s251final(su03) - Answers to Summer 2003 Final exam Q:1 A A...

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Answers to Summer 2003 Final exam. Q#:1 A A D.E. that satisfies these requirements is y 0 = ( y - 1)( y - 2)( y - 3). B A D.E. that satisfies these requirements is y 00 + 4 y 0 + 4 y = 3 cos t - 4 sin t . C A D.E. that satisfies these requirements is 2 xy + e y + ( x 2 + xe y ) dx dy = 0. Q#:2 The answer, for both methods, is y ( t ) = t 2 e t - 1 4 e t + 5 4 e - t Q#:3 A There is an equilibrium solution at v = 15, which is stable, and another at v = - 15, which is unstable. B The solution is t + c = 1 120 [ln(30 + 2 v ) - ln(30 - 2 v )]. Q#:4 A The D.E. is y 00 + 7 y 0 + 10 y = cos t - u 3 π ( t ) cos t , with initial conditions y (0) = 0 and y 0 (0) = 2. B y ( t ) = 9 130 cos t + 7 130 sin t + 8 15 e - 2 t - 47 78 e - 5 t + u 3 π ( t ) - 9 130 cos t - 7 130 sin t - 2 15 e - 2( t - 3 π ) + 5 78 e - 5( t - 3 π ) . C y ( π ) = - 9 130 + 8 15 e - 2 π - 57 78 e - 5 π and y (4 π ) = 8 15 e - 8 π - 57 78 e - 20 π - 2 15 e - 2 π + 5 78 e - 5 π D The natural frequency is ω o = 10 rad s Q#:5 A The solution is X ( t ) = - e - t 5 - 3 + e - 3 t 1 - 1 . B This critical point, the only one, is a stable node. It is assymptotically stable. Q#:6 A The critical points are P = (1 , - 1), Q - ( - 1 , - 1), R = (2 , - 2) and S = (2 , 2). B The linearized matrix is A ( x, y ) = 2 x 2 y y + 1 x - 2 . C Point P: X 0 = 2 - 2 0 - 1 X . It is a saddle point, which is unstable.
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