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Unformatted text preview: September 21, 2010 81 8. Reduction of Order and more on complex roots Reduction of Order: Suppose we are given a general homogeneous second order d.e. L ( y ) = y 00 + p ( t ) y + q ( t ) y = 0 . (1) We know that, in order to find the general solution, it suffices to find two linearly independent solutions. It turns out that, if we can find one nonzero solution, then a second independent solution can always be found as usual up to integration), by a method called reduction of order. Here is how it works, Suppose y 1 is one nonzero solution to (1). Let us try to find a second solution y 2 = y 1 v where v is a nonconstant function. For y 2 to be a solution, we have ( y 1 v ) 00 + p ( y 1 v ) + qy 1 v = 0 or y 00 1 v + 2 y 1 v + y 1 v 00 + py 1 v + py 1 v + qy 1 v = 0 September 21, 2010 82 v ( y 00 1 + py 1 + qy 1 ) + v (2 y 1 + py 1 ) + y 1 v 00 = 0 . v (2 y 1 + py 1 ) + y 1 v 00 = 0 (2) since L ( y 1 ) = 0. Now, y 1 and p are known, so we get a first order linear d.e. for v . We solve this for v , then integrate to get v , and then go back to get an actual solution y 2 = y 1 v of L ( y ) = 0. . Since v is not constant, we clearly get that y 2 = y 1 v and y 1 are linearly independent functions. Let us be more specific here. Equation (2) becomes v 00 v = 2 y 1 p y 1 y 1 (3) d log ( v ) = 2 y 1 p y 1 y 1 d log ( v ) = 2 y 1 y 1 p log ( v ) = Z  2 y 1 y 1 p dt September 21, 2010 83 v = exp ( Z  2 y 1 y 1 p dt ) v = Z exp ( Z  2 y 1 y 1 p ) dt dt (4) Example 1: The function y ( t ) = t 3 + t is a solution to the d.e....
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This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.
 Fall '11
 STAFF

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