Chapter 2(d)By dividing both sides byy2we lost the solutiony≡0.31.Following the proof of Theorem 2, we come to the expression (10) on page 63 of the text forg0(y), that isg0(y)=N(x, y)−∂∂yxZx0M(s, y)ds(2.15)(where we have replaced the integration variabletbys). 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 additiveconstant and, in Theorem 2, such a constant can be chosen arbitrarily (that is,g(y)canbeanyantiderivative), we chooseg(y) that vanishes aty0. According to fundamental theoremof calculus, this function can be written in the formg(yyZy0g0(t)dt=yZy0N(x, t)−∂∂txZx0M(s, t)dsdt=yZy0N(x, t)dt−yZy0∂xZx
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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.