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Case 2: w – c² = 0. Show that this is the critically damped case of Section 4.4. Compute the inverse Laplace transform of Y (s) to show that the solution in this case is given by y(t) = yoe¬t + (vo + byo)te¬e".

Case 2: w – c² = 0. Show that this is the critically damped case of Section 4.4. Compute the inverse Laplace transform of Y (s) to show that the solution in this case is given by y(t) = yoe¬t + (vo + byo)te¬e".

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please answer #40. Thank you.

38. When there is damping, but still no forcing term, the initial value problem becomes y" + 2cy' + wy = 0, y(0) = yo, y'(0) = vo, (4.17) where yo and vo are the initial displacement and velocity of the mass, respectively. Show that the Laplace transform of the solution can be written Yos + vo + 2cyo Y (s) (4.18) (s + c)² + (w – c²)' A moment's reflection will reveal that taking the inverse transform of Y (s) will depend upon the sign of w – c². In Exercises 39–41, we will examine three cases. Notice that each has a counterpart in Section 4.4. 39. Case 1: w – c² > 0. Show that this is the underdamped case of Section 4.4. Compute the inverse Laplace trans- form of Y (s) to show that the solution in this case is given by y(t) = yoe -ct cos Yoc + vo + sin 40. Case 2: w – c2 damped case of Section 4.4. Compute the inverse Laplace transform of Y (s) to show that the solution in this case is given by 0. Show that this is the critically y(t) = yoe¬et + (vo + byo)te¬ct. 41. Case 3: w – c² < 0. Show that this is the overdamped case of Section 4.4. To simplify calculations, let's set the intitial displacement as y(0) = yo = 0. Show that with this assumption the transform in (4.18) becomes vo Y (s) = (s + c)² – (c² – w;) - (4.19) vo (++c+ Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).
Transcribed Image Text:38. When there is damping, but still no forcing term, the initial value problem becomes y" + 2cy' + wy = 0, y(0) = yo, y'(0) = vo, (4.17) where yo and vo are the initial displacement and velocity of the mass, respectively. Show that the Laplace transform of the solution can be written Yos + vo + 2cyo Y (s) (4.18) (s + c)² + (w – c²)' A moment's reflection will reveal that taking the inverse transform of Y (s) will depend upon the sign of w – c². In Exercises 39–41, we will examine three cases. Notice that each has a counterpart in Section 4.4. 39. Case 1: w – c² > 0. Show that this is the underdamped case of Section 4.4. Compute the inverse Laplace trans- form of Y (s) to show that the solution in this case is given by y(t) = yoe -ct cos Yoc + vo + sin 40. Case 2: w – c2 damped case of Section 4.4. Compute the inverse Laplace transform of Y (s) to show that the solution in this case is given by 0. Show that this is the critically y(t) = yoe¬et + (vo + byo)te¬ct. 41. Case 3: w – c² < 0. Show that this is the overdamped case of Section 4.4. To simplify calculations, let's set the intitial displacement as y(0) = yo = 0. Show that with this assumption the transform in (4.18) becomes vo Y (s) = (s + c)² – (c² – w;) - (4.19) vo (++c+ Use the technique of partial fractions to decompose (4.19), and then find the solution y(t).
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