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Exam2Practice

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

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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 0 + 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 ϕ 1 ( x ) = cos(2 log | x | ) and ϕ 2 ( x ) = sin(2 log | x | ) . J 2. Solve the initial value problem (1 + x 2 ) y 00 + 2 xy 0 - 2 y = 0 , ϕ (0) = 2 , ϕ 0 (0) = - 1 in terms of a power series in x . I Solution. The solution will have the form ϕ ( x ) = n =0 c n x n with c 0 = ϕ (0) = 2, and c 1 = ϕ 0 (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

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Exam II Practice Problems Math 4027 Using c 0 = 2, c 1 = - 1 we conclude: c 0 = 2 c 1 = - 1 c 2 = c 0 = 2 c 3 = 0 c 4 = - 1 3 c 2 = - 2 3 c 5 = 0 c 6 = - 3 5 c 4 = 2 5 c 7 = 0 c 8 = - 5 7 c 6 = - 2 7 c 9 = 0 . . . . . . By induction, we see that c 2 m = ( - 1) m +1 2 2 m - 1 for all m 0 and c 2 m +1 = 0 for all m 1. Hence we have the following formula for ϕ ( x ) ϕ ( 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 ϕ ( 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 ϕ ( 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 0 - 1 x 3 y = 0 given that ϕ 1 ( x ) = x is a solution. I Solution. Use the reduction of order technique. Thus, assume that ϕ 2 ( x ) = xu is a second solution where u is an unknown function that needs to be determined. Then 0 = ϕ 00 2 ( x ) + 1 x 2 ϕ 0 2 ( x ) - 1 x 3 ϕ 2 ( x ) = ( xu ) 00 + 1 x 2 ( xu ) 0 - 1 x 3 xu = (2 u 0 + xu 00 ) + 1 x 2 ( u + xu 0 ) - 1 x 3 xu = xu 00 + 2 + 1 x u 0 . Thus u 00 = - 2 x + 1 x 2 u 0 2
Exam II Practice Problems Math 4027 so that u 0 = Ce R - ( 2 x + 1 x 2 ) dx = Ce - 2 log | x | + 1 x = C e 1 /x x 2 . Now integrate to get u = - Ce 1 /x + B = Ae 1 /x + B where A and B are arbitrary constants. Thus, the general solution of our equation is ϕ 2 ( x ) = ux = Axe 1 /x + Bx. J 4. Find all quadratic polynomials p ( x ) = a + bx + cx 2 that are solutions of the differential equation (1 - x 2 ) y 00 - 2 xy 0 + 2 y = 6(1 - x 2 ) .

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

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