Exam 3: Solutions for Laplace Transforms and Partial Fraction Expansion - Prof. William Ad, Exams of Mathematics

Download Exam 3: Solutions for Laplace Transforms and Partial Fraction Expansion - Prof. William Ad and more Exams Mathematics in PDF only on Docsity! Name: Exam 3 Instructions. Answer each of the questions on your own paper. Put your name on each page of your paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. A table of Laplace transforms has been appended to the exam. The following trigonometric identities may also be of use: sin(θ + ϕ) = sin θ cos ϕ + sin ϕ cos θ cos(θ + ϕ) = cos θ cos ϕ− sin θ sin ϕ 1. Solve: [12 Points] t2y′′ + 7ty′ + 9y = 0. 2. Solve: [16 Points] y′′ + 4y′ + 3y = 9t. 3. [20 Points] You may assume that S = {e−2t, te−2t} is a fundamental set of solutions for the homogeneous equation y′′ + 4y′ + 4y = 0. Use variation of parameters to find a particular solution of the nonhomogeneous dif- ferential equation y′′ + 4y′ + 4y = t−3e−2t. 4. [14 Points] Find the Laplace transform of the following function: f(t) = { t2 − 4t if 0 ≤ t < 4, 0 if t ≥ 4. 5. [16 Points] Find the inverse Laplace transform of the following functions: (a) F (s) = 1 (s + 1)3 e−2s + 2 s4 e−4s (b) G(s) = 2 s2 + 9 e−2πs 6. [22 Points] Solve the following initial value problem: y′′ + 16y = h(t− π)− h(t− 3π), y(0) = 1, y′(0) = 0. Math 2065 Section 1 November 24, 2008 1 Name: Exam 3 Exam III Supplementary Sheets A Short Table of Laplace Transforms 1. L{af(t) + bg(t)} (s) = aF (s) + bG(s) 2. L{eatf(t)} (s) = F (s− a) 3. L{f(t− c)h(t− c)} = e−scF (s) 3’. L{g(t)h(t− c)} = e−scL{g(t + c)} 4. L{−tf(t)} (s) = d ds F (s) 5. L{1} (s) = 1 s 6. L{tn} (s) = n! sn+1 7. L{eat} (s) = 1 s− a 8. L{tneαt} (s) = n! (s− α)n+1 9. L{cos bt} (s) = s s2 + b2 10. L{sin bt} (s) = b s2 + b2 11. L{eat cos bt} (s) = s− a (s− a)2 + b2 12. L{eat sin bt} (s) = b (s− a)2 + b2 13. L{f ′(t)} (s) = sF (s)− f(0) 14. L{f ′′(t)} (s) = s2F (s)− sf(0)− f ′(0) 15. L {∫ t 0 f(x) dx } (s) = F (s) s 16. L{(f ∗ g)(t)} (s) = F (s)G(s) Math 2065 Section 1 November 24, 2008 2 Name: Solutions Exam 3 where B = −3s− 12 (s + 1)(s + 3) ∣∣∣∣ s=0 = −12 3 = −4. Since, p(s) = (s+1)(s+3) = s2 +4s+3 is the characteristic polynomial of the associated homogeneous equation, it follows that L−1 { p2(s) (s + 3)(s + 1) } is a particular solution of the homogeneous equation y′′ + 4y′ + 3y = 0 and hence can be included in the homogeneous part of the solution yh. Thus, for yp we get yp(t) = L−1 { 3 s2 + −4 s } = 3t− 4. Then, yg = yp + yh so yg(t) = 3t− 4 + c1e−t + c2e−3t. Alternate Solution: One may also use the alternative method of undetermined coef- ficients from Section 4.4. For this method, the characteristic polynomial of the associ- ated homogeneous equation y′′+4y′+3y = 0 is q(s) = s2 +4s+3 = (s+3)(s+1) and the Laplace transform of the right hand side is L{9t} = 9 s2 , which has denominator v(s) = s2. Thus, the polynomial q(s)v(s) = s2(s+3)(s+1) and the corresponding basis for the homogeneous equation with polynomial q(s)v(s) is Bq(s)v(s) = {e−t, e−3t, 1, t} and the basis Bq(s) = {e−t, e−3t}. Thus Bq(s)v(s) \ Bq(s) = {1, t}. It follows that a particular solution yp = A + Bt for