final_solutions_22b_07

# final_solutions_22b_07 - Ordinary Differential Equations...

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Ordinary Differential Equations Math 22B-002, Spring 2007 Final Exam Solutions 1. [20 pts.] (a) Solve the following initial value problem for y ( t ): y + 2 ty = e - t 2 , y (0) = y 0 . (b) For what initial-value y 0 is y (2) = 0? Solution. (a) This is a first-order, linear ODE, so we can solve it by the integrating- factor method. Multiplication of the ODE by the integrating factor μ ( t ) = e R 2 t dt = e t 2 gives e t 2 y = 1 . Integration of this equation and imposition of the initial condition gives y ( t ) = ( t + y 0 ) e - t 2 . (b) We have y (2) = 0 if y 0 = - 2. 1

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2. [20 pts.] (a) Solve the initial value problem yy + 1 = t, y (6) = 3 . (b) For what t -interval is the solution defined? Solution. (a) The equation is separable. Separating variables, we get y dy = ( t - 1) dt. Integrating this equation, and multiplying the result by 2, we get y 2 = ( t - 1) 2 + c. The initial condition implies that c = - 16. After solving for y , we find that the solution is y ( t ) = ( t - 1) 2 - 16 . (b) The solution is well-defined and differentiable provided that the quantity inside the square-root is positive, meaning that 5 < t < . 2
3. [20 pts.] (a) Find the equilibrium solutions of the equation y = y ( y - 2) 3 . (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) = - 1 behave as t + ? How does the solution with y (0) = 1 behave as t → -∞ ? Solution. (a) The equilibria are y = 0 and y = 2. (b) The function f ( y ) = y ( y - 2) 3 is positive if y < 0 or y > 2, and negative if 0 < y < 2. Hence, the flow on the phase line is to the right with increasing t if y < 0 or y > 2, and to the left if 0 < y < 2. The equilibrium y = 0 is asymptotically stable, and the equilibrium y = 2 is unstable. (A sketch of the phase line is omitted.) (c) If y (0) = - 1, then y ( t ) 0 as t + . If y (0) = 1, then y ( t ) 2 as t → -∞ . 3

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4. [20 pts.] Suppose that y 1 ( t ), y 2 ( t ), y 3 ( t ) are solutions of the following initial value problems: e t y 1 + y 1 = 1 1 + t , y 1 (0) = 1 , y 1 (0) = 0 , e t y 2 + y 2 = 1 1 - t , y 2 (0) = 0 , y 2 (0) = 1 , e t y 3 + y 3 = 2 1 - t 2 , y 3 (0) = 1 , y 3 (0) = 1 . (a) According to general theorems, for what t -intervals are y 1 ( t ), y 2 ( t ), y 3 ( t ) uniquely defined? (b) Express y 3 ( t ) in terms of y 1 ( t ), y 2 ( t ), and justify your answer. Solution. (a) Since e t is never zero, the continuity of the coefficient functions p ( t ) = 0 , q ( t ) = e - t , g 1 ( t ) = e - t 1 + t , g 2 ( t ) = e - t 1 - t , g 3 ( t ) = 2 e - t 1 - t 2 implies that y 1 ( t ) is defined for - 1 < t < , y 2 ( t ) is defined for -∞ < y < 1, and y 3 ( t ) is defined for - 1 < t < 1.
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