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1. (1 pt) Find the Laplace transform of the following functions: 1. f(t) = −1 √ t −6t F(s) = . 2. f(t) = 8t 3/2 −e −4t F(s) = . 3. f(t) = sin(5t) +cos(5t) F(s) = . 2. (1 pt) Find the Laplace transform of the following functions: 1. f(t) = sin(8t) cos(8t) F(s) = . 2. f(t) = cos2 (6t) F(s) = . 3. f(t) = (8−4t) 2 F(s) = . 1. (1 pt) Find the inverse Laplace transform of 4s+2 s 2 +23 s > 0 y(t) = . 2. (1 pt) Find the inverse Laplace transform of 7s+8 s 2 −4 s > 2 y(t) = . 3. (1 pt) Consider the following initial value problem: y 00 −6y 0 −7y = sin(2t) y(0) = 3, y 0 (0) = 2 Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s) = 4. (1 pt) Consider the following initial value problem: y 00 −4y 0 −32y = sin(9t) y(0) = 6, y 0 (0) = −2 Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s) = 5. (1 pt) Use the Laplace transform to solve the following initial value problem: y 00 +7y 0 −18y = 0 y(0) = 4, y 0 (0) = −3 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation = 0 Now solve for Y(s) = and write the above answer in its partial fraction decomposition, Y(s) = A s+a + B s+b where a < b Y(s) = + Now by inverting the transform, find y(t) = . 6. (1 pt) Use the Laplace transform to solve the following initial value problem: y 00 +2y 0 = 0 y(0) = 4, y 0 (0) = 8 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation = 0 Now solve for Y(s) = and write the above answer in its partial fraction decomposition, Y(s) = A s+a + B s+b where a < b Y(s) = + Now by inverting the transform, find y(t) = . 7. (1 pt) Use the Laplace transform to solve the following initial value problem: y 00 +16y = cos(7t) y(0) = 0, y 0 (0) = 0 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solving for Y: Y(s) = Find the partial fraction decomposition of Y(s) and its inverse Laplace transform to find the solution of the DE: y(t) = . 8. (1 pt) Use the Laplace transform to solve the following initial value problem: x 0 = 9x+4y, y 0 = −5x+e 4t x(0) = 0, y(0) = 0 Let X(s) = L{x(t)}, and Y(s) = L{y(t)}. Find the expressions you obtain by taking the Laplace transform of both differential equations and solving for Y(s) and X(s): X(s) = Y(s) = Find the partial fraction decomposition of X(s) and Y(s) and their inverse Laplace transforms to find the solution of the system of DEs: x(t) = y(t) = . 9. (1 pt) Find the Laplace transform of te6t sin(9t). L{te6t sin(9t)} = . 10. (1 pt) Find the Laplace transform of t 2 sin(6t). L{t 2 sin(6t)} = . 11. (1 pt) Given that L{ cos(3 √ t) √ πt } = e −2.25/s √ s find the Laplace transform of q t π cos(3 √ t). L{ q t π cos(3 √ t)} = . 1. (1 pt) Find the Laplace transform of f(t) = −5u2(t) +2u3(t) +1u6(t) F(s) = . 2. (1 pt) Find the inverse Laplace transform of F(s) = −2e −2s −1e −4s −1e −5s +5e −8s s f(t) = . (Use step(t-c) for uc(t) .) 3. (1 pt) Consider the function f(t) =    −6, 0 ≤ t < 1 3, 1 ≤ t < 5 4, t ≥ 5 ; 1. Write the function in terms of unit step function f(t) = . (Use step(t-c) for uc(t) .) 2. Find the Laplace transform of f(t) F(s) = . 4. (1 pt) Find the inverse Laplace transform of F(s) = e −9s s 2 +4s−5 f(t) = . (Use step(t-c) for uc(t) .) 5. (1 pt) Find the inverse Laplace transform of F(s) = 4e −6s s 2 +49 f(t) = . (Use step(t-c) for uc(t) .) 6. (1 pt) Find the Laplace transform of f(t) = ( 0, t < 2 (t −2) 2 , t ≥ 2 F(s) = . 7. (1 pt) Find the Laplace transform of f(t) = ( 0, t < 2 t 2 −4t +7, t ≥ 2 F(s) = . 1. (1 pt) Consider the following initial value problem: y 00 +64y = ( 6, 0 ≤ t ≤ 9 0, t > 9 y(0) = 8, y 0 (0) = 0 Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s) = 2. (1 pt) Take the Laplace transform of the following initial value problem and solve for Y(s) = L{y(t)}: y 00 −7y 0 −18y = ( 1, 0 ≤ t < 1 0, 1 ≤ t y(0) = 0, y 0 (0) = 0 Y(s) = . Now find the inverse transform to find y(t) = . (Use step(t-c) for uc(t) .) Note: 1 s(s−9)(s+2) = − 1 18 s + 1 22 s+2 + 1 99 s−9 3. (1 pt) Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: y 00 +1y = ( sin(πt), 0 ≤ t < 1 0, 1 ≤ t y(0) = 0, y 0 (0) = 0 Y(s) = . Hint: write the right hand side in terms of the Heaviside function. Now find the inverse transform to find y(t) = . (Use step(t-c) for uc(t) .) Note: π (s 2 +π 2)(s 2 +1) = π π 2 −1  1 s 2 +1 − 1 s 2 +π 2  4. (1 pt) Consider the following initial value problem: y 00 +9y = ( 3t, 0 ≤ t < 2 0, t ≥ 2 y(0) = 0, y 0 (0) = 0 Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s) = 5. (1 pt) Consider the following initial value problem: y 00 +25y = ( 4t, 0 ≤ t ≤ 7 28, t > 7 y(0) = 0, y 0 (0) = 0 Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s) = 6. (1 pt) Find the Laplace transform of f(t) =    0, t < 4 5 sin(πt), 4 ≤ t < 5 0, t ≥ 5 F(s) = . 7. (1 pt) Take the Laplace transform of the following initial value problem and solve for Y(s) = L{y(t)}: y 00 −5y 0 −24y = S(t) y(0) = 0, y 0 (0) = 0 Where S(t) = ( 1, 0 ≤ t < 1 0, 1 ≤ t < 2 , S(t +2) = S(t). Y(s) = . The graph of S(t) (a square wave function): 8. (1 pt) Take the Laplace transform of the following initial value problem and solve for Y(s) = L{y(t)}: y 00 +12y 0 +17y = T(t) y(0) = 0, y 0 (0) = 0 Where T(t) = ( t, 0 ≤ t < 1/2 1−t, 1/2 ≤ t < 1 , T(t +1) = T(t). Y(s) = . Graph of T(t) (a triangular wave function): 9. (1 pt) Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: y 00 +4y = R(t) y(0) = 0, y 0 (0) = 0 Where R(t) = sin(πt), R(t +1) = R(t). Y(s) = . Graph of R(t) (a rectified sine wave function): 1. (1 pt) Use the Laplace transform to solve the following initial value problem: y 00 −6y 0 −7y = δ(t −8) y(0) = 0, y 0 (0) = 0 Use step(t-c) for uc(t). y(t) = . y 00 +9y = 7δ(t −3) y(0) = 0, y 0 (0) = 0 Use step(t-c) for uc(t). y(t) = . 3. (1 pt) Use the Laplace transform to solve the following initial value problem: y 00 +12y 0 +45y = δ(t −7) y(0) = 0, y 0 (0) = 0 Use step(t-c) for uc(t). y(t) = . 2. (1 pt) Use the Laplace transform to solve the following initial value problem:

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Differential Equations
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