- [Voiceover] Welcome back for two more examples
of determining the region in the XY plane, for which a
first-order differential equation would have
a unique solution through a point.
When we have a first-order differential equation
in this form, and given this point by the initial condition,
if F of X comma Y and the partial derivative of F
that respects to Y, are continuous on the rectangular region
defined here, that contains the given point
from the initial condition,
then there exists an interval, centered at X sub-zero,
and a unique function, Y of X, defined on the interval
that satisfies the initial value problem.
So, again, our main goal is to find where F
and the partial derivative of F,
that respects to Y are continuous on the XY plane.
So we want to determine the region in the XY plane
for which the given differential equation
would have a unique solution through the given point.
First, recognize the differential equation
is already in standard form.
Therefore, we can see that F of X comma Y,
is going to be three X cosine Y.
Now, let's also find the partial derivative of F
with respects to Y.
So we'll treat X as a constant.
So that derivative of cosine Y is going to be negative sine Y.
So we'll have negative three X sine Y.
Well, this is actually pretty straightforward.
Again, we're trying to find where F is continuous,
and where this partial derivative is continuous.
Well, there are no restrictions for X and Y,
therefore F is continuous on the entire XY plane.
And there are also no restrictions
on our partial derivative.
Negative three X times sine Y
is continuous over the entire XY plane as well.
And since there are no restrictions on either
of these two functions, the differential equation
has unique solutions on the region of the entire XY plane.
Let's take a look at another example.
Again, our goal is to find the region in the XY plane
for which the given differential equation has a unique
solution through the given point.
Notice how this differential equation is not
in standard form, so our first step is going to be
to divide both sides by the quantity one plus Y cubed.
So we'll have Y-prime equals X
divided by the quantity one plus Y cubed.
Which means, F of X comma Y,
is equal to X divided by
the quantity one plus Y cubed.
Now, to find the partial derivative of F
with respect to Y, it'll be helpful to think of this
as F equals X times the quantity one plus Y cubed,
to the negative one.
So the partial of F, with respect to Y,
treating X as a constant would be,
we'd multiply it by a negative one,
so we'd have negative X, times the quantity one plus Y cubed
to the negative two, times the derivative Y cubed,
which would be three Y squared.
So the partial derivative would be
negative three XY squared,
divided by the quantity one plus Y cubed
to the second power.
But notice how for both F and our partial derivative,
the restriction is division by zero.
And since both contain factors of one plus Y cubed,
our restriction is we know that one plus Y cubed
can't equal zero.
So we'd subtract one on both sides,
we'd have Y cubed can't equal negative one.
Now, if we cube-root both sides,
our restriction is Y can't equal a negative one.
Therefore, the differential equation
has unique solutions on the region
when Y is less than negative one
or when Y is greater than negative one.
These would be two half-planes,
which is the region when the differential equation
is going to have a unique solution
through any point in that region.
Okay, that's going to do it for part two.
I hope you're now able to find the region
in which a first-order differential equation
would have a unique solution through a point.