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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 , . . . , y 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 ) = cos 3 x . All these functions, except p 2 ( x ), are continuous on ( −∞ , ), and p 2 ( x ) is defined and continuous on (0 , ). Thus, Theorem 1 guarantees the existence of a unique solution on (0 , ). (b) By dividing both sides of the given differential equation by x 2 1, we rewrite the equation in standard form, that is, y + sin x x 2 1 y + x + 4 x 2 1 y + 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 + 4 x 2 1 , p 3 ( x ) = e x x 2 1 , and g ( x ) = x 2 + 3 x 2 1 . Functions p 1 ( x ), p 3 ( x ), and g ( x ) are defined and continuous on ( −∞ , ) except x = ± 1; p 2 ( x ) is defined and continuous on { x ≥ − 4 , x = ± 1 } . Thus, the common domain for p 1 ( x ), p 2 ( x ), p 3 ( x ), and
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