1b-2009-exam_3_solutions - Math 1B Exam #3 Solutions Fall...

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Math 1B — Exam #3 Solutions Fall Program for Freshmen 2009 Fred Bourgoin 1. Use power series to solve ( x - 3) y 0 + 2 y = 0. Solution. Plugging y = X n =0 c n x n and y 0 = X n =1 nc n x n - 1 into the equation yields ( x - 3) X n =1 nc n x n - 1 + 2 X n =0 c n x n = 0 x X n =1 nc n x n - 1 - 3 X n =1 nc n x n - 1 + 2 X n =0 c n x n = 0 X n =1 nc n x n - X n =1 3 nc n x n - 1 + X n =0 2 c n x n = 0 X n =0 nc n x n - X n =0 3( n + 1) c n +1 x n + X n =0 2 c n x n = 0 X n =0 [ nc n - 3( n + 1) c n +1 + 2 c n ] x n = 0 . We get the recurrence relation c n +1 = n + 2 3( n + 1) c n . The first few coefficients are c 0 , c 1 = 2 3 c 0 , c 2 = 3 3 · 2 c 1 = 3 3 2 c 0 , c 3 = 4 3 · 3 c 2 = 4 3 3 c 0 . Thus, c n = n + 1 3 n c 0 , and the solution is y = c 0 X n =0 n + 1 3 n x n = C ( x - 3) 2 . 2. Solve y 00 - y 0 - 6 y = e - 2 x . Solution. First solve the complementary equation. The characteristic equation is r 2 - r - 6 = 0, which has roots - 2 and 3. So, y c = c 1 e - 2 x + c 2 e 3 x . For a particular solution to the equation, use the method of undetermined coefficients with y p = Axe - 2 x since Ae - 2 x is a solution to the complementary equation. Then y 0 p = ( A - 2 Ax ) e - 2 x and y 00 p = ( - 4 A + 4 Ax ) e - 2 x .
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1b-2009-exam_3_solutions - Math 1B Exam #3 Solutions Fall...

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