00:01
All right, so right now we're going to be looking at a boundary value problem.
00:05
In this problem, we're going to be looking at how our variables a through d correlate with one another by looking at when our problem has a unique solution, when it has no solution, and when it has infinitely many solutions.
00:20
Now, before we can delve straight into this, we are going to need to fiddle around a little bit with the equation that we are given.
00:28
So in the given equation, we have y double prime, y prime, and y.
00:37
Now even though these are the same, these are variants of the same thing, it's going to be a little bit difficult to work with them in the forms that they are currently in.
00:46
So let's go ahead and convert them into terms of r.
00:51
Our y double prime will become r squared, our y prime will become r and our y will just become one.
01:00
Because we're essentially saying that it is r to the power of zero, which just equals one.
01:08
So now that we have our new terms, let's rewrite our equation.
01:12
It's going to become r squared minus 2r plus 2 equals 0.
01:20
Now we have two ways that we can solve this auxiliary equation.
01:23
We could either factor or we could use a quadratic equation, but because of the way that this specific equation is, it's definitely going to be easier for us to just use a quadratic equation.
01:36
So let's keep in mind while we're doing this, that our coefficient for r squared will be our a value, our coefficient for r, so negative 2, will be b, and this lovely 2 right here is going to be our c value.
01:49
So let's solve for r.
01:52
R is going to equal negative b, and our b value in this scenario is negative 2.
01:59
Plus or minus the square root of b squared.
02:05
So negative 2 squared minus 4 multiplied by a, which is 1, and also multiplied by c, which is 2.
02:16
And we will divide all of this by 2 multiplied by a, or 1.
02:24
Let's go ahead and simplify this.
02:26
Negative negative 2 is just going to become a positive 2 plus or minus the square root of 2 squared negative 2 squared is going to be 4 minus 4 times 1 times 2 which will become 8 and 2 times 1 is going to be 2 so when we're simplifying now let's go ahead and separate the two parts of the numerator since we're able to since there's a plus and minus separating them so we will have two over two plus or minus the square root of four minus eight or negative four divided by two now let's go ahead before we simplify any further let's note that this discriminant the negative four is negative because this determines which equation we are going to be using which equation we will be plugging this r value into.
03:28
So now that we know that we will be using what the book calls a case three, let's finish simplifying.
03:35
2 divided by 2 is 1 plus or minus the square root of negative 4 is 2i.
03:47
We're going to have 2 i.
03:49
2 divided by 2 is 1.
03:51
So we will have 1 plus or minus 1 i.
03:58
Now, let's go ahead and note that for the specific equation we are plugging this into, this one right here will become our alpha value, and this coefficient of i, the other one, will be our beta value.
04:18
So let's plug this into our new equation.
04:23
We are going to have e to the power of alpha times x, which is one, so i'll be our new equation.
04:30
Alpha times x, multiplied by the sum of constant 1 times cosine of beta x, and our beta is 1 as well, plus constant 2 multiplied by sine of beta x.
04:55
So 1x again.
04:57
Before we set these, we set our boundaries to this equation, let's go ahead and simplify it a little bit.
05:06
So what we're going to do is distribute our e -value.
05:12
We'll go ahead and get rid of the ones because they aren't particularly important since one times anything is whatever the anything value is.
05:22
And let's also get rid of our constants because our constants are not super pertinent to what we are currently doing.
05:31
So we are going to obtain from this e to the power of x times cosine of x plus e to the power of x times sine of x.
05:49
Times sine of x.
05:51
Now let's go ahead and create the two equations that we will be using to solve a through c based upon our boundaries.
06:01
So our first boundary is going to set a, the variable a, equal to x.
06:08
And we're going to set this entire equation equal to c.
06:13
So let's go ahead and do that.
06:16
E instead to the power of x will be to the power of a, multiplied by the cosine of a, plus e to the power of a, multiplied by the sign of a will be equal to c.
06:33
And for our second boundary, we are going to plug our b value, since it is y of b, into x, and set the entire equation equal to d.
06:45
So we're going to do the exact same thing, but instead of a and c, respectively, we'll be doing b and d.
06:51
So e to the power of b, multiplied by cosine of b, plus e to the power of b multiplied by the sign of b equals d.
07:06
So now that we've successfully accomplished setting our boundaries to our equation, let's go ahead and look at a through c.
07:17
So a is requiring us to look for a unique solution.
07:26
Let's go ahead and look at what a unique solution even means.
07:31
If you think about it when we're dealing with a two -dimensional plane, it means that there's a unique solution.
07:45
Our given lines only intersect one time.
07:49
That means that our slopes are never going to be equal to one another.
07:57
Because if they're equal to one another, then that means that they're either parallel or they're the exact same line.
08:04
And we don't want that for unique solution...
