# Using partial fractions it follows that based on

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Unformatted text preview: __________________________________________________________ page 280 —————————————————————————— CHAPTER 6. —— . 31. 37 The function is periodic, with . Using the result of Prob. , . Using the result of Prob. , It follows that 32. 38 The function is periodic, with We first calculate Hence 33 39 . . ________________________________________________________________________ page 281 —————————————————————————— CHAPTER 6. —— . Let . Then . Let . Then . ________________________________________________________________________ page 282 —————————————————————————— CHAPTER 6. —— 34 . . The given function is periodic, with Based on the piecewise definition of . Using the result of Prob. , , Hence . Since satisfies the hypotheses of Theorem , . Using the result of Prob. , . We note the , hence . ________________________________________________________________________ page 283 —————————————————————————— CHAPTER 6. —— Section 6.4 2. Let be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain . Applying the initial conditions, . The forcing function can be written as Its transform is . Solving for , the transform of the solution is . First note that . Using partial fractions, Taking the inverse transform, term-by-term, . Now let . Then . Using Theorem , . Hence the solution of the IVP is ________________________________________________________________________ page 284 —————————————————————————— CHAPTER 6. —— . That is, . The solution starts out as free oscillation, due to the initial conditions. The amplitude increases, as long as the forcing is present. Thereafter, the solution rapidly decays. 4. Let be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain . Applying the initial conditions, ________________________________________________________________________ page 285 —————————————————————————— CHAPTER 6. —— . The transform of the forcing function is . Solving for , the transform of the solution is . Using partial fractions, . It follows that . Based on Theorem , . Hence the solution of the IVP is . ________________________________________________________________________ page 286 ————————————————————————— CHAPTER 6. —— Since there is no damping term, the solution follows the forcing function, after which the response is a steady oscillation about . 5. Let be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain . Applying the initial conditions, . The transform of the forcing function is . Solving for the transform, . Using partial fractions, Hence . Based on Theorem , . Hence the solution of the IVP is ________________________________________________________________________ page 287 —————————————————————————— CHAPTER 6. —— . The solution increases to a temporary steady value of the response decays exponentially to . . After the forcin...
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## This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue.

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