72_pdfsam_math 54 differential equation solutions odd

72_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 (d) By dividing both sides by y 2 we lost the solution y 0. 31. Following the proof of Theorem 2, we come to the expression (10) on page 63 of the text for g 0 ( y ), that is g 0 ( y )= N ( x, y ) ∂y x Z x 0 M ( s, y ) ds (2.15) (where we have replaced the integration variable t by s ). In other words, g ( y ) is an antideriva- tive of the right-hand side in (2.15). Since an antiderivative is de±ned up to an additive constant and, in Theorem 2, such a constant can be chosen arbitrarily (that is, g ( y )canbe any antiderivative), we choose g ( y ) that vanishes at y 0 . According to fundamental theorem of calculus, this function can be written in the form g ( y y Z y 0 g 0 ( t ) dt = y Z y 0 N ( x, t ) ∂t x Z x 0 M ( s, t ) ds dt = y Z y 0 N ( x, t ) dt y Z y 0 x Z 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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