final_22b_08_solutions

final_22b_08_solutions - Ordinary Differential Equations...

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Ordinary Differential Equations Math 22B-002, Spring 2008 Final Exam: Solutions NAME. ................................................................... SIGNATURE. ......................................................... I.D. NUMBER. ...................................................... No books, notes, or calculators. Show all your work Question Points Score 1 20 2 20 3 20 4 20 5 20 6 20 7 20 8 20 9 20 10 20 Total 200 1
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1. [20 pts.] (a) Find the solution of the initial value problem ty 0 = 1 y + 1 , y (1) = 0 . (b) For what t -interval is the solution defined? Solution. (a) The equation is separable. Separating variables, we get Z ( y + 1) dy = Z dt t . Evaluation of the integrals gives 1 2 ( y + 1) 2 = ln t + C. Imposing the initial condition, and using ln 1 = 0, we get 1 2 = C. Solving for y , we find that y ( t ) = 2 ln t + 1 - 1 . Instead, you can integrate the separated equation as 1 2 y 2 + y = ln t + C, which just differs from the previous solution in how the constant of integration is chosen. Imposing the initial condition, we get C = 0, so y 2 + 2 y - 2 ln t = 0 . Solving this quadratic equation for y , we get y ( t ) = - 2 ± 4 + 8 ln t 2 . Choosing the +-root, which gives y = 0 when t = 1, and simplifying the result, we get y ( t ) = 2 ln t + 1 - 1 as before. 2
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(b) This solution is defined and continuously differentiable provided that 2 ln t + 1 > 0, or ln t > - 1 / 2, which holds in the interval e - 1 / 2 < t < . 3
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2. [20 pts.] Suppose that a is a constant, and consider the initial value problem y 0 - y = e at , y (0) = 0 . (a) Find the solution if a 6 = 1. (b) Find the solution if a = 1. (c) Show that the solution in (b) is the limit of the solution in (a) as a 1. (Hint: use l’Hospital’s rule.) Solutions. The equation is linear and nonhomogeneous, so we use the integrating factor method. An integrating factor is μ ( t ) = exp Z ( - 1) dt = e - t . Multiplying the equation by e - t and rewriting the left-hand side as an exact derivative, we get ( e - t y ) 0 = e ( a - 1) t . (1) (a) If a 6 = 1, then an integration of this equation gives e - t y = 1 a - 1 e ( a - 1) t + C. Imposition of the initial condition gives C = - 1 a - 1 . The solution of the IVP is therefore y ( t ) = e at - e t a - 1 . If a = 1, equation (1) becomes ( e - t y ) 0 = 1 . Integration of this equation gives e - t y = t + C. 4
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Imposition of the initial condition gives C = 0 . The solution is therefore y ( t ) = te t . (c) By l’Hospital’s rule the limit of the solution in (a) as a 1 is given by lim a 1 y ( t,a ) = lim a 1 e at - e t a - 1 = lim a 1 d da [ e at - e t ] d da [ a - 1] = lim a 1 te at 1 = te t . Thus, we obtain the solution in (b).
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This note was uploaded on 03/05/2009 for the course MATH 22B taught by Professor Hunter during the Spring '08 term at UC Davis.

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final_22b_08_solutions - Ordinary Differential Equations...

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