problem_lab_06_sol_s08

problem_lab_06_sol_s08 - ECE220 Solution Lab #6 Review for...

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ECE220 Solution Lab #6 Review for Test 2 Date: Week of March 10, 2008 The goal of this lab is to review material in preparation for Test 2. Work out as many problems as you can. If you have worked on a problem before, you may skip it. Do not worry if you do not finish all problems in time. I. Classification of differential equations 1. Give two examples of the following: A linear DE (a) d 2 v dt 2 + 2 dv dt + 6 v ( t ) = 5 e - 2 t u ( t ) (b) d 4 v dt 4 + dv dt + 5 tv ( t ) = 10cos(2 π 250 t ) u ( t ) A nonlinear DE (a) ( d 2 v dt 2 ) 2 + 2 dv dt = 10 u ( t ) (b) ( d 3 v dt 3 ) 3 + 5 v ( t ) = 10 u ( t ) A DE of 3rd order (a) d 3 v dt 3 + 2 d 2 v dt 2 + tv ( t ) = e - t u ( t ) (b) d 3 v dt 3 + v ( t ) = 10 cos( t ) u ( t ) A DE with constant coefficients (a) d 2 v dt 2 + 2 dv dt + 6 v ( t ) = 5 e - 2 t u ( t ) (b) d 4 v dt 4 + dv dt + 5 v ( t ) = 10 cos(2 π 250 t ) u ( t ) A DE with two time-varying coefficients (a) t ( dv dt ) 2 + t 4 v ( t ) = 20 u ( t ) (b) d 3 v dt 3 + 2 t dv dt + 7 tv ( t ) = e - 12 t u ( t ) 1
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A homogeneous DE (a) d 2 v dt 2 + 3 dv dt + 12 v ( t ) = 0 (b) d 4 v dt 4 + v ( t ) = 0 A system of DEs (a) 2 dv 1 dt + 6 v 1 ( t ) + 12 v 2 ( t ) = 5 u ( t ) dv 2 dt + 5 v 1 ( t ) = sin(2 π 250 t ) u ( t ) (b) 2 dv 1 dt + 6 v 1 ( t ) + 12 v 2 ( t ) + v 3 ( t ) = 5 u ( t ) dv 2 dt + 5 v 1 ( t ) = sin(2 π 250 t ) u ( t ) dv 3 dt + v 2 ( t ) = 50 u ( t ) A linear DE with constant coefficients (a) d 2 v dt 2 + 2 dv dt + 6 v ( t ) = 5 e - 2 t u ( t ) (b) d 4 v dt 4 + 4 dv dt + 5 v ( t ) = 10 cos(2 π 250 t ) u ( t ) ————————————————————————————– II. Linearity and time-shifting properties Question 1. Using the results of Examples 7.13, 7.14 and 7.15 (or theorem 7.4, page 348), determine the particular solution of the DE dv ( t ) dt + 4 v ( t ) = 10 u ( t ) + 5cos(2 πt ) u ( t
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problem_lab_06_sol_s08 - ECE220 Solution Lab #6 Review for...

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