My Math Forum Differential Equation Problem 2

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 July 18th, 2015, 12:58 PM #1 Senior Member     Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 Differential Equation Problem 2 Can someone help me spot what I'm doing wrong? Problem: $\displaystyle z=x-y$ $\displaystyle \frac{dy}{dx} = \frac{x-y+3}{x-y-3}$ My attempt: $\displaystyle \frac{dz}{dx} = 1- \frac{dy}{dx}$ $\displaystyle \frac{dy}{dx} =1 - \frac{dz}{dx}$ $\displaystyle 1 - \frac{dz}{dx} = \frac{z+3}{z-3}$ $\displaystyle - \frac{dz}{dx} = \frac{z+3}{z-3} -1$ $\displaystyle - \frac{dz}{dx} = \frac{6}{z-3}$ $\displaystyle \frac{z-3}{6} * -dz = dx$ After Integrating, $\displaystyle - \frac{z^2}{12} + \frac{3z}{6} = x+A$ $\displaystyle - \frac{(x-y)^2}{12} + \frac{3x}{6} + \frac{3y}{6}$ Last edited by skipjack; July 18th, 2015 at 08:13 PM.
 July 18th, 2015, 02:11 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,377 Thanks: 2474 Math Focus: Mainly analysis and algebra You are missing the right-hand side of the final line. Last edited by skipjack; July 18th, 2015 at 08:13 PM.
 July 18th, 2015, 02:49 PM #3 Senior Member     Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 Oh right, sorry, that would be $\displaystyle = x + C$ but my final answer doesn't match with the book I am using. :P Last edited by skipjack; July 18th, 2015 at 08:14 PM.
 July 18th, 2015, 04:19 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,377 Thanks: 2474 Math Focus: Mainly analysis and algebra Subtract $x$ from both sides? What answer are you looking for? Last edited by skipjack; July 18th, 2015 at 08:11 PM.
July 18th, 2015, 08:20 PM   #5
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Quote:
 Originally Posted by BonaviaFx ... what I'm doing wrong?
In your final line, you have replaced the second occurrence of $z$ with $x + y$ instead of $x - y$.

 July 19th, 2015, 03:29 AM #6 Senior Member     Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 The answer on the book is: $\displaystyle \frac{(x-y)²}{2}+ 3x + 3y = C$
July 19th, 2015, 08:45 AM   #7
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Quote:
 Originally Posted by BonaviaFx \ $\displaystyle - \frac{(x-y)^2}{12} + \frac{3x}{6} - \frac{3y}{6} = x + C$
Multiply both sides by -6...

-Dan

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