- [Lecturer] Welcome to an example on how to solve
an initial value problem involving
a linear second order nonhomogeneous differential equation
using the method of undetermined coefficients.
So to solve the initial value problem of a differential
equation, this form here,
we'll first find the general solution,
which is given by Y of X.
Then once we do this, we'll find the value of C sub one
and C sub two using the initial conditions.
So for a quick review, to use the method
of undetermined coefficients, we start by solving
the corresponding homogeneous differential equation
given here, where the solution to the homogeneous
differential equation from number one give us
the first two terms of the general solution.
This is often called the complementary function.
And then, we're going to use steps two and three
to determine big Y sub P a particular solution.
So what we'll do is start by guessing a form
of a particular solution to differential equation
with undetermined coefficients based upon the
form of G of X.
Then by using our guess and its derivatives,
we'll perform substitution into differential equation
and solve for the undetermined coefficients.
Then finally, we use initial conditions
to find C sub one and C sub two.
So let's take a look at our example.
Therefore, saying we should notice that
we do have a linear second order,
nonhomogeneous differential equation.
And since the right side of the equation consists of
polynomial functions and exponential functions,
the method of undetermined coefficients
is an appropriate strategy.
So for the first step, we'll solve the corresponding
homogeneous differential equation,
which would be Y double prime plus
6Y prime plus 8Y equals zero.
And because we have constant coefficients,
meaning our differential equation now fits this form,
we can find the solution using a characteristic equation,
which should be AR squared plus BR plus C equals zero.
So notice that A is one, B is six and C is eight.
So the characteristic equation would be R squared
plus 6R plus eight equals zero, which is factorable.
The factors of eight that add to six are positive four
and positive two and therefore, the solution
to this equation would be R equals negative four
or R equals negative two.
And now these two values of R help us determine
the general solution to the homogeneous
differential equation given here by Y of X.
But these two terms also give us the first two terms
in the general solution to the nonhomogeneous differential
equation, which is actually our goal.
So from here, we can conclude that the general solution
to the nonhomogeneous differential equation
would be Y of X equals C sub one times
E raised to the power of negative 4X plus C sub two times
E raised to the power of negative 2X.
And then plus a particular solution,
or plus big Y sub P.
So now we'll make a guess for a particular solution,
or big Y sub P based upon G of X,
the right side of our differential equation.
Notice how we have a linear function
and an exponential function.
So we're going to go ahead and let Big Y sub P,
a particular solution, be a linear function
with undetermined coefficients where the form AX plus B,
that would be linear function with unknown coefficients,
and then for the exponential, we'll have plus C times E
raised to the power of negative X.
Some constant times our exponential function.
So from here, we'll find our first and second derivative
and then perform substitution
into the differential equation.
So big Y sub P prime would be equal to A
and then we'd have zero plus C times the derivative
of E to the the negative X, which should be
E to the negative X times negative one.
So a minus C times E to the negative X.
And now the second derivative would be zero
and then minus C times E to the negative X
times negative one or C times E to the negative X.
So now we'll take our three functions here
and perform substitution into
the given differential equation.
And then from here, we should be able to solve
for A, B and C.
Let's go and do this on the next slide.
So we'll have Y double prime, which is CE to the negative X
plus six times Y prime, or six times A minus
CE to the negative X and then plus eight times Y
or eight times AX plus B plus CE to the negative X.
This must equal 2X plus 1E to the negative X.
And now we'll equate the coefficients.
But before we do this, we need to clear the parentheses.
So we'll have CE to the negative X plus
6A minus 6CE to the negative X plus
8AX plus 8B plus 8CE to the negative X
equals 2X plus 1E to the negative X.
So now we're going to group the constant terms,
the X terms and the exponential terms.
Let's first identify the constant terms.
We have 6A and 8B.
Now we'll look for the X terms.
There's only one.
One here and the remaining terms are exponential terms.
Here, here, and here.
So again, our constant terms.
We have 6A plus 8B and then we have plus 8A times X,
which I'll write like this.
