This time, we now consider second order linear homogeneous equation. Possible variable coefficient y double prime + p(x) times y-prime + q(x)y = 0, okay? Where the p(x) and q(x) are just a continuous functions on an interval I, okay? I'm taking the deferential equation one is in it's standard form, in other words it's leading coefficient that is a coefficient of y double prime is identically 1 but is only for the simplicity, okay? Then because this is a second order linear homogenous differential equation, solving differential equation (1) means finding two linear independent solution over the differential equation. Then we completely determine the whole family of solutions of the differential equation (1), right? Sometimes, one solution, say y sub 1, is easy to find and then we need to find another solution, say y sub 2 which is linearly independent of y1, okay? Then since y1 and y2 are linearly independent, their ration y2 over y1 cannot be constant on the interval I, okay? So that suggest that we may set the second solution y sub 2 of x is = u(x)y1(x) where the nonconstant function u(x) is to be determined. Then substituting this our choice of y sub 2 in the equation 2, in the equation 2 what I mean is I set the second linear independent solution, y sub 2 as u(x) times the unknown solution of y over x, right? Substitute this one into the differential equation one here just to remind you the differential equation one, okay? That is y double prime + p(x) y-prime + q(x)y and that = 0, right? Simple computation then we get y double prime now, when y is equal to y sub 2, it will be u double prime times y1 + 2 times u prime times y1 prime, = u times y1 double prime and + p of x times the y2 prime. y2 prime would be u prime times y1 + u times the y1 prime right? Then + q of y2, y2 is equal to the u times the y1, okay? So plug in this expression, y sub 2 is equal to ux times the y1(x) into this differential equation, okay? So left-hand side gives you this first line, then we organize this line in terms of u, okay? Of course, the collector terms are collecting u double prime that is y1 times u double prime. Then collector terms containing u prime is coefficient to variable 2y1 prime from this term. And plus we have one more such term here, p times y1 times u prime, okay? Then the remaining terms of each contains u, okay? Then is coefficient to variable, y1 double prime + p times y1 prime + q times y sub 1, right? What can you say about this quantity, y1 double prime + py1 prime + qy1. That should be = 0 because y1 is known to be a solution of this homogeneous differential equation, okay? So in fact, this last term disappear, okay? So finally we get y1u double prime + 2 times y1 prime + p times y1 times u prime, okay? There's a second order, the differential equation for u but if you put u prime = v, okay? If u prime is equal to v, then it becomes y1 times v prime + 2y1 prime + py1 and v, that is = 0, okay. So now we have first order, okay? First order linear homogeneous differential equation for the unknown function v, where v = u prime, okay? So that equation, the last equation which I called equation 3. Which is simple enough and we know how to handle it, okay? For example solving this equation 3, differential equation 3. Let's remind it, that is here. 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.