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Unformatted text preview: Math 201 Fall 2011 Midterm Exam 2 (A) October 28, 2011 1. [5 marks] Find the solution of the initial value problem y 00 4 y +5 y = 0 , y (0) = 1 , y (0) = 5 The auxiliary equation 2 4 + 5 = 0 , = 2 i . So the two linearly independent solutions are y 1 = e 2 x cos x , y 2 = e 2 x sin x The general solution is y = c 1 y 1 + c 2 y 2 = c 1 e 2 x cos x + c 2 e 2 x sin x We use the initial values to determine c 1 and c 2 y (0) = c 1 = 1 y = 2 c 1 e 2 x cos x c 1 e 2 x sin x + 2 c 2 e 2 x sin x + c 2 e 2 x cos x , y (0) = 2 c 1 + c 2 = 5 , c 2 = 5 2 c 1 = 5 2 = 3 So, y = e 2 x cos x + 3 e 2 x sin x 2. [5 marks] Find the general solution to y 00 + y 2 y = 3 e 2 x The general solution to the homogeneous part: Auxiliary equation: 2 +  2 = 0 , = 1 , = 2 Two linearly independent solutions: y 1 = e x , y 2 = e 2 x Guess for a particular solution: y p = axe 2 x (we need x multiplied to e 2 x because e 2 x is a solution to the homogeneous part).is a solution to the homogeneous part)....
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This note was uploaded on 01/15/2012 for the course MATH 201 taught by Professor Steacy during the Winter '10 term at University of Victoria.
 Winter '10
 STEACY
 Math

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