test2-0 - y = e-t , with y (0) = 1, y '(0) = 1. Answer:...

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MATH 315 FALL 1998 TEST 2 SOLUTION KEY 1. (15 pts.) Consider the differential equation y '' +2 y ' + y = 0. (a) Find the general solution for the equation. Answer: Characteristic polynomial is r 2 +2 r +1 = ( r +1) 2 , with roots -1, -1. The general solution is therefore c 1 e - t + c 2 te - t . (b) Determine the Wronskian for fundamental solution set. Answer: 2. (15 pts.) Find the general solution for the differential equation y ''' + 2 y '' - y ' - 2 y = 0. Answer: Characteristic polynomial is r 3 +2 r 2 - r -2 = ( r -1)( r +1)( r +2), with roots 1, -1, -2. The general solution is therefore c 1 e t + c 2 e - t + c 3 e -2 t . 3. (15 pts.) Find the general solution for the differential equation y '''' + 3 y '' - 4 y = 4 e - t , assuming that the general solution to the homogeneous equation is given by Answer: A particular solution should have the form Y ( t ) = Ate - t .
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4. (20 pts.) Solve the differential equation y '' + 2 y ' + 2
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Unformatted text preview: y = e-t , with y (0) = 1, y '(0) = 1. Answer: Characteristic polynomial is r 2 +2 r +2, with roots , so . A particular solution should have the form Y ( t ) = Ae-t . 5. (15 pts.) Suppose the general solution to an n th order homogeneous equation L [ y ]=0 is Determine the general form that would be used with the method of undetermined coefficients for the solution to the nonhomogeneous equation Answer: 6. (20 pts.) Use the variation of parameters method to solve the differential equation y ''+3 y '+2 y = 4 e t . Answer: Characteristic polynomial is r 2 +3 r +2 = ( r +1)( r +2), with roots -1, -2; y c ( t ) = c 1 e-t + c 2 e-2 t . Let Y ( t ) = u 1 ( t ) e-t + u 2 ( t ) e-2 t ; then Therefore y ( t ) = y c ( t ) + 2 e 2 t e-t- 4 e 3 t e-2 t /3 = y c ( t ) + 2 e t /3....
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This test prep was uploaded on 04/06/2008 for the course MATH 302 taught by Professor Goldberg during the Spring '08 term at Johns Hopkins.

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test2-0 - y = e-t , with y (0) = 1, y '(0) = 1. Answer:...

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