Have a look at our second example.
We mentioned lots of times
through this course
that differentiation helps you find the
growth and decay of different objects.
So be it a bacteria or chemical or a tumor,
you can model accurately the growth
and the decay of those objects.
So we’ve been given the
growth of a certain bacteria,
and we’ve been told that the growth
rate is dN by dt equals to 2N over T.
So if I just write that
here, we have dN by dt
equals to 2N over t.
And it says T is the
time in minutes,
but the question is asking
us to find N in terms of T.
Now, let’s just understand
this for a minute.
Now that you’ve done
a calculus course,
you can actually analyze
what this is saying.
So if you look at this
it says dN by dt is giving
you some type of a value.
Now, if you look at this closely, dN by dt
is giving you the rate of change.
So you’re saying that the growth
rate, N, is changing over time,
and it’s changing
with that function.
So you can say that it’s almost proportional
to N because you could see here,
if I just use this,
you can say that dN by dt is
proportional to this N value here.
Which really is saying that
the growth rate is growing.
So the rate of growth is
increasing as N is increasing.
So you can do some analysis
on this, just looking at it.
But the question is asking
us to find N in terms of t.
Let me just explain
what they’re wanting.
We know the differential equation so it’s
telling you the rate of change of N with t.
So we know what the rate of change is.
But the question is asking you
to find the original equation.
So where does this rate of
change actually come from?
It’s quite common in real life for us
to start with differential equations,
to start with rate of change and then go
back to finding the original equation
for ourselves or for
And this is exactly that
kind of a question.
It’s asking you to solve
a differential equation.
Now, hopefully will you remember
that the opposite of differentiation
or in order to solve differentiation
or differential equations,
we need to do the opposite
And if you think really hard,
you’ll remember that the opposite method
of differentiation is integration.
We’ll have to use a slightly
different technique on this.
And this technique is known
as separation of variables,
which you can later develop
by just studying integration.
But it’s fairly straightforward, so
I’ll just show that to you here.
We have dN by dt if I just
rewrite it so we can see
as 2N over t.
We have a slight problem here
because you have dN on this side,
but you have the N
function on this side.
And then you have the dt on this side, but
you have the t function on the other side.
This problem can easily be resolved.
All you have to do is swap places.
Take all your t's to one side and
take all your N's to one side.
And the process is known as
separation of variables.
You’ve got 2 variables, you can separate
them out on each side of the equation.
Let me show you what I mean.
So I’ll bring that 2 –
Not 2, I’ll leave the 2 there.
I’ll bring the N down so I’ll
rewrite this as dN over N.
It was multiplying here, so
you can divide on this side.
And on the other side,
I still have 2t left,
and you can take the dt to the other side.
So the dt is dividing here,
so when you move it to the
other side, it’s multiplying.
You can see that what you’ve done here is
created two integrals for yourselves.
So if I put this little
integration signs on both sides,
you’ll be able to see that you
have an integration question here
and an integration
Each with their own respective variables.
So if I rewrite this in a
way that we can see this,
we have 1 over N dN
equals to 2 over t dt.
And all we have to do now is
integrate them separately.
So you do one integral and you do the other
integral with respect to their variables.
Think back at what this is now.
1 over N dN.
Can you remember what
result this gave you?
If I remind you that when you have
ln of x, and you differentiate it,
you get 1 over x.
So think of what’s going to happen
when you have to integrate 1 over X,
that gives you ln of x.
So 1 over x integrates
to give you ln of x.
And so all you have to do
is change the variable.
Rather than X, let’s use N.
1 over N dN integrates
to ln of N.
On the other side, you can
take the 2 out of the integral
and write it as T dt.
And you’ll see that the other side
also gives you an ln function.
So we have ln of N
equals to 2 ln of t.
And don’t forget to add a plus C there.
The question said express N in terms of t,
so we don’t quite have N by itself here.
So what we can do is solve this equation.
In order to get rid of ln’s,
you can E both sides,
so exponentials are the opposite of ln’s.
So we can do e to the ln of N
equals to e to 2ln of t,
plus e to the power of C.
So I’ve exponentiated
every term here.
You could now rewrite this
because e and ln cancel out.
You could write this side
as N, which is good news.
This side doesn’t quite cancel out
yet because you have a 2 at front.
Let me just show that to you on the side.
So we’ve got e 2ln of t.
Now, if it was e and ln next to each
other, you could cancel it out,
but you have this 2.
Think of the rules of logs that we
spoke about through this course
and this is the third rule of log,
which allows you to take the power up.
So you could rewrite this as
e to the ln of t squared.
You can take the 2 up and at
which case, the e in ln will
just be the inverse of each
other so they just disappear.
So this leaves you with
t squared plus e to C.
e to the C is just any number,
it’s just a constant,
so you could just write it as N equals
to t squared plus let’s just call it C.
And so we found our equation.
So the equation that this
differential must have come from
would look like this, we have
N equals to t squared plus C.
C, remember we mentioned
before that you can
find out when you know
some initial conditions.
So if I was told what the time
was when N is a certain number,
I can substitute it into this
equation and calculate my C.
But if you’re not given a C or a
constant, you just leave it there.
It could be zero or it
could be another number,
but we don’t want to risk just
removing it and assuming it’s zero.