- [Narrator] Welcome to a lesson on determining
the intervals for which a linear first order
differential equation would have unique solutions
as well as the interval for which the differential equation
would have a unique solution containing an initial value.
If we have a first order differential equation
in this form here with this initial condition
if P of x and f of x are continuous on an open interval
from a to b, then there exists a unique solution
for every x in the interval.
So it follows that if the interval contains x sub zero
from the initial condition then there exists
a unique solution on the interval that satisfies
the initial value problem.
So a couple things to notice here...
This is function P of x.
This is function f of x and x sub zero is the x value
from the initial condition.
Also notice this does not tell us what the solution is
or how to find it, it just tells us
if there's a unique solution.
Also y sub zero from the initial condition does not affect
the interval and the interval containing x sub zero
is sometimes called the interval of validity.
So the main idea is we want to find the interval
on which both P of x and f of x are continuous and this
will be the interval for which unique solutions exist.
So let's take a look at some examples.
We want to find the intervals for which
the DE has unique solutions.
Then state the interval containing the initial condition.
The first step is to recognize that we do have
a linear first order differential equation.
However, it's currently not in the correct form
or the form given here below.
We want our first term here to be dy dx.
So we're going to divide everything by x to begin with.
So we'll have the differential equation dy dx plus,
I'm going to write this as three divided by x times y
so it fits this form equals four x.
Now we should recognize that P of x is equal
to three divided by x and f of x is equal to four x.
Now we'll find where P of x is continuous,
find where f of x is continuous,
and then find the intersection of those two intervals.
Well, for P of x, we know x can't equal zero
because we'd have division by zero.
So P of x is continuous from negative infinity
to zero or from zero to infinity.
F of x is continuous for all real values of x.
Therefore, we can say that f of x is continuous
from negative infinity to positive infinity.
So now our main goal is to find the interval
for which both of these functions are continuous
or the intersection of these two intervals.
Well, the intersection of these two intervals
would just be the interval for which P of x is continuous
or this interval here.
Which means this is the interval for which
the original differential equation
would have unique solutions for all values of x
in this interval.
So that's the first part of the question
and now if we look at the given initial condition,
we have y of one equals two.
So we want to find the interval containing x equals one
which would be the second interval here
and because it's in this interval,
we know that this initial value problem is going
to have a unique solution.
Remember these intervals represent x values only.
Let's take a look at another example.
Same question, different differential equation.
So the first step is to put the differential equation
in the correct form.
So here we're going to divide everything
by the quantity x minus three.
So P of x is going to be equal to natural log x
divided by the quantity x minus three
and f of x is going to be equal to two x
divided by the quantity x minus three.
Now we'll start by determining where P of x is continuous.
Now there are two things to consider here.
We know x can't equal three because we'd have division
by zero but the domain for natural log x is when x
is greater than zero or the interval from zero to infinity.
So P of x will be continuous on this interval
as long as x doesn't equal three.
So P of x is continuous on the open interval
from zero to three or from three to infinity.
So it's where natural log x is continuous
except we must exclude three.
Now, looking at f of x, the numerator is a linear function
which is always continuous but notice
we would have division by zero when x equals three.
So we know x can't equal three
if we want f of x to be continuous.
So f of x is continuous from negative infinity
to three or from three to infinity
which means the differential equation
will have unique solutions on the intersection
of these two intervals or when both P of x
and f of x are continuous.
So comparing these intervals, notice that the intersection
of these two intervals would just be the interval
for P of x.
We would have to exclude the interval
from negative infinity to zero.
So we'd have the open interval from zero to three
or the open interval from three to infinity.
Now going back to our initial condition,
we have y of one equals two.
So we're going to find the interval
that contains x equals one which would be
the open interval from zero to three
and because it's in this interval,
we know that this initial value problem
has a unique solution.
Okay, let's take a look at one more.
Same question, different differential equation.
Notice how this time, the differential equation
is in the correct form so we should recognize
that P of x is equal to tangent x and f of x
is equal to sin x.
For P of x, remember tangent theta is equal
to y over x on the unit circle.
So if we look at the coordinate plane of the unit circle,
P of x will be undefined or discontinuous
whenever the angle has a terminal side where x equals zero.
So x equals zero is actually the y axis.
So when x equals pi over two or three pi over two,
P of x will be discontinuous but it will be continuous
in between these two angles.
So we could start by saying that P of x is continuous
on the open interval from pi over two to three pi over two
but for the same reason why P of x would be discontinuous
at these two angles, it's going to be discontinuous
at any co-terminal angle to these two as well.
Notice how these angles are just pi over two plus
or minus multiples of pi.
So we'd also have to include the interval
from three pi over two to five pi over two and so on.
There's going to be an infinite number of intervals.
Moving to the left, we could also start
at negative pi over two and go to pi over two.
Again, and so on.
So these are the intervals for which P of x
would be continuous.
For f of x equals sin x, there are no restrictions.
X can be any real number.
So f of x is continuous from negative infinity
to positive infinity which means the differential equation
is going to have unique solutions on the intersection
of these two intervals which would just be the interval
where P of x is continuous.
So I'm going to go ahead and state this as
the intervals where P of x is continuous
and then looking at our initial condition,
we have y of pi equals zero.
So we're going to find the interval
that contains x equals pi which would be this interval here
and because it's in this interval,
we know this initial value problem has a unique solution.
And I'll go ahead and write the interval down here.
Pi over two, three pi over two and this is an open interval.
Okay, I hope you found this helpful.