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\[x(x+y)y \prime-y(x-y)=0\]

what? lol

i tried substituting u = y/x

im sort of messing up in the algebra part i think.

i got \[(y/x)((y/2x)+1)=c\]

\[x(x+y)y'-y(x-y)=0\\y'=\frac{y(x-y)}{x(x+y)}\\\quad=\frac{y(1-y/x)}{x(1+y/x)}\]

now substitute u = y/x

oh ok thanks!