And now for the exponential terms,
we're going to factor out E to the negative X.
So we'll have plus, this would be C.
And then minus 6C and then plus 8C
E to the power of negative X must equal,
I'm going to go ahead and put a zero here for the constant term
plus 2X plus 1E to the negative X,
which means 6A plus 8B must equal zero.
8A must equal two and finally,
C minus 6C plus 8C must equal positive one.
Let's go ahead and right this as a system of equations.
Again, we have 6A plus 8B equals zero.
8A equals two.
And this would be 1C minus 6C plus 8C.
That's going to be 3C must equal one.
We're now looking at this equation here.
If we divide both sides by eight,
we have A equals 2/8 or A equals 1/4.
Here we have C equals 1/3.
And then here, we can see that 8B would be equal
to negative 6A or negative six times 1/4,
since A is equal to 1/4.
So we have 8B equals, this would be
negative 6/4 or negative 3/2.
Multiply both sides by 1/8
and we have B equals to, this would be negative 3/16.
So now we'll take the values of A, B and C
and substitute them into our particular solution
or Big Y sub P here, here and here.
Once we do this, we have our general solution.
Then we can go back and fine the values of
C sub one and C sub two based upon initial conditions.
So our particular solution, big Y sub P
is going to be equal to AX or 1/4 X plus B,
which is minus 3/16 plus C times E to the X
or plus 1/3 E to the negative X,
which now means we have the general solution
to the given differential equation where we had
Y of X here, which is found Y sub P.
So the general solution would be the sum of these two terms
plus Big Y sub P or plus 1/4 X minus 3/16
plus 1/3 E to the negative X.
Now we've made some really good progress.
We have the general solution here.
So now we use initial conditions here
to find the values of C sub one, C sub two
to solve the initial value problem.
So let's start with the fact that we know that
Y of zero equals zero.
So if Y of zero equals zero, we'll substitute zero for X.
So here we'd have E to the zero.
That'd be one.
So we have C sub one plus another E to the zero.
That's C sub two.
And again, if X is zero, this would be zero.
You'd have minus 3/16.
This would be plus 1/3.
Again, we know this must equal zero.
So if we add these two fractions
and move it to the right side of the equation,
we would have C sub one plus C sub two equals negative 7/48.
And now we need to find the first derivative of Y
to use the second initial condition
that Y prime of zero equals zero.
So we need to remember this equation here
and then we'll form another equation with C sub one
and C sub two from the second initial condition.
So let's find Y prime.
We would have negative 4C sub one E to the negative 4X
minus two C sub two E to the negative 2X
plus 1/4 minus zero, and this would be minus 1/3
E to the negative X.
So if Y prime of zero equals zero,
then we would have negative 4C sub one minus 2C sub two
plus 1/4 minus 1/3 equals zero.
So if we subtract these two fractions,
and then move it to the right side of the equation,
that would give us negative 4C sub one minus 2C sub two
equals 1/12 and again, from the previous slide,
we found that C sub one plus C sub two equals negative 7/48.
So C sub one plus C sub two equals negative 7/48.
So now, we want to take the system of equations
and solve for C sub one and C sub two.
Now due to time, I'm going to go ahead and let you verify this.
I already found that C sub one must be equal to 5/48
and C sub two must be equal to negative 1/4.
Again, to solve the system, we can use either
substitution or the elimination method.
But now that we've found the method of C sub one
and C sub two, we have the solution
to the initial value problem.
It's going to be Y of X, after we substitute
the value of C sub one here and the value of C sub two here.
So the solution to our initial value problem would be
the function Y of X equals C sub one, which would be 5/48
times E raised to the power of negative 4X
plus C sub two, which is negative 1/4.
So minus 1/4 E to the negative 2X
and then plus 1/4 X minus 3/16
plus 1/3 E to the negative X.
So as you can see, that was quite a bit of work.
But this would be the solution to the initial value problem.
I hope you found this explanation helpful.