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Unformatted text preview: Exercise Set 3 Math 4027 Due: March 15, 2005 1. Verify that the function 1 ( x ) is a solution of the given differential equation, and find a second linearly independent solution 2 ( x ) on the interval indicated. (a) y 00 2 x 2 y = 0 1 ( x ) = x 2 < x < (b) y 00 4 xy + (4 x 2 2) y = 0 1 ( x ) = e x 2 < x < (c) (1 x 2 ) y 00 2 xy + 2 y = 0 1 ( x ) = x < x < 1 (a) I Solution. 00 1 ( x ) 2 x 2 1 ( x ) = 2 2 = 0 so 1 ( x ) is a solution to y 00 2 x 2 y = 0 . If we define 2 = 1 u , then 2 ( x ) = x 2 u 2 ( x ) = 2 xu + x 2 u 00 2 ( x ) = 2 u + 4 xu + x 2 u 00 . If 2 is to be a solution of the differential equation y 00 2 x 2 y = 0, then we must have 0 = 00 2 ( x ) 2 x 2 2 ( x ) = 2 u 2 u + 4 xu + x 2 u 00 . This means that v = u satisfies the first order linear equation x 2 v + 4 xv = 0. Writing this in standard form gives v + 4 x 1 v = 0 which means that there is an integrating factor I ( x ) = x 4 . That is ( x 4 v ) = 0 so that x 4 v = C where C is a constant. Then u = v = Cx 4 so that u = C 3 x 3 + B for B a constant. Choosing C = 3 and B = 0 gives one choice for u , namely u = x 3 which gives a second solution 2 ( x ) = x 2 u = 1 x . J (b) I Solution. Computing the derivatives of 1 ( x ) gives 1 ( x ) = e x 2 1 ( x ) = 2 xe x 2 00 1 ( x ) = (2 + 4 x 2 ) e x 2 , which gives 00 1 ( x ) 4 x 2...
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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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