{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

m238sp05t3 - Mathematics 238 test three due Friday 1 Given...

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

View Full Document Right Arrow Icon
Mathematics 238 test three due Friday, May 27, 2005 1 . Given the differential equation ( x 2 + 4) y + 3 xy + 4 y = 0 list the singular points of the equation. for which x -values will a power series solution of the form k =0 a k ( x - 3) k converge? 2 . For which values of x does the each of following power series converge? ( a ) k =1 ( x - 4) 2 k k 9 k ( b ) k =1 a 2 k x 2 k where a 2 = 1 & ka 2 k +2 = 3( k + 1) a 2 k for k 1 3 . Find a power series solution about x = 0 (through the degree four term) to the first-order initial value problem dy dx = x 2 + y 2 and y (0) = 1 extra credit parts of problem (3) compare your solution to a numerically obtained solution from a math software package or website. (plot the two solutions together and compare) compare your solution to the first few Picard iterates. 4 . Given the second-order differential equation y + xy + y = 0 for which values of x will power series solutions of this differential equation converge? find the recursion relation for the power series solution about x
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
Image of page 2
This is the end of the preview. Sign up to access the rest of the 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