2302-practice-final-soln

# Solution first we solve the homogeneous equation y 4

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Solution: First we solve the homogeneous equation, y ′′ - 4 y + 5 y = 0. This equation has characteristic polynomial r 2 - 4 r + 5, which has roots r = 4 ± 16 - 20 2 = 2 ± i.

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MAP 2302, Fall 2011 — Final Exam Review Problems 3 The homogeneous solution is therefore y h = e 2 t ( C cos t + D sin t ) . Now we use the method of undetermined coefficients to find a particular solution y p . Our first guess for y p is y p = Ae 3 t . As this guess does not overlap with the homogeneous solution y h , it will work. We then compute that y = Ae 3 t , y = 3 Ae 3 t , y ′′ = 9 Ae 3 t . Plugging this into the equation, we have y ′′ - 4 y + 5 y = 9 Ae 3 t - 4(3 Ae 3 t ) + 5( Ae 3 t ) = 2 Ae 3 t . Since we want this to equal 4 e 3 t , we take A = 2. Our general solution is therefore y ( t ) = e 2 t ( C cos t + D sin t ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright y h + 2 e 3 t bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright y p . Now we need to choose C and D to match the initial conditions. We are given that y (0) = 2. On the other hand, plugging t = 0 into the formula for y ( t ) we get y (0) = C + 2, so C = 0, and thus y = De 2 t sin t + 2 e 3 t , y = 2 De 2 t sin t + De 2 t cos t + 6 e 3 t . Plugging t = 0 into our formula for y we see that y (0) = D + 6. Since we want this to equal 7, we set D = 1 and get the final solution y = e 2 t sin t + 2 e 3 t .
MAP 2302, Fall 2011 — Final Exam Review Problems 4 4 Find the general solution of the differential equation y ′′ + 4 y + 4 y = 1 t 2 e 2 t . Solution: First we solve the homogeneous equation, y ′′ + 4 y + 4 y = 0. This equation has the characteristic polynomial r 2 + 4 r + 4 = ( r + 2) 2 , so it has a repeated root at r = - 2, and thus the solution y h = C e 2 t bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright y 1 + D te 2 t bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright y 2 . The righthand side of the given (nonhomogeneous) equation is not a candidate for the method of undetermined coefficients (its derivatives go on forever), so we must use variation of parameters, i.e., look for a solution of the form y ( t ) = C ( t ) y 1 + D ( t ) y 2 . First we compute the denominator in this formula, y 1 y 2 - y 2 y 1 = ( e 2 t )( - 2 te 2 t + e 2 t ) - ( te 2 t )( - 2 e 2 t ) = e 4 t . Now we use the formula to find C ( t ) and D ( t ). First, we have C ( t ) = integraldisplay - fy 2 y 1 y 2 - y 2 y 1 dt, = integraldisplay - ( 1 t 2 e 2 t ) te 2 t e 4 t dt, = - integraldisplay 1 t dt, = - ln t + E. Next we get that D ( t ) = integraldisplay fy 1 y 1 y 2 - y 2 y 1 dt, = integraldisplay ( 1 t 2 e 2 t ) e 2 t e 4 t dt, = integraldisplay t 2 dt, = - 1 t + F. Therefore the final answer is y ( t ) = C ( t ) y 1 + D ( t ) y - 2 = ( - ln t + E ) e 2 t + parenleftbigg - 1 t + F parenrightbigg te 2 t .

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MAP 2302, Fall 2011 — Final Exam Review Problems 5 5 Suppose that the Laplace transform of y ( t ) is given by 2 s 3 + s 2 + 8 s + 6 ( s 2 + 1)( s 2 + 4) . What is y ?
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