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

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