2 (s + 1) s + 1 (s + 1)2 so that y(t) = e−t + te−t . 3

Differential Equations Quiz 6 Solutions

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