I know this is a year late reply but he does use separation of variables. First thing he does. He then does partial fraction decomposition to help find the anti derivative on the LHS of the equation.
pretty much he linearized the whole integral that was created by using partial fractions and broke that big integral up into two smaller ones like this: 1/n + (1/k)/(1-n/k)dn/dt = r Is the same as ↳ ∫1/n +(1/k)/(1-n/k)dn = ∫r dt
Connor Mooneyhan he wasn't. He was just showing you that you CAN take antiderivatevs with respect to t as well because T is a dependent variable of N. He showed you both options.
Well at 6:33, I didnt use the partial fraction like you, I used 1/(K-N) instead of using (1/K)/(1-(N/K)). Here is the problem ; mathematically they are exatly the same, but I got different answer, my calculations are right just like yours. Aren't they same expressions, if not, can you tell me why not? I found the equation as follows : N(t)=((KN.)/(e^(-rt)K+N.)) yours is N(t)=((KN.)/((K-N.)e^(-rt)+N.)) only different thing between our functions is you have coefficent (K-N.) to e^(-rt), mine has a coefficient K
Great video. Just have one question: why don't you use separation way just like what you did in the previous videos? Thanks.
I know this is a year late reply but he does use separation of variables. First thing he does. He then does partial fraction decomposition to help find the anti derivative on the LHS of the equation.
was completely with you until 7.20m havent a clue after that!
pretty much he linearized the whole integral that was created by using partial fractions and broke that big integral up into two smaller ones like this: 1/n + (1/k)/(1-n/k)dn/dt = r
Is the same as ↳ ∫1/n +(1/k)/(1-n/k)dn = ∫r dt
@@marioleon4128 I did help! even though this is two years later haha
Holy cow this is loooong and complicated.
if we regulate 1/(1-N/K) as K/(K-N) then we can see K/N(K-N) = 1/N - (-1)/(K-N). And the solution will be easier.
Yeah. The problem can be solved way faster this way.
helal meltem as bayrakları
11:15 The integral of (-1/k)/(1-n/k) with respect to n is 1/k * Ln(1-n/k), note the 1/k coefficient. Why does yours not have a coefficient?
The integral of (-1/k)/(1-n/k) IS -ln(1-n/k). If you use u = 1-n/k you should get integral of - du/u = - lnu
I was just about to say. He’s missing it.
@@athenovae I have since realised that what I said is wrong. the 1/(1-n/k) produces a factor of k at the front. This cancels the -1/k.
11:21 I think you’re missing 1/k in front of the second ln.
Geat job how you did this partial fractions expansion.
Hans Peter Roth
yep. I had completely forgotten that was a thing
We did a video (featuring the logistic equation) about where equations come from in social science: th-cam.com/video/zXY6LEwHgaI/w-d-xo.html
how is he taking derivative of terms with no "t" in them w.r.t "t" and not making them 0???
At 7:30, I'm very confused. Why are we taking the antiderivative with respect to N and then taking the antiderivative of that with respect to t?
Connor Mooneyhan he wasn't. He was just showing you that you CAN take antiderivatevs with respect to t as well because T is a dependent variable of N. He showed you both options.
thank you once again sir....sorry I haven't said that in a while.
goooood video
Is separable equation with an implicit or explicit solution not applicable in this situation??
1337, nice
Thanks, it helps me lot.
can you help me with (computing the poincare map )?
and thank you for this video its helpful
What is demographic explanation of K
nft haha
mathematics is so versatile, is like water taking the shape of its vessel.
Im stuck at 8:44, I don't get that Ln derivative.
derivative of ln x is 1/x
Thank you!
haha part 1 isn't showing in my inbox, yet here it is :)
Well at 6:33, I didnt use the partial fraction like you, I used 1/(K-N) instead of using (1/K)/(1-(N/K)). Here is the problem ; mathematically they are exatly the same, but I got different answer, my calculations are right just like yours. Aren't they same expressions, if not, can you tell me why not? I found the equation as follows :
N(t)=((KN.)/(e^(-rt)K+N.)) yours is N(t)=((KN.)/((K-N.)e^(-rt)+N.)) only different thing between our functions is you have coefficent (K-N.) to e^(-rt), mine has a coefficient K
Facing the same problem after more than 4 years man.
I'm in pain
terrible
brilliant
Much easier to solve if you know how to solve bernoulli equations.
Calculus sucks