I am trying to evaluate integral of the form $dN/(N(k-N))$, where $N(t)$ is function of $t$, I know that without factor of $N$, result will be $\ln(t)$, but what can I do with another factor $N(t)$? please help me to solve it
solution of differential equation
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integration
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0See also http://math.stackexchange.com/questions/78560/how-do-you-solve-the-initial-value-probelm-dp-dt-10p1-p-p0-0-1 – 2012-01-25
2 Answers
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From ${dN\over dt}={rN(k-N)\over K}$ you get $\int\,dt=\int{K\;dN\over rN(k-N)}$ On the left, that's just $t$, and on the right, use Davide's comment to get $t=\int{K\;dN\over rkN}+\int{K\;dN\over rk(k-N)}={K\over rk}\left(\int{dN\over N}+\int{dN\over k-N}\right)$ Now, can you do those two integrals?
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Use partial fractions: $ \frac{1}{N(k-N)} = \frac {1/k}{N} + \frac{1/k}{k-N}. $ (And don't forget to apply the chain rule when you find the antiderivative of the second fraction.)
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0(It was correct up to a constant.) – 2012-01-25