# solsamp2 - MAP 2302 Solutions to Sample Second Exam Please...

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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 cos 4 x + B sin 4 x ) . Setting y (0) = 1 yields A = 1, so y = e - x (cos 4 x + B sin 4 x ) . Differentiating, y = e - x ( - 4 sin 4 x + B cos 4 x ) - e - x (cos 4 x + B sin 4 x ) . Setting y (0) = - 1 yields - 1 = B - 1, so B =0. Thus, y = e - x cos 4 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 . 2. We use the method of variation of paramenters. (This problem is very similar to an example on p. 216.) The auxiliary equation has imaginary roots ± 2 i , so y 1 = cos 2 x and y 2 = sin 2 x are fundamental solutions. We must solve the equations

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