- [Instructor] Welcome to a lesson on how to find
a second solution and the general solution
to a linear second order homogeneous differential equation
using the method of reduction of order.
To begin, a differential equation must fit this form here
if it's a linear second order
homogeneous differential equation.
And since p of x, q of x, and r of x,
the coefficients here,
are not going to be constants,
we cannot use a characteristic equation
in order to solve the differential equation.
One limitation to the method of reduction of order
is that we must be given
one solution to the differential equation.
So, if we're not given one solution,
we can't use this method.
And this technique is a method for reducing
a second order differential equation
to a linear first order differential equation
to help solve the original differential equation.
The way this works is that we're going to assume
that the second solution will be in the form of
y sub two of x is equal to u of x, some function of x,
times y one of x,
where y one of x is the given solution.
Now I do want to mention before we begin,
there is a shortcut formula that we can use
to find y sub two of x given here
if the differential equation is in this form.
We'll take a look at the shortcut method in the next lesson.
So, looking at our first example,
notice how the given differential equation
does fit the form, and therefore it is
a linear second order homogeneous differential equation
and that we're also given one solution, y one of x.
So again, for the reduction of order method,
we start by assuming that the second solution,
which we'll call y, or y of x,
is equal to some function u of x
times y one of x,
which in this case would be x squared.
Let's go ahead and write this
as just y equals u times x squared.
And because we're assuming this is a solution
to the differential equation,
we'll now find y prime and y double prime
and then perform substitution
into the differential equation.
So if y is equal to u times x squared,
y prime will require the product rule.
So we'd have u times the derivative of x squared,
which would be two x.
So two x u
plus the second function x squared times u prime.
And now we'll find y double prime,
which again is going to require the product rule.
So for this first term, we'll have two x times u prime
plus u times the derivative of two x,
which would be two.
So we have two u plus x squared
times u double prime
plus u prime times the derivative of x squared.
That would be two x.
So two x u prime.
Notice how we do have two like terms, here and here.
So y double prime is equal to four x u prime
plus two u plus x squared u double prime.
So now using y,
y prime,
and y double prime,
we'll perform substitution
into the given differential equation.
So we'd have x squared times y double prime.
That would be four x u prime.
Plus two u plus x squared u double prime
minus three x times y prime.
Y prime is two x u.
Plus x squared u prime.
And then plus four times y.
Let's write this as x squared u.
Equals zero.
Now we'll clear the parentheses
by distributing and combining like terms.
So we'll distribute here
and here.
So we'll have four x cubed u prime
plus two x squared u
and then plus x to the fourth u double prime
and then minus six x squared u
minus three x cubed u prime
plus four x squared u
equals zero.
Now let's first focus on the u terms.
Here we have two x squared u minus six x squared u.
That's negative four x squared u.
And then plus four x squared u,
which is equal to zero.
So all the u terms simplify out.
And this is key
for making the reduction of order method work.
So when applying this method,
all the u terms should simplify to zero.
So now looking at the u double prime terms,
we have x to the fourth
u double prime.
Then looking at the u prime terms,
we have four x cubed u prime
minus three x cubed u prime.
So that would just be plus x cubed u prime equals zero.
So because we only have u double prime and u prime terms,
we're now going to perform a substitution
to reduce the order of this differential equation
to a linear first order differential equation.
For example, if we let w equal u prime
and w prime equal u double prime,
we can write the differential equation as
x to the fourth w prime
plus x cubed w equals zero.
And now we can solve this differential equation for w
using separation of variables
and then integrate w to find u.
Once we find u,
remember our second solution is u times x squared.
Let's begin by replacing w prime with d w d x.
So we'll have x to the fourth d w d x.
Let's move this term to the other side
so we have equals negative x cubed w,
which means x to the fourth d w
equals negative x to the third w d x.
And now we'll both sides by w
and also divide both sides by x to the fourth.
So if we divide both sides by w
we'd have one divided by w d w.
And now if we decide both sides by x to the fourth,
notice how we'd have negative one divided by x d x.
Now integrate both sides.
This would give us natural log w
equals negative natural log x plus a constant C.
Now we'll exponentiate both sides of the equation,
meaning e raised to the power of natural log w
must equal e raised to the power
of negative natural log x plus C.
Notice on the left side we just have w.
And on the right side,
we can write this as e to the natural log,
moving this negative one to the exponent of x,
we'd have natural log x to the negative one
times e to the C.
Well, e to the C is just a constant
and e raised to the power
of natural log x to the negative one
is just x to the negative one.
So we have w equals,
let's call it C sub two,
times x to the negative one,
or C sub two divided by x.
And now that we have w,
which is really equal to u prime,
we can integrate this to find u.
Again, remember we let w equal u prime,
so this is really just u prime.
And our goal is to find y equals u times x squared.
So let's go ahead and replace w with u prime,
or d u d x,
equals C sub two divided by x,
which means differential u is equal
to C sub two divided by x d x.
Integrate both sides.
And we have u of x equals,
this would be C sub two natural log x,
plus another constant, which we'll call C sub one.
Which means y of x is equal to u,
which is C sub two natural log x
plus C sub one
times x squared.
So y of x is equal to
C sub two x squared natural log x
plus C sub one x squared.
So notice how when we found y,
this is actually the general solution,
because notice how we have the sum of multiples of two
with near-independent functions.
So this would be the general solution.
Notice how the general solution also contains
the given solution, y one of x equals x squared
when C sub one equals one.
So to give a second solution,
we can let C sub two be any constant.
Let's let C sub two equal one.
And therefore, y sub two of x, our second solution,
could be x squared natural log x.
That's going to do it for this example.
Remember in the next lesson,
we'll talk about how we can use a shortcut formula
to find y of x and y sub two of x.
I hope this was helpful.