-Let's now attempt to find a solution for
the logistic differential equation.
And we already found some constant solutions,
we can think through that a little bit just
as a little bit of review from the last few videos.
So if this is the t-axis and this is the N-axis
we already saw that if N of zero, if a time equals
zero, or a population is zero, there is no one
to reproduce and this differential equation is
consistent with that, because if N is zero, this thing
is going to be zero, and so our rate of change
is going to be zero with respect to time,
so our population just won't change.
It'll just stay at zero.
Which is nice because that's what would
actually happen in a real population.
That gives us the constant solution that N
of t is equal to zero, that's one solution
of this differential equation, not that interesting.
A zero population will never grow or change.
The other constant solution is what if our
population started at the maximum of what
the environment could sustain, and in that
situation this term is going to be K over K,
which is one.
One minus one is zero, and so in there, the
population would also not change.
It would just stay at K, and so the rate of change
would just stay at zero.
So that's another constant solution, that
we start at the maximum population and then
this differential equation tells us
a scenario that never changes.
But we said, hey look, there could be something
interesting that happens if our initial condition,
which we called N sub-not, that's our N of zero.
Time equals zero if we started someplace in between,
maybe closer, much below our eventual ceiling
I guess, because if we're a lot below our
eventual ceiling this thing is going to be
a small fraction, this thing is going to be
close to one, and so the rate of change is going
to be pretty close to being proportional to N.
When we're thinking that, it might look something
like that, as N increases, our rate of change
increases, but then as N approaches K this thing
is going to approach one, this thing is going
to approach zero, and so it's going to overwhelm
this and our rate of change is going
to approach zero.
So we can imagine a scenario where we
asymptote towards K.
Let's see if we can solve this to find this,
this N of t to find the actual analytic
expression for this N of t right over here.
Because this is interesting, this is what could
be used to model populations that would be
more consistent with a Malthusian mindset.
So let's see if we can do that.
To do that we just have to realize this is a
separable differential equation, and we're
assuming is a function of d, we're going to
solve for an N of t that satisfies this.
So we just have to separate it from the explicit
ts, but there are no explicit ts here, so
it's quite easy to do.
So what I'm going to do is I'm going to take
this part right over here, and I'm going to divide
both sides by that.
I'm going to leave the r on the right hand side,
it'll make things, I think, a little bit easier
as we try to solve for N of t.
So let me just do that.
So this is going to be equal to one over N
times one minus N over K.
One minus N over K times dN dT, times dN dT
is equal to r.
Another way we could think about it, well
actually, let me just continue to tackle it this way.
So we get that, and now what I want to do is
take the anti-derivative of both sides with
respect to t.
Well, this is pretty straightforward.
That's just going to be rt times some constant,
but what's this going to be?
This is kind of this messy thing.
And maybe if we could break this out, if we
can expand this into two fractions, do a little
bit of partial fraction expansion, maybe
we can come up with an expression that's
a little bit easier to find the anti-derivative of.
I'm hoping that I can find an A and a B
where A over N plus B over one minus N over K
is equal to this business.
Is equal to one over N times one minus N over k.
Let's see if we can find an A and a B.
This is just partial fraction expansion, if this
looks unfamiliar to you I encourage you to
review that part on Khan Academy.
Do a search for partial fraction expansion
on Khan Academy.
So how do we do it?
We'll just add these two right over here.
If we take the sum, right over here, this is
going to be A times this, which is A minus
A over K N plus B times this, plus BN, over
the product of these two.
So it's just going to be N times one
minus N over k.
One way to think about it, I just multiplied
the numerator and the denominator of this one
by one minus N over k, so A times one minus
N over k is that, and times one minus N
over k is that, and I multiplied the numerator
and the denominator of this both by N, BN
and N times that, and then I added the two
now that I had the same denominator.
So that's just adding fractions
with unlike denominators.
It's going to be equal to this, one over
N times one minus N over K.
Now we can try to think about what is
A and B going to be equal to?
What can they be equal to?
Well I have a constant term here.
I don't have any N term here.
I could say that maybe I have a zero times
an N, and that actually helps things a little bit,
because maybe we could say that this thing,
that this plus this, which are the coefficients
on N, are going to sum up to zero, and that
this, which is rA, is going to be equal to one.
That's pretty nice.
So we can say that A is equal to one, A is
equal to one, and that if A is equal to one
we have negative one over K, negative one over K,
plus B is equal to zero.
Well what's B going to be?
