{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam 2 - Name Exam 2 Instructions Answer each of the...

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

View Full Document Right Arrow Icon
Name: Exam 2 Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. 1. [18 Points] Solve the initial value problem 2 x 2 y 00 + xy 0 - y = 0; y (1) = 1 , y 0 (1) = - 1 . I Solution. This is an Euler equation, so we first find the indicial polynomial q ( r ) = 2 r ( r - 1) + r - 1 = 2 r 2 - r - 1 = (2 r + 1)( r - 1). The roots are r 1 = 1 and r 2 = - 1 / 2 so the general solution has the form y = c 1 x + c 2 x - 1 / 2 and the coefficients c 1 and c 2 are determined from the initial conditions: 1 = y (1) = c 1 + c 2 - 1 = y 0 (1) = c 1 - 1 2 c 2 . Subtracting the second equation from the first gives 2 = (3 / 2) c 2 so that c 2 = 4 / 3 and substituting in the first equation gives c 1 = - 1 / 3. Hence y = - 1 3 x + 4 3 x - 1 / 2 . J 2. [18 Points] Solve the differential equation x 2 y 00 + 3 xy 0 + y = x - 1 ( x > 0) . You may find the integral formula R x k log x dx = ( x k +1 k +1 log x - x k +1 ( k +1) 2 + C if k 6 = - 1 1 2 (log x ) 2 if k = - 1 of use. I Solution. This is a nonhomogeneous equation so we will use variation of parame- ters. First solve the associated homogeneous equation, which is an Euler equation, by finding the indicial polynomial q ( r ) = r ( r - 1) + 3 r + 1 = r 2 + 2 r + 1 = ( r + 1) 2 , which has a single root r 1 = - 1 of multiplicity 2. Hence, the associated homogeneous equation has the solution y h = c 1 x - 1 + c 2 x - 1 log x. Note that we do not need log | x | since we are only interested in the interval x > 0. To apply variation of parameters, we need to first put the equation in normal form with the coefficient of y 00 equal to 1: y 00 + 3 x y 0 + 1 x 2 y = x - 3 .
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