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- StudyBlue
- Alabama
- University of Alabama - Tuscaloosa
- Mathematics
- Mathematics 238
- Zhao
- test 3 Sample Questions

Drew A.

Test 3 Sample Questions Questions: 1. Use the elimination method to find a general solution for the given linear system. (a) dxdt = x − 4y, dydt = x + y; (b) dxdt = 3x− 2y + sint, dydt = 4x− y − cost. 2. Find the Laplace transform of the given functions. (a) f(t) = t3 − tet + e4t cost; (b) f(t) = tcos(bt); (c) f(t) = t − u(t − 1)(t − 1); (d) f(t) = etu(t− 2). 3. Find the inverse Laplace transform of the given functions. (a) F(s) = s + 1s2 + 2s + 10; (b) F(s) = 5s 2 + 34s + 53 (s + 3)2(s + 1); (c) F(s) = 7s2 + 23s + 30 (s − 2)(s2 + 2s + 5); (d) F(s) = 2(s − 1)e −2s s2 − 2s + 2 . 4. Use the Laplace transform to solve the given initial value problem: yprimeprime + 6yprime + 9y = 0; y(0) = −1, yprime(0) = 6 5. Express the given function using step functions and compute its Laplace transform: g(t) = 0, 0 < t < 1 t, 1 < t < 2 1, 2 < t 6. Use the Laplace transform to solve the given initial value problem: yprimeprime + y = t− (t− 4)u(t− 2); y(0) = 0, yprime(0) = 1 7. Use the Laplace transform to solve the given initial value problem: yprimeprime + y = −δ(t −pi) + δ(t− 2pi); y(0) = 0, yprime(0) = 1 Answers: 1. (a) First solve x = yprime −y. Plugging back, we have yprimeprime − 2yprime + 5y = 0. y = C1et cos2t + C2et sin2t. x = 2C2et cos2t− 2C1et sin2t. (b) First solve y = 32x−12xprime+12 sint. Plugging back, we have xprimeprime−2xprime+5x = sint+3cost. x = C1et cos2t + C2et sin2t − 110 sint + 710 cost. y = (C1 − C2)et cos2t + (C1 + C2)et sin2t + 710 sint + 1110 cost. 2. (a) F(s) = 6s4 − 1(s − 1)2 + s − 4(s − 4)2 + 1. (b) F(s) = (−1)· dds parenleftbigg s s2 + b2 parenrightbigg = s 2 −b2 (s2 + b2)2. (c) F(s) = 1s2 −e−s 1s2. (d) F(s) = e2 1s − 1e−2s = e2−2s 1s − 1. 3. (a) f(t) = L−1{ s + 1(s + 1)2 + 32} = e−t cos3t. (b) 5s 2 + 34s + 53 (s + 3)2(s + 1) = A S + 3 + A (S + 3)2 + C S + 1. Then A = −1, B = 2, C = 6. f(t) = −e−3t + 2te−3t + 6e−t. (c) F(s) = 7s 2 + 23s + 30 (s − 2)(s2 + 2s + 5) = A s − 2 + B(s + 1) + 2C (S + 1)2 + 22 . Then A = 8, B = −1, C = 3. f(t) = 8e2t − e−t cos(2t) + 3e−t sin(2t). (d) f(t) = 2et−2 cos(t − 2)u(t− 2) 4. Y = −s(s + 3)2. Partial fraction decomposition: −s(s + 3)2 = As + 3 + B(s + 3)2. Then A = −1 and B = 3. Finally y = −e−3t + 3te−3t. 5. g(t) = tu(t− 1) − (t − 1)u(t− 2). G(S) = (1s + 1s2)(e−s −e−2s) 6. y = t + [4 − t + sin(t − 2) − 2cos(t− 2)]u(t− 2) 7. Y = 1s2 + 1 − 1s2 + 1e−pis + 1s2 + 1e−2pis. y = sint−sin(t−pi)u(t−pi)+sin(t−2pi)u(t− 2pi) = sint + sint· u(t−pi) + sint· u(t− 2pi). sample3.dvi

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