how do you integrate dy/(y(1-y))?

I got ln|y| + 1/y

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- Oct 28th 2008, 02:57 PMUnt0tIntegrating help
how do you integrate dy/(y(1-y))?

I got ln|y| + 1/y - Oct 28th 2008, 03:07 PMHallsofIvy
Excellent! How did you get that? I would be inclined to use "partial fractions" writing $\displaystyle \frac{1}{x(1-x}$ as $\displaystyle \frac{A}{x}+ \frac{B}{1-x}$. But integrating that should involve a "ln(1-x)" and I don't see that in your answer.

- Oct 28th 2008, 03:16 PMUnt0t
lol I don't get it. Could you please show me how to work it out and give me the answear

- Oct 28th 2008, 03:37 PMUnt0t
I'll actually give you the full question and hopefully someone can help me.

dy/dt= y(1-y) and they want us to find the general solution... I know its a seperable equation so I get dy/(y(1-y)) = 1 dt and we're ment to integrate it... though I dont know how. And the answer they give is y(t)= ke^t/(ke^t +1) - Oct 28th 2008, 03:57 PMMathstud28
$\displaystyle \frac{1}{x(x+1)}=\frac{x+1-x}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$

$\displaystyle \therefore\int\frac{dx}{x(x+1)}=\int\bigg[\frac{1}{x}-\frac{1}{x+1}\bigg]dx=\ln\left(\frac{x}{x+1}\right)+C$ - Oct 28th 2008, 04:13 PMKrizalid