- [Instructor] Welcome to an example on
how to use the variation of parameters method
to solve a linear second order
nonhomogeneous differential equation.
If we have a differential equation that fits this form here,
where G of X is nonzero, and P, Q, and G
are continuous on an open interval, and Y sub one,
and Y sub two are linearly independent solutions
to the corresponding homogeneous differential equation,
then a particular solution to the
nonhomogeneous differential equation is given
by big Y sub P, and this formula here.
Therefore, we can use Y sub one, Y sub two, and big Y sub P
to form the general solution to the
nonhomogeneous differential equation,
as we see here, given by Y of X.
Now, we should also notice that the sum
of these first two terms here,
which would be the general solution
to the homogeneous differential equation,
is often also called the complementary function, or Y sub C.
To apply the method of variation of parameters,
we'll take three steps.
Number one, we will solve the corresponding
homogeneous differential equation,
which will give us Y sub C,
or the first two terms of the general solution,
and then we will use the formula given here
to find big Y sub P which we did derive
in a previous lesson, and then
we will use the results of one and two to
form the general solution, in this form here.
Let's take a look at our example.
The first thing that we should notice is that we do have
a linear second order nonhomogeneous differential equation,
so to apply the method of variation of parameters,
step one is to solve the corresponding
homogeneous differential equation,
which would be Y double prime minus Y prime,
minus two Y equals zero.
Notice, in this case, the function on the right, G of X
is equal to E raised to the power of three X.
Now, because we have constant coefficients here,
we can solve this using a characteristic equation,
where A is equal to one, B is equal to negative one,
and C is equal to negative two,
giving us the characteristic equation,
R squared minus R minus two equals zero.
We can solve this by factoring.
We have R and R.
The factors of negative two that add to negative one
would be negative two, and positive one,
so the characteristic equation has two real distinct zeros.
We have R equals two, or R equals negative one.
Remember, the nature of these solutions will
affect the form of the general solution
to the homogeneous differential equation,
or the complementary function,
again, given here by Y of C,
because we have two distinct
real zeros to the characteristic equation.
Now we know that Y sub C is equal to C sub one
times E raised to the power of two X
plus C sub two times E raised to the power
of negative one X, or just negative X.
Notice that Y sub one is equal to E to the power of two X,
and Y sub two is equal to E to the power of negative X.
Now, for the next step, or step two,
we will find big Y sub P using this formula here.
Let's begin by finding the Wronskian
of Y sub one and Y sub two.
We'll have a two by two determinant
where the first row is Y sub one, and Y sub two.
We will have E to the two X, and E to the negative X.
The second row will be the first derivatives.
The derivative of E to the two X
would be E to the two X times two,
or two E to the two X.
The derivative of E to the negative X would be
E to the negative X times negative one,
or negative E to the negative X.
Now we will have this product minus this product.
When multiplying, the bases are the same,
we add the exponents, so this
first product would be negative E to the power of two X
plus negative X, which would just be X,
and then minus this product,
which would be two E to the power of X,
giving us negative three E to the X,
which means big Y sub P is going to be equal
to negative Y sub one, or negative E to the two X,
times the integral of Y sub two times G of X.
Y sub two is E to the negative X here,
times G of X, which is E the three X,
divided by the Wronskian, which is
negative three letter E to the X DX,
and we have plus Y sub two.
Y sub two is E to the negative X,
times the integral of Y sub one times G of X.
Y sub one is E to the two X.
G of X is E to three X,
divided by a negative three E to the X.
Now, before we integrate, let's simplify.
Looking at this first part, notice
how we have two negatives that make a positive.
Let's also factor out the one third,
that would give us E to the two X divided by three.
Then, if we integrand, if we multiply the numerator,
that would be E to the two X, adding the exponents.
E to the two X divided by E to the X,
we subtract the exponents, giving us just E to the X.
Next we have a sum, but then we have a negative here,
so it's going to be minus E to the negative X.
Let's factor out the one third,
and we have the integral of --
The numerator would be E to the five X
divided by E to the X, that would be E to the four X DX.
Now we have big Y sub P is equal to --
The integral of E to the X is just E to the X,
so we would have one third.
Each of the two X times E to the X,
that would be E to the three X.
Then we have minus E to the negative X
divided by three, times this integral.
Notice how this does require I use substitution,
where U would be equal to four X,
differential U is equal to four DX,
so DX is equal to one fourth DU,
so this integral would be one fourth times E to the four X.
Let's go ahead and simplify this.
We have one third E to the three X.
Now here, we'll have minus one twelfth,
but then we multiply the exponentials
and we add the exponent,
we get another E to the three X.
These are like terms,
so big Y sub P would be equal to --
One third is equal to four twelfths.
Four twelfths minus one twelfths is three twelfths,
and three twelfths is one fourth,
so we have one fourth E to the power of three X.
Finally, now that we have Y sub C, and big Y sub P,
we can form a general solution
to the original nonhomogeneous differential equation.
To do this, we will use this form here,
where the general solution Y of X is
equal to Y sub C plus big Y sub P.
Let's finish this on the next slide.
The general solution, Y of X, is equal to Y sub C,
C sub one E to the two X plus C sub two E to the negative X,
plus big Y sub P, or plus one fourth
E to the power of three X.
This would be the general solution
to our nonhomogeneous differential equation
using the method of variation of parameters.
I hope you found this helpful.