Differential Equations 6.3 Homework Notes

Section 6.3 Homework Notes 2. To sketch the graph, try first rewriting what is given as a piecewise defined function. The function: (t − 3)u2 (t) − (t − 2)u3 (t) depends on the value of t: Function Interval 0≤t<2 0 t−3 2≤t<3 t≥3 (t − 3) − (t − 2) = −1 Now it’s easy to graph. 3. Shift the right half of the graph of t2 to the right π units. 6. Break it up:      0 (t − 1) (t − 1)u1 (t) − 2(t − 2)u2 (t) + (t − 3)u3 (t) =  (t − 1) − 2(t − 2)    (t − 1) − 2(t − 2) + (t − 3) if if if if t<1 1≤t<2 2≤t<3 t≥3 Rewriting this, we get:      0 t−1 =  −t + 3    0 if if if if t<0 1≤t<2 2≤t<3 t≥3 8. We could do this one piecemeal: • Start things off at 1. • Next, we need to go to -1 (2 units down), so take 1 − 2u1 (t) • Next, get back to 1 (2 units up) at t = 2: 1 − 2u1 (t) + 2u2 (t) • At time 3, back to −1 (2 units down): 1 − 2u1 (t) + 2u2 (t) − 2u3 (t) • At time 4, back to zero (1 unit up): 1 − 2u1 (t) + 2u2 (t) + 2u3 (t) + u4 (t) Alternative: Use the on-off switch and simplify: (1 − u1 (t)) − (u1 − u2 (t)) + (u2 (t) − u3 (t)) − (u3 (t) − u4 (t)) 11. In this case, best to use the on-off switch and simplify: t(1 − u1 (t)) + (t − 1)(u1 (t) − u2 (t)) + (t − 2)(u2 (t) − u3 (t)) = t − u1 (t) − u2 (t) − (t − 2)u3 (t) 1 14. To use the table, we need that: u1 (t)f (t − 1) = u1 (t)(t2 − 2t + 2) so that f (t − 1) = t2 − 2t + 2. This means that f (t) = (t + 1)2 − 2(t + 1) + 2 = t2 + 2t + 1 − 2t − 2 + 2 = t2 + 1 Now the table entry says: f (t) F (s) uc (t)f (t − c) e−sc F (s) so we see that F (s) = L(t2 + 1) = so that our final answer is: e−s  2 s3 + 1 s 2 1 + s3 s  15. Same idea as 14: f (t) = (t − π)(uπ (t) − u2π (t)) = (t − π)uπ (t) − (t − π)u2π (t) Inverting term by term, the first is: uπ (t)(t − π) = uπ (t)g(t − π) ⇒ g(t) = t ⇒ e−πs s2 and for the second term, think of u2π (t)(t − π) = uπ (t)g(t − 2π) ⇒ g(t) = t + π ⇒ e−2πs πe−2πs + s2 s 19. Use Table Entry 11. 20. Find the inverse Laplace transform of F (s) = e−2s · s2 1 +s−2 This is of the form e−sc F (s), so we need to find the inverse transform of 1/(s2 + s − 2). We’ll do this by Partial Fraction Decomposition. This F refers to the table, not the original F : F (s) = s2 1 A B 1 1 1 1 = + =− + +s−2 s+2 s−1 3 s+2 3 s−1 and 1 1 f (t) = − e−2t + et 3 3 So our overall answer is u2 (t)f (t − 2), or: 1 1 u2 (t) − e−2(t−2) + et−2 3 3  2  21. Find the inverse Laplace transform of G(s) = 2e−2s (s − 1) s2 − 2s + 2 (I changed the notation of the original function so as not to confuse F (s) in the table with F (s) in the original question). We will rewrite this expression, keeping the table in mind: G(s) = 2e−2s s2 s−1 s−1 = 2e−2s = 2e−sc F (s) − 2s + 2 (s − 1)2 + 1 We see that, given this F (s), then f (t) = et cos(t) and our overall inverse Laplace transform is: 2u2 (t)f (t − 2) = 2u2 (t)et−2 cos(t − 2) 25. Good practice to work from the definition, and also to look at where we want to go: We want to show that L(f (ct)) = Z ∞ −st e 0 1 Z ∞ −s/c f (ct) dt = e f (w) dw c 0 So, it looks like we want to do a substitution: w = ct, so dw = c dt and dt = 1c dw. For part (b), use part (a)f (ct) = L−1  1 F (s/c) c  ⇒ cf (ct) = L−1 (F (s/c)) Then let c = k1 . For part (c), use these ideas: • From part (b), L−1 (F (as)) = a1 f (t/a) • Note that as + b = a(s + b/a) • Use the table to see that L(ect f (t)) = F (s − c) 27. We want to try to use the result of Exercise 25 to solve this one, so put it in the form F (s) = G(as + b). F (s) = where G(s) = 2s + 1 2s + 1 2s + 1 = = = G(2s + 1) 2 + 4s + 5 (4s + 4s + 1) + 4 (2s + 1)2 + 4 4s2 s s2 +4 and so g(t) = cos(2t). Using the results of 25, with a = 2 and b = 1, 1 1 L−1 (G(2s + 1)) = e−t/2 cos(2t/2) = e−t/2 cos(t) 2 2 3 28. Similar to 27- Put this in the form G(as+b). We might note that (3s−2)2 = 9s2 −12s+4. Therefore, 1 1 = = G(3s − 2) 2 9s − 12s + 3 (3s − 2)2 − 1 so that G(s) = 1/(s2 − 1) and g(t) = sinh(t) (You can use the hyperbolic functions from the table). Therefore, with a = 3 and b = −2, 1 L−1 (G(3s − 2)) = e2t/3 sinh(t/3) 3 32. For this problem, we’ll need to have a sum for a finite geometric series. Here’s one way to get that formula: S = 1 + r + r2 + · · · + rm rS = r + r2 + · · · + rm+1 (1 − r)S = 1 − rm+1 ⇒ S= 1 − rm+1 1−r Now, back to the text. Note that, if t > 0, then u0 (t) = 1. The Laplace transform is a linear operator, so: L 2n+1 X ! k (−1) uk (t) = k=0 2n+1 X 2n+1 X k=0 k=0 (−1)k L(uk (t)) = (−1)k k X  e−ks 1 2n+1 = −e−s s s k=0 This is a geometric series with r = −e−s , so now apply the formula we just derived for the sum of a geo series to get: 1 1 − (−e−s )2n+2 · s 1 + e−s 33. Pretty much the same idea as 32, except we take the infinite sum: L ∞ X ! k (−1) uk (t) = ∞ X k (−1) L(uk (t)) = (−1)k k=0 k=0 k=0 ∞ X e−ks s −s This reminds us of a geometric series with r = −e , except now it is an infinite sum. 35. Use the results of Problem 34- If f is periodic with period T , then L(f (t)) = R T −st e f (t) dt 0 1 − e−sT In this case, T = 2, and Z 2 e−st f (t) dt = 0 Z 1 0 1  1 e−s 1 1 e−st dt = − e−st = − + = 1 − e−s s s s s 0 Put this into the formula: − e−s ) 1 − e−2s This may not seem to be the same answer as in Problem 33; However note that: L(f (t)) =  1 − e−2s = 12 − e−s 4 2 1 (1 s  = 1 − e−s  1 + e−s