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Prelim_1_Solns

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

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MATH 293 Prelim 1 Problem 2 Solution We are to find the general solution to the DE dy dx = 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 1 v + 1 dv = dx = x + C. To evaluate the integral 1 v +1 dv we use the substitution u = v , i.e. u 2 = v , which gives 2 udu = dv . Thus 1 v + 1 dv = 2 u u + 1 du = 2 ( u + 1) - 1 u + 1 du = 2 1 - 1 u + 1 du = 2( u - log( u + 1)) = 2( v - log( v + 1)) = 2( x + y + 1 - log( x + y + 1 + 1)) . Therefore the general solution is given implicitly by the equation 2( x + y + 1 - log( x + y + 1 + 1)) = x + C where C is some constant. 1
Method 2: Set v = x + y + 1. Then by differentiating both sides we get dv dx = d dx x + y + 1 = 1 2 x + y + 1 1 + dy dx . Therefore the DE for v is dv dx = v + 1 2 v which is a seperable equation. We get 2 v v + 1 dv = dx = x + C. Note that the integral on the left hand side of the equation is the same integral we encountered in method 1, where we derived 2 v v +1 dv = 2( v - log( v + 1)). The solution to the original DE is therefore given implicitly by 2( x + y + 1 - log( x + y + 1 + 1)) = x + C. 2

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MATH 293, Prelim – 1 Partial Marking scheme with solution for Problem 5
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Prelim_1_Solns - MATH 293 Prelim 1 Problem 2 Solution We...

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