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#### find_the_fun

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- Feb 1, 2012

- 166

\(\displaystyle \frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}\)

So separate x and y terms

\(\displaystyle (xy-2x+4y-8) dy = (xy+3x-y-3)\) ugh I'm stuck

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- #1

- Feb 1, 2012

- 166

\(\displaystyle \frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}\)

So separate x and y terms

\(\displaystyle (xy-2x+4y-8) dy = (xy+3x-y-3)\) ugh I'm stuck

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- Feb 1, 2012

- 166

I can factor to \(\displaystyle \frac{dy}{dx}=\frac{(x-1)(y+3)}{(x+4)(y-2)}\) and rewriting gives \(\displaystyle \frac{(y-2)}{(y+3)} dy = \frac{(x-1)}{(x+4)} dx\). Am I on the right track? I don't know how to integrate this.

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- Feb 1, 2012

- 166

Is an alternative to \(\displaystyle \frac{y+3-5}{y+3}\) doing polynomial division and seeing y+3 goes into y-2 once with a remainder of 5? I'm not super clear on the thought process of getting \(\displaystyle 1-\frac{5}{y+3}\).Yes, you are correct. For the left side, consider:

\(\displaystyle \frac{y-2}{y+3}=\frac{y+3-5}{y+3}=1-\frac{5}{y+3}\)

Do the same kind of thing on the right side, and you should be able to integrate now.

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Yes, although the remainder is actually -5, but then you get the same result. I just find it simpler to do as I did above. To make what I did more clear, consider:Is an alternative to \(\displaystyle \frac{y+3-5}{y+3}\) doing polynomial division and seeing y+3 goes into y-2 once with a remainder of 5? I'm not super clear on the thought process of getting \(\displaystyle 1-\frac{5}{y+3}\).

\(\displaystyle \frac{y-2}{y+3}=\frac{(y+3)+(-2-3)}{y+3}=\frac{y+3}{y+3}-\frac{5}{y+3}=1-\frac{5}{y+3}\)