- [Voiceover] Welcome to an intro lesson
to boundary value problems.
The objectives of this video are to define
a boundary value problem,
and also to solve a boundary value problem
given a general solution to a differential equation.
Notice in these examples, we will be given
the general solution to the differential equation.
A boundary value problem is just a special
initial value problem.
So, for review, a differential equation that has
given conditions, allows us to find the specific function
that satisfies a given DE,
rather than just a family of functions.
And these types of problems
are called initial value problems.
If the conditions are given at more than one point
and the differential equation is of order two or higher,
it is called a boundary value problem.
Abbreviated by BVP.
A BVP problem can have none, one, or many solutions.
Let's take a look at our first example.
We want to find a solution to
the given BVP problem.
The second derivative of Y with respect to X
minus Y equal zero
Y of zero equals zero.
Y of one equals one.
If we know that Y of X equals C sub-one,
times E to the X, plus C sub-two times E to the negative X
is a general solution to the differential equation.
So, given the general solution, here,
and these two conditions,
we want to find a particular solution.
So, if Y of zero
is equal to zero,
and if we substitute zero for X into this function here,
the function value must be zero.
So, C sub-one times E to the zero,
plus C sub-two times E to the negative zero,
which is still just zero, must equal zero.
So, C sub-one plus C sub-two must equal zero.
Which means that C sub-one must equal
negative C sub-two or vice-versa.
So, they're opposites of one another.
So we're going to keep track of this.
Now, we'll use the fact that we know Y of one equals one.
Well, if Y of one equals one,
then C sub-one E to the one,
plus C sub-two E to the negative one,
must equal one.
Again, we're trying to find the values of C sub-one
and C sub-two.
Notice how this equation still has two unknowns,
but we do know that C sub-one is equal
to negative C sub-two.
So let's form up substitution here for C sub-one
by replacing it with negative C sub-two.
So, negative C sub-two
times E plus C sub two, times E to the negative one
must equal one.
Let's go ahead and factor out
our common factor of C sub-two.
That's going to leave us with negative E
plus E to the negative one.
Which I'm going to write as E to the negative one minus E
must equal one.
We'll go ahead and divide by this quantity here,
to solve for C sub-two.
This simplifies to one.
So now we know that C sub-two is equal to:
one divided by the quantity E to the negative one minus E.
And since C sub-one and C sub-two are opposites,
we also know that C sub-one would be equal to negative one
all over E to the negative one minus E.
Remember, this is all the information we need
to solve this BVP problem.
We now know C sub-one
and we now know C sub-two.
So let's go ahead and give our solution on the next slide.
So, Y of X is equal to C sub-one times E to the X,
or negative one, all over E to the negative one minus E
times E to the X
plus C sub-two, which is positive one
all over E to the negative one minus E
times E to the negative X.
Notice how we do have a common denominator, here.
So we can write Y of X as,
this would be negative E to the X plus E to the negative X,
I'm going to go ahead and put the positive term first.
So we have E to the negative X, minus E to the X,
all over our common denominator of E to the negative one
minus E.
This would be a particular solution to this BVP problem.
Now, if we did want to check this solution
in the differential equation, we may want to write this
in a slightly different form.
We could write this as Y of X equals,
notice how our denominator is a constant
so we could factor out one over E to the negative one,
minus E.
This would be times a quantity, E to the negative X
minus E to the X.
It might be easier to find the derivative in this form
rather than the previous form.
Now let's take a look at a second example.
Here, we're given that Y-double-prime plus four Y
equals zero, where Y of zero equals one,
Y-prime of pi over two equals two,
and that Y of X equals C sub-one cosine two X
plus C sub-two sine two X is a general solution
to the given differential equation.
So again, we'll use this general solution
with these two initial conditions
to solve this BVP problem.
Since Y of zero equals one,
we can replace X with zero in this function here
and know the function value is positive one.
So C sub-one times cosine of two times zero,
which is zero, plus C sub-two
times the sine of two times zero, or zero,
must equal positive one.
Well, the sine of zero is equal to zero,
and the cosine of zero is equal to one.
So this tells us that C sub-one must equal one.
Now that we know that C sub-one is equal to one,
we can rewrite Y of X as:
Y of X
equals just cosine two X, since C sub-one is equal to one,
plus C sub-two sine two X.
Now, let's go ahead and find the derivative of this,
so we can use this second initial condition.
So Y-prime of X, or Y-prime can be equal to
the derivative of cosine two X,
which is going to be negative sine two X times two,
due to the chain rule.
So we have negative two sine two X,
plus the derivative of C sub-two sine two X.
That'll be C sub-two times cosine two X times two,
or two C sub-two cosine two X.
And now we'll use the fact that Y-prime
of pi over two equals two
to find C sub-two.
We'll substitute pi over two for X,
and know the function value must be equal to two.
So we'll have negative two sine of,
two times pi over two is just pi,
plus two C sub-two times the cosine of two times pi over two
which is just pi.
And this must equal positive two.
Well, sine pi is equal to zero.
Cosine of pi is equal to negative one.
So here, we're going to have negative two C sub-two equals two.
Divide both sides by negative two.
And we have C sub-two equals negative one.
Now that we know C sub-one and C sub-two,
we can solve this BVP problem.
Let's go ahead and do that on the next slide.
Okay, we found that C sub-one was equal to one.
And C sub-two was equal to negative one.
And now we can perform this substitution
into our general solution.
So, Y of X
is equal to cosine two X
plus negative one times sine two X,
or just minus sine two X.
Okay, I hope you found these two examples helpful.
Thank you for watching.