Exam2Practice

Exam2Practice - Exam II Practice Problems Math 4027 The...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Exam II Practice Problems Math 4027 The syllabus for the second exam will consist of Chapter 3 (Sections 1 8) and Chapter 4 (Sections 1 4, 6 8). Here are a few sample problems similar to previously assigned problems. 1. Find a basis for the solution set of the differential equation x 2 y 00 + xy + 4 y = 0 consisting of real valued functions. I Solution. This is an Euler equation with indicial polynomial q ( r ) = r ( r- 1) + r + 4 = r 2 + 4, which has roots 2 i . Thus | x | 2 i and | x |- 2 i are a basis of solutions and a basis consisting of real valued functions is 1 ( x ) = cos(2log | x | ) and 2 ( x ) = sin(2log | x | ) . J 2. Solve the initial value problem (1 + x 2 ) y 00 + 2 xy- 2 y = 0 , (0) = 2 , (0) =- 1 in terms of a power series in x . I Solution. The solution will have the form ( x ) = n =0 c n x n with c = (0) = 2, and c 1 = (0) =- 1. Substituting the power series into the differential equation gives (1 + x 2 ) X n =2 n ( n- 1) c n x n- 2 + 2 x X n =1 nc n- 2 X n =0 c n x n = X n =2 n ( n- 1) c n x n- 2 + X n =2 n ( n- 1) c n x n + X n =1 2 nc n x n- X n =0 2 c n x n = X n =0 ( n + 2)( n + 1) c n +2 x n + X n =0 n ( n- 1) c n x n + X n =0 2 nc n x n- X n =0 2 c n x n = X n =0 [( n + 2)( n + 1) c n +2 + ( n ( n- 1) + 2 n- 2) c n ] x n = X n =0 [( n + 2)( n + 1) c n +2 + ( n + 2)( n- 1) c n ] x n = 0 . Since the coefficient of every power of x must be 0, this gives the following recursion relation that must be satisfied by the coefficients c n is c n +2 =- n- 1 n + 1 c n for all n 0. 1 Exam II Practice Problems Math 4027 Using c = 2, c 1 =- 1 we conclude: c = 2 c 1 =- 1 c 2 = c = 2 c 3 = c 4 =- 1 3 c 2 =- 2 3 c 5 = c 6 =- 3 5 c 4 = 2 5 c 7 = c 8 =- 5 7 c 6 =- 2 7 c 9 = . . . . . . By induction, we see that c 2 m = (- 1) m +1 2 2 m- 1 for all m and c 2 m +1 = 0 for all m 1. Hence we have the following formula for ( x ) ( x ) =- x + X m =0 (- 1) m +1 2 2 m- 1 x 2 m . While it was not part of the question asked, it is easy to see from the ratio test that the series ( x ) converges for | x | < 1. By comparing the above series with the power series expansion of the tan- 1 function, one can also observe that ( x ) =- x + 2 x tan- 1 x. Again, this is for information only. It was not part of the requested problem. J 3. Find the general solution to y 00 + 1 x 2 y- 1 x 3 y = 0 given that 1 ( x ) = x is a solution. I Solution. Use the reduction of order technique. Thus, assume that 2 ( x ) = xu is a second solution where u is an unknown function that needs to be determined. Then 0 = 00 2 ( x ) + 1 x 2 2 ( x )- 1 x 3 2 ( x ) = ( xu ) 00 + 1 x 2 ( xu )- 1 x 3 xu = (2 u + xu 00 ) + 1 x 2 ( u + xu )- 1 x 3 xu = xu 00 + 2 + 1 x u ....
View Full Document

This note was uploaded on 03/20/2011 for the course MATH 4027 taught by Professor Adkins during the Spring '06 term at LSU.

Page1 / 12

Exam2Practice - Exam II Practice Problems Math 4027 The...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online