<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>Bump......</p>
<p>This may or may not be make sense/be helpful:</p>
<p>I 1/(y-y^2) dy = I 1/x dx
I 1/[y(1-y)] dy = ln x</p>
<p>Integration by partial fractions:
1/[y(1-y)] dy = A/y + B/1-y
Multiply both sides by y(1-y):
A(1-y) + B(y) = 1
Let y=1 --> B = 1
Let y=0 --> A = 1</p>
<p>I 1/y dy + I 1/(1-y) dy = ln x
ln y ln (1-y) = ln x
y/1-y = x
y = x xy
y + xy = x
y(1+x) = x
y = x/(x+1) QED?</p>
<p>I actually figured it out-I proved it by doing this:</p>
<p>y=x/(x+C)
dy/dx= ((x+C)-(x))/((x+C)^2)
dy/dx=C/((x+C)^2)</p>
<p>Then:</p>
<p>dy/dx= ((x/((x+C)^2))-((x^2)/((x+C)^2)))
dy/dx= (x(x+C)-(x^2))/(x(x+C)^2)
dy/dx= ((x^2)+xC-(x^2))/(x(x-C)^2)
dy/dx= C/((x+C)^2)</p>
<p>Basically, I plugged in the original equation for the "y" in the derivative, and simplified it. This proved that y= x/(x+C) is the correct general equation because plugging it into the derivative equation resulted in the sam equation.</p>
<p>Your way works as well, timepiece, but this made more sense to me at the time.</p>