0:46

Then because this is a second order linear homogenous differential equation,

Â solving differential equation (1) means finding two linear

Â independent solution over the differential equation.

Â Then we completely determine the whole family of

Â solutions of the differential equation (1), right?

Â Sometimes, one solution, say y sub 1,

Â is easy to find and then we need to find another solution,

Â say y sub 2 which is linearly independent of y1, okay?

Â Then since y1 and y2 are linearly independent,

Â their ration y2 over y1 cannot be constant on the interval I, okay?

Â So that suggest that we may set the second

Â solution y sub 2 of x is = u(x)y1(x) where

Â the nonconstant function u(x) is to be determined.

Â 2:00

Then substituting this our choice of y sub 2 in the equation 2,

Â in the equation 2 what I mean is I set the second linear independent solution,

Â y sub 2 as u(x) times the unknown solution of y over x, right?

Â Substitute this one into the differential equation

Â one here just to remind you the differential equation one, okay?

Â That is y double prime + p(x) y-prime

Â + q(x)y and that = 0, right?

Â Simple computation then we get y double prime now,

Â when y is equal to y sub 2, it will be u double prime

Â times y1 + 2 times u prime times y1 prime,

Â = u times y1 double prime and + p of x times the y2 prime.

Â y2 prime would be u prime times y1 + u times the y1 prime right?

Â 3:12

Then + q of y2, y2 is equal to the u times the y1, okay?

Â So plug in this expression, y sub 2 is equal to ux times

Â the y1(x) into this differential equation, okay?

Â So left-hand side gives you this first line,

Â then we organize this line in terms of u, okay?

Â Of course, the collector terms are collecting u double prime

Â that is y1 times u double prime.

Â Then collector terms containing u prime is coefficient to

Â variable 2y1 prime from this term.

Â And plus we have one more such term here, p times y1 times u prime, okay?

Â Then the remaining terms of each contains u, okay?

Â Then is coefficient to variable,

Â y1 double prime + p times y1 prime + q times y sub 1, right?

Â What can you say about this quantity,

Â y1 double prime + py1 prime + qy1.

Â That should be = 0 because y1 is known to be a solution of

Â this homogeneous differential equation, okay?

Â So in fact, this last term disappear, okay?

Â So finally we get y1u double prime + 2

Â times y1 prime + p times y1 times u prime, okay?

Â There's a second order, the differential equation for

Â u but if you put u prime = v, okay?

Â If u prime is equal to v, then it becomes

Â y1 times v prime + 2y1 prime + py1 and

Â v, that is = 0, okay.

Â So now we have first order, okay?

Â First order linear homogeneous differential equation for

Â the unknown function v, where v = u prime, okay?

Â So that equation, the last equation which I called equation 3.

Â Which is simple enough and we know how to handle it, okay?

Â For example solving this equation 3, differential equation 3.

Â Let's remind it, that is here.

Â 5:51

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

Â divide the whole equation by y1,

Â then you are going to get v prime + (2

Â times over y1 prime over y1 + p and

Â v, and that is = 0, right?

Â That's the first order homogeneous linear differential equation for

Â the unknown v, right?

Â But what is the integrating factor of this equation?

Â Okay, let me right it in that way here.

Â The integrating factor for

Â that first order differential equation for we will be okay?

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

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

Â Let's do the part of it that is

Â e to the 2 times the log of y1 +

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

Â So that is equal to,

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

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

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

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

Â homogeneous linear differential equation, okay?

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

Â then what you are going to get is y1 squared times

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

Â the quotient derivative then must be = 0, right?

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

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

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

Â Then c1, times 1 over that quantity, so

Â that means 1 over the quantity y1 squared, and

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

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

Â That's what I'm writing here, okay?

Â Solving this differential equation 3, right?

Â This is the differential equation 3, okay?

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

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

Â Since the v = u prime,

Â this is equal to u prime.

Â Take one more integration on both side is 10.

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

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

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

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

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

Â That's a general solution of this

Â differential equation 3 for u, right?

Â We now have two arbitrary constants, c1 and c2.

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

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

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

Â constant then u is a a nonconstant function, okay?

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

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

Â u( x) is given by any nonzero constant times

Â the anti-derivative of the function.

Â They are linear independence of solutions of our original

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

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

Â original differential equation which is suborder two, right?

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

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

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

Â is to reduce the from 2 to 1.

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

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

Â