- [Instructor] In the previous video, you learned
what boundary value problems are.
In this video, you'll focus on a common type
of boundary value problems,
which are two-point boundary value problems.
Two-point boundary value problems
are when you have exactly two points
as in boundary conditions you are looking at
for your initial conditions.
These often represent the top
and bottom bounds of integration
to solve the differential equation over a range of values.
Let's take a look at a few examples of this type
of boundary value problem.
Let's solve the equation y double prime
plus nine multiplied by y equal to zero.
Let's have your boundary conditions be y of zero
equal to negative two and y of pi divided by four
equal to eight.
In this case, you will have the characteristic equation
of r squared plus nine equal to zero, which if you solve it,
you get two roots of r equal to plus
and minus three multiplied by i,
meaning that they are both complex roots.
Since you are working with complex roots,
you'll then plug it into your general solution
of y of x equals C1 multiplied by cosine of three,
multiplied by x plus C2 multiplied by a sine
of three multiplied by x.
Now, let's start applying your boundary conditions.
Starting with your first condition,
you have y of zero equals negative two.
So if you plug this into your equation,
you get y of zero equals negative two,
which equals C1 multiplied by cosine of three,
multiplied by zero plus C2,
multiplied by sine of three multiplied by zero.
Once you simplify this, you get C1 equals negative two.
Now, let's plug this into your equation to get y of x
equals negative two, multiplied by cosine of three,
multiplied by x plus C2 multiplied by sine of three
multiplied by x.
Let's plug in your next boundary condition
of y of pi divided by four equal to eight.
So this time you have y of pi divided by four equals eight,
which equals negative two, multiplied by cosine of three,
multiplied by pi divided by four plus C2,
multiplied by sine of three,
multiplied by pi divided by four.
Once you solve this, you get a result of C2
equal to 16 multiplied by the square root of two
subtracted by four.
Let's plug this in to get your final solution
of y of x equals negative two,
multiplied by cosine of three, multiplied by x plus 16
multiplied by the square root of 2 minus four,
and you'll multiply all of that by sine of three
multiplied by x.
Let's look at another example.
This time, let's solve the equation
y double prime minus four multiplied by y equals zero.
In this case, y of zero equals one
and y of one equals zero.
Let's look at your characteristic equation like last time,
which in this case is r squared minus four equal to zero,
which gives you two roots of r equal to plus and minus two.
Let's plug this into your general solution
to get y of x equals C1 multiplied by e
to the two multiplied by X plus C2 multiplied by e
to the minus two multiplied by x.
Let's apply your first condition, y of zero equals one.
Plugging this into your equation,
you get y of zero equals one, which equals C1,
multiplied by e to the two multiplied by zero
plus C2 multiplied by e to the minus two multiplied by zero.
Once you simplify this, you get one equal to C1 plus C2.
In this case, you'll not substitute a value
into either C1 or C2 for your constant values yet,
so let's continue plugging in the next condition.
This time, you'll be plugging in y of one equal to zero,
which going into your equation is y of one equals zero,
which equals C1 multiplied by e to the two multiplied by one
plus C2 multiplied by e to the minus two multiplied by one.
Using your previous result of one equals C1 plus C2,
you can then conclude
that C2 equals one minus C of one.
Let's plug in one minus C1 for C2
into your second result to get y of one equal to zero
equals C1 multiplied by e to the two
multiplied by one plus one minus C1
multiplied by e to the minus two multiplied by one.
This will make it easier for you
to focus on solving just one constant
and then you can gather the second constant after that.
So this equation simplifies to negative e
to the minus two equals C1
multiplied by e squared minus e minus two,
which if you divide both sides, gets you C1
equals to negative e to the minus two
divided by e squared minus e to the minus two.
And in this case, your denominator e squared
minus e to the minus two is actually equal
to the hyperbolic function sine h of two.
Let's substitute this into your equation
to get C1 equals negative e to the negative two
divided by sine h of two,
and let's now get your second constant of C2,
which as you saw before, is equal to one minus C1.
So plugging this in, you get C2 equals
one minus negative e to the minus two
divided by sine h of two, which simplifies
to C2 equals one plus e to the minus two
divided sine h of two.
Let's plug in these two constant values
to get your final result of y of x equals
negative e the minus two divided by sine h of two,
multiplied by e to the two multiplied by x
plus one plus e to the minus two over sine h of two,
multiply that by e to the minus two multiplied by x.
As you saw, the process is fairly similar
to when you're working with an initial value problem,
just that you are now plugging in two conditions
instead of one for each of the levels accordingly.
There are many other two-point boundary value problems
you can solve, but this should give you a general idea
of how to work with them.
As you could tell, they range from easy to complicated,
depending on the problem you are solving.
In the next video, let's take a look at periodic functions.