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I'm trying to find the general solution for

$(x+2)y' = 3-\frac{2y}{x}$

This is what I've done so far:

$y'+\frac{2y}{x(x+2)}=\frac{3}{x+2}$

$(\frac{x}{x+2}y)'=\frac{3x}{(x+2)^2}$

$\frac{x}{x+2}y=3\int \frac{x}{(x+2)^2}dx$

$\frac{x}{x+2}y= 3\int \frac{1}{x+2}dx - 6\int\frac{1}{(x+2)^2}dx$

$\frac{x}{x+2}y= 3 ln|x+2| + \frac{6}{x+2}+c$

I tested this solution for when c=0, but it failed. Can anyone spot my mistake?

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    Loo$k$s fine. Did you finish solving for y?2012-09-19

0 Answers 0