Well B would be equal to one over K.
So we can re-write this as, we can re-write it
as one over N plus one over K,
plus one over K over,
let me just do this,
over one minus N over K,
and then times dN dt, dn dt
is equal to r.
So I just did a little partial fraction expansion,
and I like giving myself a little bit of
real estate, let me clear this out.
You can rewind the video and review that,
if you find it necessary,
so let's do that.
So how does this help us?
Well, hey, you probably might be recognizing
the anti-derivative of one over N, and you might
even see this.
So let's just think through this a little bit.
We know the anti-derivative of one over N
is a natural log, just did a new color,
is a natural log of the absolute value of N.
We can see the derivative of that with respect
to N is equal to one over N, and if we were
to find the derivative of this with respect
to t, derivative with respect to t of the natural
log of the absolute value of N is equal to,
that's going to be the derivative of this
with respect to N times the derivative of N
with respect to t, times dN dt.
We could do the same thing over here.
Notice, I have an expression.
What would be the derivative of this expression
right over here?
It would be negative one over K, if I'm
talking the derivative with respect to N,
it would be negative one over K.
I have a positive one over K here, and I could
even make it negative.
I can turn this, I can clear this out of the
way, instead of having a positive there I
could have, I guess I could say a double negative.
So a double negative, I haven't changed the
value of if, and notice, now this is the derivative
of this, which makes it very nice for u-substitution,
or you might be already used to doing this thing
in your head, and so we know that the derivative
with respect to N.
Let me write this in a new color.
We know that the derivative with respect to N of the
natural log of one minus N over K, once again this
comes straight out of the chain rule, it's
going to be the derivative of this with respect
to N, which is negative one over K times
the derivative of this whole thing with respect
to this thing, which is times one over
one minus N over K, which is exactly what
we have over here.
But if I were to take the derivative of this
with respect to t, so the derivative with
respect to t of the natural log of one minus
N over K, it's going to be the derivative
of this with respect to N, which is that,
which is what we just found.
It's going to be this.
So copy and paste.
It's going to be that, it's going to be that.
Times the derivative of N with respect to t,
just straight out of the chain rule, dN dt.
Well notice, we have a dN dt, we can multiply
it times each of these things.
Actually, why not, let's just do that
just to make it clear, cause this really
isn't so much differential equations,
but sometimes some of the calculus that we
learned not too long ago, and even frankly
some of the algebra, it's nice to not skip steps.
So if I distribute this, edit, paste.
Woops, I wanted to copy and paste.
So, copy and paste, so I have that, and then
I have, I can copy and paste, so that I'm just
distributing this business right over here.
So I get that, now let me clean it up a little bit.
So I have that right over here, and of course
we have this being equal to r.
If I take the anti-derivative of this with respect
to t, well I'm just going to get this
minus this right over that.
So let's do that.
Let's take the anti-derivative with respect to t,
so the left hand side I'm going to get the
natural log, the anti-derivative of that
with respect to t is the natural log of
the absolute value of N, and then minus
the anti-derivative of this with respect to t
is the natural log of the absolute value of
one minus N over K is equal to,
oh, and let's not forget we're going to
have some constant here, I want to find a pretty
general solution, let's call it C one, is
equal to the anti-derivative of this with respect
to t is r times t, maybe plus some other constant,
plus some other constant,
just like that.
Now I'm going to assume that my N of t meets
this assumption right over here.
So I'm going to assume,
assume N of t is going to be less than K
and greater than zero.
That means that this thing, this N is always going
to be positive, and if N is always between
zero and K, that means that this thing is
always going to be positive.
So that helps me clear things up a little bit.
Actually let me do it in this color.
That helps me get rid of that,
helps me get rid of that.
I can instead throw some parentheses here,
and why don't I subtract the C one from both sides.
That would get rid of it here, so edit, cut,
and paste, so I have the C one there.
I know I'm overwriting my own work, which
is not making it look as clean as possible.
I'm just trying to clean it up a little bit.
I have an arbitrary constant that I haven't really
solved for yet minus another constant.
Let me just call this some other constant,
let me just call it a general C.
So let me just call this C,
so clear, and I'm just going to call that C.
I am just going to call that C.
And I could even simplify this a little bit,
but actually I just realized I'm thirteen
minutes into this video, which is longer than
I like to do.
So let's continue this in the next video.
I was getting excited because I'm so close.
I'm so close to solving for an N of t
that satisfies our logistic differential equation.
Very, very exciting.