final_22b_sample_solutions - Ordinary Differential...

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Ordinary Differential Equations Math 22B-002, Spring 2007 Sample Final Exam Solutions 1. (a) Solve the following initial value problem for y ( t ) in t > 0: y - 2 t y = t, y (1) = 2 . (b) How does y ( t ) behave as t 0 + ? Solution. (a) This is a first-order linear ODE. The integrating factor is 1 /t 2 : 1 t 2 y - 2 t 3 y = 1 t , 1 t 2 y = 1 t , 1 t 2 y = ln t + c, y = t 2 ln t + ct 2 . Imposing the initial condition, we get c = 2, so the solution is y ( t ) = t 2 ln t + 2 t 2 . (b) Since t ln t 0 as t 0 + , y ( t ) 0 as t 0 + . 1
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2. (a) Solve the initial value problem y + (cos t ) y 2 = 0 , y (0) = y 0 , where y 0 is an arbitrary constant. (b) For what values of the initial data y 0 is your solution defined for all -∞ < t < + ? Solution. (a) The equation is separable. We have y = - (cos t ) y 2 , dy y 2 = - cos t dt, - 1 y = - sin t + c. Imposing the initial condition, assuming that y 0 = 0, we get c = - 1 y 0 , Solving for y , we get y ( t ) = 1 1 /y 0 + sin t if y 0 = 0. If y 0 = 0, then the solution is y ( t ) = 0. (b) The solution is defined for all t if | y 0 | < 1. 2
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3. Consider the ordinary differential equation y = ( y - 3) ( y 2 - 1 ) (a) Sketch a graph of the right-hand side of this equation as a function of y , and find all equilibrium solutions of the equation. (b) Sketch the phase line of the equation, and determine the stability of the equilibria you found in (a). (c) How does the solution with y (0) = 0 behave as t + ? How does the solution with y (0) = 2 behave as t → -∞ ? Solution. (a) The equilibrium points are y = - 1 , 1 , 3. (b) Trajectories move to the left with increasing t if y < - 1 or 1 < y < 3; trajectories move to the right if - 1 < y < 1 or y > 3. The equilibrium y = 1 is asymptotically stable. The equilibria y = - 1 , 3 are unstable. (Sketch of phase line omitted.) (c) If y (0) = 0, then y ( t ) 1 as t + . If y (0) = 2, then y ( t ) 3 as t → -∞ .
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