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Unformatted text preview: MAP 2302, Fall 2010 Midterm 2 Review Solutions 1 Please note: I do sometimes make mistakes. Please send me an email ( vatter@ufl.edu ) if you think youve found a mistake. 1 Solve the following initial value problems. a. y  4 y + 8 y = 0; y (0) = 1; y (0) = 0. Solution: This is a homogeneous equation with characteristic polynomial r 2 4 r + 8 . With the quadratic formula, we see that r = 2 2 i , so the general solution is y = e 2 t ( C cos 2 t + D sin2 t ) . Now we need to choose C and D to match the initial conditions. We see that y (0) = e ( C + 0) = C = 1 , so C = 1, and y = e 2 t ( 2 C sin2 t + 2 D cos 2 t ) + 2 e 2 t ( C cos 2 t + D sin2 t ) , so plugging in t = 0, y (0) = e (0 + 2 D ) + 2 e ( C + 0) = 2 D + 2 C = 0 . Since C = 1, we see that D = 1. Putting it all together, we get y ( t ) = e 2 t (cos 2 t sin 2 t ) as the final answer. b. y + 2 y  3 y = 0; y (0) = 9; y (0) = 3. Solution: The characteristic polynomial here is r 2 + 2 r 3 = ( r 1)( r + 3) , so r = 1 , 3, and our general solution is y = Ce t + De 3 t . MAP 2302, Fall 2010 Midterm 2 Review Solutions 2 To find C and D we plug t = 0 into y and y : y (0) = C + D = 9 , y ( t ) = Ce t 3 De 3 t , y (0) = C 3 D = 3 . From the first equation, we see that D = 9 C . Plugging this into the last equation shows that C 3(9 C ) = 3, so 4 C = 24. We conclude that C = 6 and D = 3, giving y ( t ) = 6 e t + 3 e 3 t as the final solution. c. y  4 y = 0; y (0) = 2; y (0) = 98. Solution: The characteristic polynomial is r 2 4 = ( r 2)( r + 2) , so the roots are r = 2, and thus the solution is y = Ce 2 t + De 2 t . To find C and D we plug t = 0 into y and y : y (0) = C + D = 2 , y ( t ) = 2 Ce 2 t 2 De 2 t , y (0) = 2 C 2 D = 98 . From the first equation, D = 2 C . Plugging this into the last equation gives us that 2 C 2(2 C ) = 98, so 4 C = 94, C = 47 / 2 , D = 51 / 2 , and the final solution is y ( t ) = 47 2 e 2 t + 51 2 e 2 t . MAP 2302, Fall 2010 Midterm 2 Review Solutions 3 2 What form of particular solution y p would you guess in order to solve the following differential equations using the method of undetermined coefficients? Do not solve these problems. a. y  4 y  21 y = 2 e 7 t . Solution: First we have to solve the corresponding homogeneous equation, y  4 y  21 y = 0 . This equation has characteristic polynomial r 2 4 r 21 = ( r 7)( r + 3) , so the roots are r = 7 , 3 and the the homogeneous solution is y h = Ce 7 t + De 3 t Our first guess for y p is to match the function on the righthand side and all its derivatives....
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This note was uploaded on 12/10/2011 for the course MAP 2302 taught by Professor Tuncer during the Fall '08 term at University of Florida.
 Fall '08
 TUNCER

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