a. First, compute the Laplace transform of f(t). -2s -6s L{f(t)}(s) = b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L { -2s -6s -e -6s – 58 L {y(t)}(s) = + s(s? + 10s+ 29) s2 + 10s + 29 c. We will need to take the inverse Laplace transform in order to find y(t). To do so, let's first rewrite L {y(t)} as L {u(?}{(+) = ( - F(s) 1 -28 e e - F(s) - 6F(s) + 2G(s) 1 -6s 29 where s+10 F(s) = and G(s) = s2 + 10s + 29 s2 + 10s + 29 d. Part c. indicates we will need L-1{F(s)} and L-1{G(s)}. Let's go ahead and find those now. L-'{F(s)}(t) = and L-1{G(s)}(t) e. Use your answer in Part d. to compute L-1{e=2*F(s)} and L-1 {e-6*F(s)}. L-'{e-»F(s)}(t) = and C-1{e-6*F(s)}(t) = f. Finally, combine all the previous steps to write down y(t).

a. First, compute the Laplace transform of f(t). -2s -6s L{f(t)}(s) = b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L { -2s -6s -e -6s – 58 L {y(t)}(s) = + s(s? + 10s+ 29) s2 + 10s + 29 c. We will need to take the inverse Laplace transform in order to find y(t). To do so, let's first rewrite L {y(t)} as L {u(?}{(+) = ( - F(s) 1 -28 e e - F(s) - 6F(s) + 2G(s) 1 -6s 29 where s+10 F(s) = and G(s) = s2 + 10s + 29 s2 + 10s + 29 d. Part c. indicates we will need L-1{F(s)} and L-1{G(s)}. Let's go ahead and find those now. L-'{F(s)}(t) = and L-1{G(s)}(t) e. Use your answer in Part d. to compute L-1{e=2F(s)} and L-1 {e-6F(s)}. L-'{e-»F(s)}(t) = and C-1{e-6*F(s)}(t) = f. Finally, combine all the previous steps to write down y(t).

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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