AP Calc question

<p>differential equation- dy/dx = (y-y^2)/x for all x cannot = 0</p>

<p>Verify that y = x/(x+C) , x cannot = -C is a general solution for the given differential equation and show that all solutions contain (0,0).</p>

<p>I understand you must separate the variables to find the original equation, but how do you do this with the y in its current form?</p>

<p>first you seperate:</p>

<p>dy = (1/x) * (y - y^2) *dx
(dy)/(y - y^2) = (dx)/(x)
rewrite:
dy * (y - y^2)^-1 = dx * (x)^-1
Integrate both sides</p>

<p>ln|y - y^2| = ln|x| + C
e^(ln|x| + C) = (y - y^2)
e^ln|x| * e^C = (y - y^2)
C * x = -y^2 + y
-(C * x) = y^2 - y
C<em>x - .5 = y^2 - y -.5
C</em>x -.5 = (y - .5)^2
sqrt(C<em>x -.5) = y - .5
y = sqrt(C</em>x - .5) + .5</p>

<p>This is the answer that I got but I think I went wrong somewhere hope this helps.</p>