{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw4_solution

hw4_solution - a dc dt =-kc 3(6 Separate c t d c c 3 =-k...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Numerical Analysis in Engineering ME 140A, Fall 2007 HW #4 solutions 1. Problem 1. Consider the first order linear equation λ dy dt = 1 - y (1) SOLUTION: y 0 + y λ = 1 λ (2) Multiply with an integrating factor μ ( t ) : μy 0 + μ y λ = μ λ (3) Add and subtract μ 0 y ( μy 0 + μ 0 y ) + ( μ λ - μ 0 ) y = μ λ (4) Need ( μ λ - μ 0 ) = 0 μ 0 μ = 1 λ ln ( μ ) = t λ + c 1 μ ( t ) = e t λ + c 1 = c 2 e t λ Having found μ ( t ), now solve (4), ( μy 0 + μ 0 y ) + 0 = μ λ ( μy ) 0 = c 2 λ e t/λ μy = c 2 e t/λ + c 3 μy = c 2 e t/λ + c 3 y = c 2 μ e t/λ + c 3 μ y = 1 + c 4 e - t/λ 1
Image of page 1

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

View Full Document Right Arrow Icon
Initial condition: y (0) = 0 c 4 = - 1 y ( t ) = 1 - e - t λ (5) b) find time t * when y ( t ) = 0 . 99 : e - t * λ = 0 . 01 e t * λ = 100 t * λ = ln (100) t * = λln (100)
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a) dc dt =-kc 3 (6) Separate c, t : d c c 3 =-k dt (7) Integrate:-1 2 c-2 =-kt + c 1 c-2 = 2 kt + c 2 c = 1 √ 2 kt + c 2 Initial condition: c (0) = 2 : → c 2 = 1 4 → c = 1 √ 2 kt + 1 4 b) The equation is nonlinear, homogeneous, constant coeFcient. c) 1 √ 2 t * + 1 4 = 10-3 → q 2 t * + 1 4 = 10 3 → 2 t * + 1 4 = 10 6 t * = 499 , 999 . 875 2...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern