Today we're going to be talking about how to solve a boundary value problem.
And in this particular problem we've been given the equation why double prime or the second derivative
of y is equal to y prime or the first derivative of y.
We've also been given two initial conditions wives equals 1 and y of one equals two.
Now you might be wondering what is a boundary value problem and how is it different from an initial
value problem.
Well it's really the same except that a boundary value problem isn't guaranteed to have a solution when
we have an initial value problem.
Given enough initial conditions and the right information will always be able to find a solution and
a boundary value problem.
That's not necessarily the case.
What we're trying to do with the boundary value problem is solve for the constant CSUB one and CSUB
two.
But given these initial conditions that may not be possible.
So what we're going to do today is try to solve this boundary problem.
Given this second order differential equation and two initial conditions.
So let's see how it goes.
The first thing we need to do with this equation is convert it into an equation in terms of our instead
of 1 in terms of why the silly trick that I like to use to do that is I count the prime marks on each
Y variable to determine what the new exponent is it's going to be on the variable.
So every time we have a Y variable we just change it to our and the exponent on that.
Our variable is going to depend on how many hashmarks we have on that particular y variable.
So for example this is why double prime.
It's going to change into our squared because there's two hashmarks is the new exponent is going to
be a 2.
So this why double prime becomes r squared.
We're going to set that equal to y prime over here so this y prime is going to become R to the one which
of course is just r.
Now we need to solve this for the roots of the equation.
What we'll do is we'll subtract r from both sides and we'll get our squared minus R is equal to zero.
And now we can try to factor the left hand side.
We'll factor out an R and we'll get our times are minus one equals zero.
When we pull the R out from this equation.
Now we have two factors that we can set independently equal to zero.
We have the first factor here will set that equal to zero and we'll get our equals zero.
Then we have our second factor here or minus one.
We'll set that equal to zero will add 1 to both sides and we'll get our equals positive one.
These are the two roots of our equation of our second order differential equation and what we noticed
right away is that they are distinct real roots.
If you have practiced at all with second order differential equations you know that you can get three
types of roots.
When you go through this process distinct real roots equal real roots and complex conjugate roots.
These are distinct real roots because they are real numbers 0 and 1 are real numbers but they are distinct
from one another.
In other words they're different.
Zero is not equal to one.
These are different numbers.
They're are distinct real roots.
You can also get equal real roots for example if we had gotten here are equals 1 and R equals 1.
Those would both be real numbers.
They would be equal to each other so we'd have equal real roots.
You can also have complex conjugate roots which will always involve the imaginary number I.
And they're real easy to spot.
These are clearly distinct real roots.
When you have distinct real roots you're going to be using this equation here for y of X for the general
solution to the second order differential equation.
When you have equal real roots or complex conjugate roots you use a different formula than the one shown
here for the general solution.
But since we have distinct real roots this is the formula that we use.
So what we do is we plug these roots in for our sub one and our sub too.
And it doesn't matter which route you plug in for which variable here are somewhat are our sabtu.
It'll work out either way.
I usually just plug the route I find first in for Arsa one and the one I find sikat in for our sub 2.
So the equation for our general solution will be y of X equals CSUB one easy to the zero x plus CSUB
to E to the 1 x.
And of course we can simplify that to say y of x is equal to either the zero X is either the zero either
the 0 is 1.
C sub one times one is just see someone so we get see see one and then each to the one X is the same
as either the X so we get plus C Sabtu E to the X..
So this is the general solution for our second order differential equation.
Now because we've been given initial conditions we want to try to use those to solve for see sub one
and CSUB to what we'll do is we'll use our first initial condition Y of 0 equals 1 to try to solve for
either of these constants.
So we'll set the right hand side equal to the value we've been given here so we'll say one is equal
to and then whatever we get when we plug 0 into this right hand side.
So we'll get to see some one.
Plus CSUB to eat to the zero we're plugging in zero here for x.
When we do that when we simplify this even zero was just one.
So we get one is equal to CSUB one plus CSUB sabtu.
Now we can try to solve this for either see someone or CSUB too.
Clearly we're not going to get a real or or finite value we're going to need the other initial condition
hopefully to solve for both constants.
But let's go ahead and solve this for see sub one so that we have something to plug in to the other
equation for see someone once we get it.
So we'll solve this for someone by subtracting CSUB two from both sides and we'll get see someone is
equal to 1 minus CSUB 2.
Now let's use the second initial condition Y of one equals two.
Again we'll plug that into our general solution here and we'll get to is equal to see sub 1.
Plus see sub to E to the one.
So when we get either the one here that just simplifies to E and we get 2 equals C sub 1 plus C sub
2.
Now notice both of our equations still involve see someone and CSUB to what we want to do is take the
value that we found here for CSUB 1 1 minus CSUB sub 2 and plug that in here for CSUB one to get this
equation in terms of CSUB to only.
So we'll get to is equal to 1 minus CSUB 2 plus CSUB to see if we subtract 1 from both sides we'll get
one is equal to negative CSUB 2 plus see some 2.
And if we reorder our terms to avoid starting with the negative sign here we'll get one is equal to
see sub 2.
E minus CSUB 2.
And now we can go ahead and factor out a CSUB to sew up here we'll get one is equal to CSUB 2 times
the minus one when we factor out that CSUB to dividing both sides by minus 1 gives us CSUB 2 is equal
to 1 divided by the minus one.
Now we have a value for CSUB too which is great.
So this is going to be one of our constants here CSUB 2.
Now we just need to solve for see someone and of course we already have a formula for that.
So we'll go ahead and plug in the value we just found Prissie's up to we'll get see someone is equal
to 1 minus 1 over the minus 1.
And now we just need to simplify this.
We'll go ahead and find the lowest common denominator which is going to be e minus 1 and we'll multiply
this one here by e minus 1 over e minus 1.
So we'll get CSUB one is equal to minus 1 over the minus 1 minus 1 over the minus 1.
When we combine these fractions we'll get six of one is equal to the minus 1 minus 1 all over minus
1.
And of course that gives us a value for seis of one of the minus 2 over the minus 1.
And there's our second value the value of the constant CSUB one.
So now it's not just enough to find these two values.
The goal is to find these two and then plug them back into the general solution for our differential
equation which we had here.
So when we do that we'll get our final answer which is going to be why x is equal to.
Notice here we have CSUB one.
So we're going to go ahead and grab that value.
E minus 2 over the minus 1 plus CSUB to so we're going to say plus here we have this value proceeds
up to 1 over the minus one.
And then we want to multiply that by to the X..
So to the X like that now the only thing you can consider doing is just simplifying this by moving this
either the ex into the numerator of this fraction getting me to the X appear in the numerator and having
this one go away right here but that's it.
This is our final answer for the general solution of this second order differential equation.
Given our two initial conditions.