lab06_sol_su09

lab06_sol_su09 - ECE220 Problem Lab #6 Solution...

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Unformatted text preview: ECE220 Problem Lab #6 Solution Differential Equations Date: Week of June 29, 2009 The main goal of this lab is to give you practice on solving differential equations. There may be more problems than you can finish during the lab period, and any remaining problems can be used as practice problems. Answering Questions: Open a word processor window and answer any questions below in that window. This is your Lab Text File (LTF). Later, you will submit this file via Wolfware. Feel free to experiment with the MATLAB scripts and GUIs in any way you want. I. Classification of Differential Equations 1. Give two examples of the following: A linear DE d 2 y ( t ) dt 2 + 10 dy ( t ) dt + 5 y ( t ) = 0 (1) A non-linear DE dy ( t ) dt 2 + t 3 10 dy ( t ) dt + 5 y ( t ) = 0 (2) A DE of 3rd order d 3 y ( t ) dt 3 + 3 d 2 y ( t ) dt 2 + 3 dy ( t ) dt + y ( t ) = 0 (3) A DE with constant coefficients d 2 y ( t ) dt 2 + 4 dy ( t ) dt 2 + 3 y ( t ) = 0 (4) 1 A DE with two time-varying coefficients t 2 dy ( t ) dt + 3 t ( y ( t )) 2 = 0 (5) A homogeneous DE d 2 y ( t ) dt 2 + 2 dy ( t ) dt + y ( t ) = 0 (6) A system of DEs dv ( t ) dt + 3 y ( t ) = 10 sin ( t ) u ( t ) dy ( t ) dt + 5 tv ( t ) = 5 t 2 sin 2 (10 t ) u ( t ) (7) A linear DE with constant coefficients d 2 y ( t ) dt 2 + 4 dy ( t ) dt + 3 y ( t ) = 0 (8) II. Linearity and time-shifting properties 1. Using the results of Examples 7.13, 7.14 and 7.15 (or Theorem 7.4, page 290), determine the particular solution of the DE dv ( t ) dt + 4 v ( t ) = 10 u ( t ) + 5 cos(2 t ) u ( t ) + 6 e- 2 t u ( t ) . Solution: We can find the three particular solutions and then use the linearity prop- erty to combine all the particular solutions: v p 1 ( t ) = 10 4 u ( t ) v p 2 ( t ) = 5 4 2 + 16 cos (2 t- arctan (2 / 4)) u ( t ) = . 671 cos (2 t- 1) u ( t ) v p 2 ( t ) = 6- 2 + 4 e- 2 t u ( t ) (9) Therefore, the particular solution is: v p ( t ) = 2 . 5 + 0 . 671 cos (2 t- 1) + 3 e- 2 t u ( t ) (10) 2 2. The particular solution of the DE d 2 v ( t ) dt 2 + 2 dv ( t ) dt + 7 v ( t ) = 40 cos( t ) u ( t ) was found in Example 7.23, page 361. Use the time shifting property to find the particular solution of the following DE d 2 v ( t ) dt 2 + 2 dv ( t ) dt + 7 v ( t ) = 40 sin( t ) u ( t ) . Verify your answer by checking out Example 7.24, page 362 (note, this example does not use the time shifting property). Solution: The particular solution to Example 7.23 is; v p ( t ) = 2 sin ( t ) u ( t ) + 6 cos ( t ) u ( t ) (11) The forcing function for the second DE is 40 sin( t ) u ( t ) which is a / 2 shift in time from the first DE. Therefore, the shifted DE can be written as: d 2 v ( t ) dt 2 + 2 dv ( t ) dt + 7 v ( t ) = 40 sin( t ) u ( t ) d 2 v ( t ) dt 2 + 2 dv ( t ) dt + 7 v ( t ) = 40 cos( t- / 2)...
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lab06_sol_su09 - ECE220 Problem Lab #6 Solution...

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