392_pdfsam_math 54 differential equation solutions odd

# 392_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 W 2 ( x )=( 1) 3 2 ± ± ± ± ± xx 1 1 x 2 ± ± ± ± ± =2 x 1 , W 3 ( x 1) 3 3 ± ± ± ± ± 5 15 x 4 ± ± ± ± ± =4 x 5 . Now we divide both sides of the given equation by x 3 to obtain an equation in standard form, that is, y 0 2 x 1 y 0 5 x 2 y 0 +5 x 3 y = x 5 . Hence, the right-hand side, g ( x ), in formula (1) on page 339 of the text equals to x 5 . Applying formula (11), page 340 of the text, yields v 1 ( x )= Z x 5 ( 6 x 3 ) 48 x 2 dx = 1 8 Z x 4 dx = 1 24 x 3 , v 2 ( x Z x 5 (2 x 1 ) 48 x 2 dx = 1 24 Z x 8 dx = 1 168 x 7 , v 3 ( x Z x 5 (4 x 5 ) 48 x 2 dx = 1 12 Z x 2 dx = 1 12 x
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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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