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# Logistic equations (Part 1)

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.H (LO)
Finding the general solution of the general logistic equation dN/dt=rN(1-N/K). The solution is kind of hairy, but it's worth bearing with us!

## Want to join the conversation?

• Why did't Sal separate dN/dt as any other separable differential equation, would't that be easier ?
• It is not so much simpler as it is a notational preference. Some texts leave the derivative intact, whereas others separate the differentials. Either way gets the job done.
• I've simplified the fraction as Evan has done ((1/k)/(1-N/k) to 1/(k-N)) what gives me -ln(k-N). But ln(k-N) =/= ln(1-N/k) (is it ?). Could someone explain it ?
• Just stick with it. You're fine. Once you solve for the constant C with your initial conditions, your equation will look different but will be equivalent to Sal's.
• Couldn't you simplify the fraction (1/k)/(1-N/k) to 1/(k-N)? Wouldn't that be simpler to integrate?
• It's a matter of preference, but (1/k)/(1-N/k) is almost f'(N)/f(N), which is a commonly known integral (ln |f| + C). All you need to do is multiply by -1 and you can just use the f'/f integral directly. You can indeed just simplify to 1/(k-N), which produces the same result, but someone who integrates a lot probably remembers the f'/f integral by heart and spots it easily.
• I don't understand why the dN/dt terms just disappear when you integrated.
• Sal first recognized that population is a function of t, so you could write it as N(t). Then he said that it wold appear that (1/N)*(dN/dt) comes from ln|N|. To check this, he used implicit differentiation and the chain rule. The derivative of the outside function (the natural log function) is one over its argument, so he go 1/N. Then he had to multiply this by the derivative of the inside function (which is N(t) ) with respect to time, which is dN/dt. Using the chain rule you get (d/dt) ln|N| = (1/N)*(dN/dt). Sal used similar logic to find what the second term came from. So Sal found two functions such that, when you took their derivatives with respect to t, you found the terms that were on the left side of the differential equation. Since the left side of the differential equation came from taking the derivative of these two functions with respect to time, by taking the anti-derivative (the inverse of the derivative) with respect to time, Sal worked back to the "original" two functions.
• why is dln(N)/dt equal to dln(N)/dN * dN/dt?
when I enter it in wolframalpha i just get zero, what makes sense to me since the t^0 (that is basically next to the ln(N) turns into t^-1*0.
EDIT: I see you just multiplied the left side with dN/dN and crossed the parts of the fraction
another thing that i dont get is why at he replaces 1/N*dN/dt with ln(N) while on the right side it says 1/N*dN/dt ) = ln(N)* d/dt. how can the d/dt just be left out oO?
• Why is it frowned upon to algebraically manipulate differentials like any other variable?
• Was it necessary to go through all the steps at (on the right side with all the "respect to N" and "respect to t" stuff)? Couldn't Sal have simply moved the dt to the right hand side of the equation in the very beginning to make it "1 / (N * (1 - N/K) ) dN = r dt"? I mean we still got to the same solution but maybe people could have understood it better?
• At about , what happened to the -1/K in the numerator of the second term inside of parenthesis? Shouldn't that have become a constant when he took the anti-derivative of N with respect to t?
• -1/k is needed for the du of the u-substitution.
• For those having trouble, I found it much easier to first simplify 1/k / (1-N/K) into -1 / (N - K) because then the logarithm integration becomes immediately apparent. But this seems to lead me to a different answer to the equation than Sal solved for.
• 1/k / (1-N/K) simplified is actually 1 / (K-N), not -1 / (N - K) :)