348_pdfsam_math 54 differential equation solutions odd

# 348_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 17. Writing the given diferential equation, x 3 y 0 3 x 2 y 0 +6 xy 0 6 y =0 , in standard Form (17), we see that its coeﬃcients, 3 /x ,6 /x 2 ,and 6 /x 3 are continuous on the speci±ed interval, which is x> 0. Next, substituting x , x 2 x 3 into the diferential equation, we veriFy that these Functions are indeed solutions. x 3 ( x ) 0 3 x 2 ( x ) 0 x ( x ) 0 6( x )=0 0+6 x 6 x , x 3 ( x 2 ) 0 3 x 2 ( x 2 ) 0 x ( x 2 ) 0 6( x 2 6 x 2 +12 x 2 6 x 2 , x 3 ( x 3 ) 0 3 x 2 ( x 3 ) 0 x ( x 3 ) 0 6( x 3 )=6 x 3 18 x 3 +18 x 3 6 x 3 . Evaluating the Wronskian yields W ± x, x 2 ,x 3 ² ( x )= ³ ³ ³ ³ ³ ³ ³ ³ xx 2 x 3 12 x 3 x 2 026 x ³ ³ ³ ³ ³ ³ ³ ³ = x ³ ³ ³ ³ ³ 2 x 3 x 2 26 x ³ ³ ³ ³ ³ ³ ³ ³ ³ ³ x 2 x 3 x ³ ³ ³ ³ ³ = x ( 6 x 2 ) ( 4 x 3 ) =2 x 3 . Thus W [ x, x 2 3 ]( x ) 6 =0on(0 , )andso { x, x 2 3 } is a Fundamental solution set For the given diferential equation. We involve Theorem 2 to conclude that
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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