Welcome to an example on how to solve
a second order nonhomogeneous Cauchy-Euler equation.
A second order nonhomogeneous Cauchy-Euler equation
must fit this form here,
where we start to recognize that this is a second order
differential equation, and it would be nonhomogeneous
as long as g of x is not equal to zero.
But what makes this a Cauchy-Euler equation
is that for every term on the left side,
the degree of the coefficient is equal to
the order of the derivative.
For example, looking at this first term,
a x squared has degree two and the term also has
y double prime, or the second derivative.
Notice how the next term has a coefficient of b x,
so the degree is one, term also contains
the first derivative,
and then finally the third term, we have a constant
times the function y.
So once we know our differential equation fits this form,
to find the general solution there will be two major steps.
The first step will be to find the complimentary function,
which we can do by finding the solutions
to this auxiliary equation where the values of a, b, and c
come from the differential equation,
and then based upon the types of solutions
to this quadratic, it will affect the form
of the complimentary function as we see here.
So once we find the complimentary function,
we'll then find a particular solution,
big Y sub p, using a variation of parameters method,
which is outlined here.
One of the most important things to remember
about this method is that before applying it,
the coefficient of y double prime must be one.
Let's take a look at our example.
Our first step is to recognize that the given
differential equation does fit this form here,
and therefore we do have a second order nonhomogeneous
Cauchy-Euler equation.
Now for the next step, we'll find the complimentary function
of the general solution by setting up
this auxiliary equation.
So we need to notice that a is equal to one,
b is equal to negative two,
and c is equal to positive two.
So the auxiliary equation
would be m times the quantity m minus one,
and since b is negative two we have minus two m,
c is positive two,
and now we'll solve, let's first clear the parentheses
and the combined like terms.
So we'll have m squared minus one m minus two m,
that's three m plus two equals zero.
We can solve this by factoring.
Factors of positive two that add the negative three
are negative two and negative one.
So we have m equals positive two or m equals positive one.
So because the auxiliary equation has two distinct
real zeroes or real roots, the form of the complimentary
function will be in this form here.
Remember, if we had...
Two real equal roots, this would be the form
of the complimentary function, and if our auxiliary
equation had complex roots, this would be the form
of the complimentary function.
So we now know that y sub c
is equal to c sub one x squared plus c sub two x.
This also tells us for the variation of parameters method
y sub one is equal to x squared,
and y sub two is equal to x.
Now before we find big Y sub p a particular solution,
let's review the method of variation of parameters.
As mentioned earlier, the most important thing to remember
is the coefficient of y double prime must be one.
When in this form, the right side of the equation
will be g of x.
So we already found y sub one and y sub two,
and once we identify g of x and find the Wronskian
of y sub one and y sub two, we'll simply apply
this formula here for the variation of parameters formula,
and this will give us big Y sub p.
So going back to our example, let's begin by dividing
everything by x squared.
So our differential equation would be y double prime,
this would be minus two divided by x times y prime,
this would be plus two divided by x squared y equals,
this would be two x e to the x.
So we know that y sub one was equal to x squared,
y sub two is equal to x,
and now we also know that g of x
is equal to two x e to the x.
For the next step, let's go ahead and find
the Wronskian of y sub one and y sub two.
We'd have a two by two determinate,
where the first row would be y sub one and y sub two,
and the second row would be the derivatives,
so two x and one.
So the Wronskian is going to be equal to
x squared
minus two x squared,
or negative x squared.
So the Wronskian is equal to negative x squared.
Now we have all the information we need,
we'll simply perform substitution for big Y sub p.
Let's do this on the next slide.
So we'll have big Y sub p equals negative y sub one,
that's going to be negative x squared times
the integral of y sub two times g of x,
that'll be x times two x e to the x, that's two x squared
e to the x divided by the Wrokskian which is negative
x squared d x,
and then plus y sub two, plus x
times the integral of y sub one times g of x,
that'll be x squared times two x e to the x,
that's two x cubed e to the x,
divided by the Wronskian, d x.
Now let's simplify the integrands.
Let's factor out the two, and also notice that
we have two negatives here, so that'll be two x squared
times the integral of, here we have x squared
over x squared, that simplifies to one,
so we just have e to the x d x.
Here we have a negative, it's going to be minus,
we'll factor out the two,
and then two factors of x will simplify out,
so we have x e to the x d x.
Notice how to integrate this we'll have to apply
integration by parts,
which means the integral of u d v is equal to u v
minus the integral of v d u.
So for this integral we'll let u equal x,
so differential u is equal to d x,
so if u was equal to x then d v must be equal to
e to the x d x.
So if we integrate this we'd have v equals e to the x.
So now we have big Y sub p equals two x squared
times integral of e to the x, which is just e to the x,
so we have two x squared e to the x minus two x
times this integral.
When we apply integration by parts would be u times v,
or x e to the x minus the integral of
v d u, which would just be e to the x d x.
So now we have two x squared e to the x
minus two x squared e to the x.
Since the integral to e to the x is just e to the x,
we'll have negative two x times negative e to the x,
so we have plus two x e to the x.
And notice how these two terms simplify to zero,
so finally, the particular solution is just
two x e to the x.
So now that we have our particular solution,
and we just found the complimentary function
here,
the sum of these two functions will be
our general solution.
So again, the general solution is equal to
the complimentary function plus the particular solution,
or in our case, y of x is equal to c sub one x squared
plus c sub two x plus big Y sub p, which was two x
e to the x.
This would be our general solution.
Hope you found this helpful.