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Unformatted text preview: MAP 2302 Solutions to Sample Second Exam Please notify me by email if you find any errors or misprints. 1. (a) The auxiliary equation is r 2 +2 r +17 r = 0, with complex roots r 1 = 1 = 4 i , r 2 = 1 4 i , so the general solution is y = e x ( A cos4 x + B sin4 x ) . Setting y (0) = 1 yields A = 1, so y = e x (cos4 x + B sin4 x ) . Differentiating, y = e x ( 4sin4 x + B cos4 x ) e x (cos4 x + B sin4 x ) . Setting y (0) = 1 yields 1 = B 1, so B =0. Thus, y = e x cos4 x. (b) The auxiliary equation is r 2 4 r + 4 = 0 so r = 2 is a double root. Therefore the general solution is y = ( Ax + B ) e 2 x . Differentiating, we obtain y = Ae 2 x + 2( Ax + B ) e 2 x = e 2 x (2 Ax + ( A + 2 B )) Putting in the initial conditions y (1) = 1, y (1) = 1 yields the equations 1 = ( A + B ) e 2 , 1 = (3 A + 2 B ) e 2 , and solving these gives A = 1 /e 2 , B = 2 /e 2 , so the solution of the initial value problem is y = ( x + 2) e 2 x 2 ....
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This note was uploaded on 02/21/2010 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.
 Spring '08
 TUNCER

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