final_22b_08_solutions

final_22b_08_solutions - Ordinary Differential Equations...

• Notes
• 17
• 100% (1) 1 out of 1 people found this document helpful

This preview shows pages 1–6. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

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
(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

This preview has intentionally blurred sections. Sign up to view the full version.

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
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). Remark. This problem is an example of how the ‘resonant’ solution te t for a nonhomogeneous term that solves the homogeneous equation arises as a limit of the nonresonant solutions.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Spring '08
• Hunter

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern