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July 18th, 2015, 12:58 PM   #1
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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.
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July 18th, 2015, 02:11 PM   #2
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You are missing the right-hand side of the final line.

Last edited by skipjack; July 18th, 2015 at 08:13 PM.
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July 18th, 2015, 02:49 PM   #3
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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.
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July 18th, 2015, 04:19 PM   #4
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Subtract $x$ from both sides? What answer are you looking for?

Last edited by skipjack; July 18th, 2015 at 08:11 PM.
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July 18th, 2015, 08:20 PM   #5
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Quote:
Originally Posted by BonaviaFx View Post
... what I'm doing wrong?
In your final line, you have replaced the second occurrence of $z$ with $x + y$ instead of $x - y$.
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July 19th, 2015, 03:29 AM   #6
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The answer on the book is:

$\displaystyle \frac{(x-y)²}{2}+ 3x + 3y = C $
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July 19th, 2015, 08:45 AM   #7
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Quote:
Originally Posted by BonaviaFx View Post
\
$\displaystyle - \frac{(x-y)^2}{12} + \frac{3x}{6} - \frac{3y}{6} = x + C$
Multiply both sides by -6...

-Dan
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