Prelim_1_Solns - MATH 293 Prelim 1 Problem 2 Solution We...

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Unformatted text preview: MATH 293 Prelim 1 Problem 2 Solution We are to find the general solution to the DE dy dx = p x + y + 1 . Since this DE is neither linear nor seperable we will need to use a clever substitution. There are several different (but equivalent) ways: Method 1: Set v = x + y + 1. Then by differentiating both sides we get dv dx = d dx ( x + y + 1) = 1 + dy dx . Therefore the DE for v is dv dx = v + 1 which is a seperable equation. We get Z 1 v + 1 dv = Z dx = x + C. To evaluate the integral R 1 v +1 dv we use the substitution u = v , i.e. u 2 = v , which gives 2 udu = dv . Thus Z 1 v + 1 dv = Z 2 u u + 1 du = 2 Z ( u + 1)- 1 u + 1 du = 2 Z 1- 1 u + 1 du = 2( u- log( u + 1)) = 2( v- log( v + 1)) = 2( p x + y + 1- log( p x + y + 1 + 1)) . Therefore the general solution is given implicitly by the equation 2( p x + y + 1- log( p x + y + 1 + 1)) = x + C where C is some constant....
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This note was uploaded on 11/02/2008 for the course MATH 2930 taught by Professor Terrell,r during the Fall '07 term at Cornell University (Engineering School).

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Prelim_1_Solns - MATH 293 Prelim 1 Problem 2 Solution We...

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