How to prove dn/dt=r(1-n/k)n, n(0)=No equal to n=n(t)=Nok/No+(k-No)e^-rt

Separate variables to transform the equation to

[1 / (n*(1 - (n/k))] dn = r dt and then integrate both sides.

For the LHS we have

integral((k / (n*(k - n))) dn). Now note that k/(n*(k - n)) = (1/n) + (1/(k - n)),

so the integral becomes

integral[((1/n) + (1/(k - n))) dn] =

(ln(n) - ln(k - n)) + C = ln(n / (k - n)) + c.

The RHS integrates to just r*t + b, so after combining constants b - c = C we have

ln(n / (k - n)) = r*t + C. Now n(0) = k*ln(No / (k - No)) = C, so we now have

(ln(n / (k - n)) = r*t + k*ln(No / (k - No)) ----->

ln(No / (k - No)) - ln(n / (k - n))] = - r*t ----->

ln[No * (k - n) / ((k - No) * n)] = - r*t ----->

No * (k - n) / ((k - No) * n) = e^(-r*t) ---->

No*k - No*n = n*(k - No)*e^(-r*t) ---->

No*k = n*(No + (k - No)*e^(-r*t)) ----> n = No*k / (No + (k - No)*e^(-r*t)).

Sorry, I messed up those two lines while I was editing my answer. they should read

"ln(n / (k - n)) = r*t + C. Now ln(No / (k - No)) = r*0 + C = C, so we now have

ln(n / (k - n) = r*t + ln(No / (k - No)) -----> " and then the rest of the solution

remains as it is. Good catch. :)

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