The Logistic Equation and the Analytic Solution

Tin Tức

Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! 🙂 !! The Logistic Equation and the Analytic Solution. In this video, I find the analytic solution to the logistic differential equation.


38 thoughts on “The Logistic Equation and the Analytic Solution

  1. Okay I found your channel when I was in algebra 2 and now I’m in calc and you still have clear and helpful videos for that, that’s pretty amazing. Thank you!

  2. Great derivation, but representation of the Logistic formula is hard to reconcile with other
    generally presented version of the formula.

  3. Hey man, thanks for the video! I know its old but could you explain why you remove the absolute and then say the A can be + or -?

  4. Thank you for this simple explanation! Unfortunately textbooks fail to understand that if they explained things like you just have – that it would build our confidence and help get our minds around more complex problems.

  5. Dude, At Khan Academy Sal tries to show how to solve the thing, too. However, for some reason he is not being all that clear about it and, according from what can be picked up from the comments, many people are left confused. You are doing so much better of a job here, so I suggest you send in this clip as a basis for improvement to K.A.!

  6. Great and very helpful but… you should learn how to write with your right hand…
    That would solve the problem of covering what has been just done 🙂

  7. Thank you for yet another brilliantly explained, concise video. So well explained – I had never done partial fractions before and it made everything so simple. Keep up the good work!!

  8. He did that to help with separating P from the equation at a later stage. But an answer can be obtained by various ways.
    I tried solving for P(t) WITHOUT multiplying by (-1) and initially got this: [P/(K – P) = Ae^kt] –> Now in order to completely separate P, I'd have to take the inverse of both sides of this equation: [(K-P)/P = A.e^-kt] –> [(K/P) – 1 = A.e^-kt] –> P = K/(A.e^-kt+1), which is Patrick's solution. Hope that clears things out (:

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