- [Instructor] Welcome to an example
in which we'll use the variation of parameters method
to find a particular solution to a linear second order
nonhomogeneous differential equation.
So for example, we're first asked to verify that y sub one
and y sub two satisfy the corresponding
homogeneous differential equation.
And then find a particular solution
to the nonhomogeneous differential equation.
Well the first thing we should notice about
the given differential equation is that the coefficient
of y double prime is not one, it's x squared,
and when applying the variation of parameters method
we do want to coefficient of y double prime to be one.
So let's start by dividing everything by x squared.
We don't actually have to do this to verify that y sub one
and y sub two satisfy the corresponding homogeneous
differential equation, but let's go ahead
and do this now anyway.
So now we have y double prime...
Minus, let's write this as two divided by x squared y
and then equals three minus one divided by x squared.
So this is the same nonhomogeneous differential equation
in a different form and therefore the corresponding
homogeneous differential equation would be y double prime
minus two divided by x squared times y equals zero.
Let's first verify that y sub one is a solution
to this homogeneous differential equation.
So if y sub one is equal to one over x
or x to the power of negative one then y sub one prime
is equal to negative one times x to the negative two
or negative x to the negative two.
And therefore y sub one double prime is equal to
two x to the negative three.
And I'll perform substitution into the
homogeneous differential equation to make sure
the left side is equal to zero.
So y double prime is going to be two x to the power
of negative three
minus two divided by x squared times y
where y is x to the negative one.
And this does look good, remember we can write this as
two x to the negative three minus, this is the same as
two x to the negative two times x to the negative one.
And so this would be two x to the negative three
minus two x to the negative three, which does equal zero.
So now that we've verified that y sub one
does satisfy the homogeneous differential equation,
let's now verify that y sub two does also.
So y sub two is equal to x squared.
So y sub two prime is equal to two x.
And y sub two double prime is equal to two.
And I'll perform substitution into our
homogeneous differential equation.
So we have y double prime, which is two
minus two divided by x squared times y equals zero
where y is x squared.
And this simplifies nicely here.
So we have two minus two equals zero,
verifying that y sub two is also a solution
to the homogeneous differential equation.
And now to find that particular solution,
we'll use this formula here for big y sub p,
which we derived in a previous lesson.
So let's start by finding the Wronskian of y sub one
and y sub two, which we can see
will be the denominator of each integrand.
So the Wronskian of, let's try this as x to the negative one
and then x squared will be a two by two determinant
where the first row is y sub one and y sub two
and the second row will be the first derivatives.
So we'll have negative x to the negative two and two x.
So this will be equal to this product minus this product.
Notice when multiplying here we'll add the exponents
on the variable x
so we'll have x to the negative one times two x would be
two x to the zero power or just two.
And then minus, this would just be negative one.
So this ends up being positive three.
Which means our particular solution, big y sub p
is going to be equal to negative y sub one,
which is negative x to the negative one times integral of
y sub two, which is x squared
times g of x.
So let's go back and take a look at our
nonhomogeneous differential equation
when we divide it through by x squared.
G of x would be equal to three
minus one divided by x squared.
It is important to remember that when applying
this formula here the leading coefficient of y double prime
must be one and therefore this would be g of x,
not three x squared minus one.
And the Wronskian is equal to three.
And then we have plus y sub two,
which is x squared times integral
of y sub one times g of x,
which will be x to the negative one times three
minus one divided by x squared divided by three dx.
But simplify the integrand before we integrate.
Let's factor out 1/3 to clear the denominator
from the integrand, and let's write negative x
to the negative one as negative one divided by three x.
Then we'll have the integral of, let's go ahead distribute,
we would have three x squared and then minus one dx plus.
Let's go ahead and factor out the 1/3, so this should be
x squared divided by three times integral of,
again we'll distribute, so we have
three x to the negative one or three divided by x.
And then we have minus x to the negative one times,
this is x to the negative two,
so we have x to the negative three: dx.
Now let's go ahead and integrate.
So we'll have negative one divided by three x times,
here we have three times x to the third divided by three
or x to the third minus x
plus x squared divided by three times,
here we'll have three natural log absolute value of x.
But since x is greater than zero
we don't need the absolute value.
Then we'll have minus x to the power of negative two
divided by negative two, which would be plus
1/2x to the negative two or plus x to the negative two
divided by two.
Now we'll clear the parentheses,
multiplying here one factor of x will simplify out.
So we have negative 1/3x squared,
and then we have plus 1/3
plus this would be x squared natural log x.
And then here, notice how we have x to the zero,
that's going to be one, so we have plus 1/6.
So we have big y sub p equals negative 1/3x squared
plus x squared natural log x
and then 1/3 plus 1/6 would be 1/2.
But there is one more thing you notice about this problem.
Notice this first term in big y sub p negative 1/3x squared
would be absorbed in the complementary function
of the general solution to the
nonhomogeneous differential equation.
Remember that the complementary function y sub c
would be equal to c sub one times y sub one,
which would be x to the negative one plus c sub two times
y sub two, which is x squared.
So notice that negative 1/3x squared would be absorbed
in this second term of the complementary function,
and therefore we can leave off this term
in the particular solution.
So we'll just go ahead and say that big y sub p
is equal to x squared natural log x plus 1/2.
It wouldn't necessarily be wrong to include this term,
but it would be a duplication of terms
because of this term here in the complementary function
of the general solution.
I hope you found this explanation helpful.