1b-2009-exam_3_solutions

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

This preview shows pages 1–2. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online