388_pdfsam_math 54 differential equation solutions odd

388_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 =( 1) n + k W [ y 1 ,...,y k 1 ,y k +1 n ]( x ) . Finally, ( 1) n + k 1) ( n k )+(2 k ) 1) n k . REVIEW PROBLEMS: page 344 1. (a) In notation of Theorem 1, we have p 1 ( x ) 0, p 2 ( x )= ln x , p 3 ( x x , p 4 ( x ) 2, and g ( x )=cos3 x . All these functions, except p 2 ( x ), are continuous on ( −∞ , ), and p 2 ( x ) is de±ned and continuous on (0 , ). Thus, Theorem 1 guarantees the existence of a unique solution on (0 , ). (b) By dividing both sides of the given di²erential equation by x 2 1, we rewrite the equation in standard form, that is, y 0 + sin x x 2 1 y 0 + x +4 x 2 1 y 0 + e x x 2 1 y = x 2 +3 x 2 1 . Thus we see that p 1 ( x sin x x 2 1 ,p 2 ( x x x 2 1 3 ( x e x x 2 1 , and g ( x x 2 x 2 1 . Functions p 1 ( x ), p 3 ( x ), and g ( x ) are de±ned and continuous on ( −∞ , ) except x = ± 1; p 2 ( x ) is de±ned and continuous on { x ≥− 4 ,x 6 = ± 1 } . Thus, the common domain for p 1 ( x ), p 2 ( x ), p 3 ( x ), and g ( x )is { x 4 6 = ± 1 } , and, in addition, these functions are
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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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