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# 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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