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Exam 2

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

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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 .

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