So we have y1 v prime + (2y 1 prime + py1) v = 0,

divide the whole equation by y1,

then you are going to get v prime + (2

times over y1 prime over y1 + p and

v, and that is = 0, right?

That's the first order homogeneous linear differential equation for

the unknown v, right?

But what is the integrating factor of this equation?

Okay, let me right it in that way here.

The integrating factor for

that first order differential equation for we will be okay?

The exponential of integral of 2 times y1 prime over y1 and

+ p and the d(x) through this integration, right?

Let's do the part of it that is

e to the 2 times the log of y1 +

the integral of p of x and d of x right?

So that is equal to,

in fact this is log of y1 you might need a absolute value sine there, right?

This is log of y1 square, exponential log of y1 squared,

that is a y1 squared, times e to the interior log of p(x) dx, right?

That's the so called integrating factor of this first order

homogeneous linear differential equation, okay?

If you multiply this v(x) on this equation,

then what you are going to get is y1 squared times

e to the integral of p(x)dx times v(x) and this,

the quotient derivative then must be = 0, right?

That's the easiest thing to solve, so what you are going to get is

v(x) equal to, the bracket should be equal to constant, okay?

If we solve it for v of x, then that constant will then be denoted by c1.

Then c1, times 1 over that quantity, so

that means 1 over the quantity y1 squared, and

e to the negative integral of p(x)dx, right?

Okay, that's the solution for v(x), okay?

That's what I'm writing here, okay?

Solving this differential equation 3, right?

This is the differential equation 3, okay?

Solving this equation for v then, you're going to get v = c1 over y1

squared times exponential minus integral p(x)dx, all right?

Since the v = u prime,

this is equal to u prime.

Take one more integration on both side is 10.

Solving for u through integration then you are going to get down there.

U is equal to some arbitrary constants c sub 1 times integration

of 1 over y 1 squared times exponential negative into a p(x) dx, okay?

And we need anti derivative of whole this quantity, this function, okay?

And plus some integral constant, c sub 2,okay?

That's a general solution of this

differential equation 3 for u, right?

We now have two arbitrary constants, c1 and c2.

They're arbitrary constant because if we want u of x to be a non-constant function.

So let me take c1 to be any nonzero constant and c2 = 0, okay?

Take c2 = 0, okay, and c1 to be any nonzero

constant then u is a a nonconstant function, okay?

And then, we can say that given y sub 1, and

y sub 2 which is equal to y1 times u(x),

u( x) is given by any nonzero constant times

the anti-derivative of the function.

They are linear independence of solutions of our original

differential equation 1 since u(x) is not constant, okay.

We call this process as the reduction of order because we start from

original differential equation which is suborder two, right?

In fact that we voided down the whole problem, in solving this,

due to differential equation for unknown u, which is first order, okay?

So actually the order of the differential equation we need to solve,

is to reduce the from 2 to 1.

So it's quite a natural to call this method a reduction of order, okay?

I'd like to confirm the reduction of order method by the concrete example.