some constants A and B to be determined by substitution into the original equation. Computing y′p = B and y ′′ p = 0 it follows that y′′p + 4y ′ p + 3yp = 0 + 4B + 3(A + Bt) = 3Bt + (4B + 3A) = 9t. Hence, by equating the constant terms and the coefficients of t on the two sides of the last equality, we get 3B = 9 and 4B + 3A = 0. Hence, B = 3 and substituting in the second equation gives A = −4. Thus yp = 3t − 4, and as with the first method, yg = yp + yh so yg(t) = 3t− 4 + c1e−t + c2e−3t. J 3. [20 Points] You may assume that S = {e−2t, te−2t} is a fundamental set of solutions for the homogeneous equation y′′ + 4y′ + 4y = 0. Use variation of parameters to find a particular solution of the nonhomogeneous dif- ferential equation y′′ + 4y′ + 4y = t−3e−2t. Math 2065 Section 1 November 24, 2008 2 Name: Solutions Exam 3 I Solution. The particular solution yp(t) has the form yp = u1y1 + u2y2 = u1e −2t + u2te−2t where u1 and u2 are unknown functions whose derivatives satisfy the following equa- tions: u′1e −2t + u′2te −2t = 0 u′1(−2e−2t) + u′2(e−2t − 2te−2t) = t−3e−2t. Adding 2 times the first equation to the second shows that u′2 = t −3, and the first equation then gives u′1 = −tu′2 = −t · t−3 = −t−2. Hence u1 = ∫ u′1 dt = − ∫ t−2 dt = t−1 u2 = ∫ u′2 dt = ∫ t−3 dt = −1 2 t−2, so that yp = t −1e−2t − 1 2 t−2te−2t = 1 2 t−1e−2t. J 4. [14 Points] Find the Laplace transform of the following function: f(t) = { t2 − 4t if 0 ≤ t < 4, 0 if t ≥ 4. I Solution. First write f(t) in terms of characteristic functions χ[a,b)(t) and unit step functions h(t− c): f(t) = (t2 − 4t)χ[0,4)(t) + (0)χ[2,∞)(t) = (t2 − 4t)(h(t)− h(t− 4)) = (t2 − 4t)− (t2 − 4t)h(t− 4). Now apply the second translation formula (number 3 or 3’ on the table): F (s) = 2 s3 − 4 s − e−4sL{(t + 4)2 − 4(t + 4)} = 2 s3 − 4 s − e−4sL{(t2 + 8t + 16− 4t− 16} = 2 s3 − 4 s − e−4sL{(t2 + 4t} = 2 s3 − 4 s − e−4s ( 2 s3 + 4 s ) . J Math 2065 Section 1 November 24, 2008 3 Name: Solutions Exam 3 5. [16 Points] Find the inverse Laplace transform of the following functions: (a) F (s) = 1 (s + 1)3 e−2s + 2 s4 e−4s I Solution. Letting F1(s) = 1 (s + 1)3 we find that f1(t) = L−1 {F1(s)} = L−1 { 1 (s + 1)3 } = 1 2 t2e−t, and letting F2(s) = 2 s4 we find that f2(t) = L−1 { 2 s4 } = 1 3 t3. Then the second translation formula, Formula 3, applied twice, gives: f(t) = L−1 {F (s)} = L−1 {F1(s)e−2s } + L−1 {F2(s)e−4s } = f1(t− 2)h(t− 2) + f2(t− 4)h(t− 4) = 1 2 (t− 2)2e−(t−2)h(t− 2) + 1 3 (t− 4)3h(t− 4). J (b) G(s) = 2 s2 + 9 e−2πs I Solution. Letting G1(s) = 2 s2 + 9 we find that g1(t) = L−1 {G1(s)} = L−1 { 2 s2 + 9 } = 2 3 sin 3t. Then the second translation formula, Formula 3, gives: g(t) = L−1 {G(s)} = L−1 {G1(s)e−2πs } = g1(t−2π)h(t−2π) = (2 3 sin 3(t−2π))h(t−2π). J 6. [22 Points] Solve the following initial value problem: y′′ + 16y = h(t− π)− h(t− 3π), y(0) = 1, y′(0) = 0. I Solution. Let Y (s) = L{y(t)} be the Laplace transform of the solution function. Applying the Laplace transform to the differential equation gives s2Y (s)− s + 16Y (s) = L{h(t− π)− h(t− 3π)} = e πss − e−3πs s . Thus, Y (s) = s s2 + 16 + e−πs ( 1 s(s2 + 16) ) − e−3πs ( 1 s(s2 + 16) ) . Math 2065 Section 1 November 24, 2008 4