As the last topic in chapter four, I'd like to introduce one more very strong way of obtaining particular solution for the linear, non homogeneous differential equation. Though the method of undetermined coefficients, it's rather simple but it has a too strong constraint. Consider differential equation L(y) = g(x) non homogeneous one, L is a linear differential operator, okay. L is a linear differential product, okay. What is that? Is equal a7(x) and D to the n plus a7 minus 1(x), D to the (n -1) and plus a1(x) and D and plus a0(x), right. That is a differential operator of all the n, okay. Consider such a differential equation, non homogeneous differential equation. For the method of one determined the coefficients to be applied, first your differential point to a must have only constant coefficient, okay. In other words, all those coefficients a sub n a sub (n-1), a sub 1 and a sub 0 they must be constant, okay. None of them can be a true function of x, okay. Second constraint is coming from the right hand side of that differential equation L(y) = g(x), right. g(x) sometimes we recorded as an input function. This input function must have differential polynomial annihilator, okay. It must have, right. Differential putting on here annihilator. We know that not every function has differential polynomial annihilated, right? For example can you remind it. For example the 1/x or the sec x, okay, things like that. They have no annihilates at all, right? So if this right hand side there's input function g(x) to be, happened to be 1(x), then the method of undetermined coefficients cannot feel the used to. So to overcome such difficulties here we introduce the second method, the method of valuation of parameters to obtain a particular solution of such a non homogeneous differential equation, okay. In this case the differential operator may have a variable coefficient and this input function g(x) can be arbitrary, okay. It may or may not have differential polynomial annihilated,okay. To make a thing that's simple, we consider only second order non homogeneous differential equation. So consider non homogeneous linear second order differential equation, possibly variable coefficient in its standard form, y double prime plus p(x) y prime plus q(x) y is equal to g(x), right. Assume that we have two linearly independent solution of responding homogeneous problem, okay, responding homogeneous problem. Okay, trivially that is equal to y double prime plus p(x) y prime plus q(x) y is equal to 0. This is okay responding homogeneous problem. We assume that y1 and y2 are two linearly independent solution of this homogeneous problem. In other words we have a complementary solution, yc = c1y1 + c2y2. And the general solution of this regional given non homogeneous equation will be, okay. The general solution of differential equation one is y is equal to complementary solution plus particular solution, okay. We already know what is this complementary solution yc, right. So only thing unknown is y sub p now, okay. The main idea of the method of variation of parameters destruction that there is a particular solution of this non homogeneous problem in the form of y sub p is equal to u1 times y1 plus u2 times y2, okay. With the u1(x) and the u2(x) are two unknown functions to be determined, okay. So that yp becomes a particular solution of this original non homogeneous problem, okay. Here in our guess for yp we have two unknowns, the u1 and the u2. So in order to determine to unknown functions the u1 and the u2, we need two conditions, right. Our guess is we are looking for a particular solution y sub p of the given differential equation in the form of yp = u1y1 + u2y2, right. Since if you have two unknowns as I said u1 and u2, we need two conditions for to determine u1 and u2, right. One condition, trivially comes from the condition that yp is a particular solution of the given differential equation. In other words yp double prime plus py p prime plus qy p is equal to g, right. Using this expression, this equation will give us one equation for two unknowns u1 and u2. Trivially, we need one more equation, right? That another equation we will take it so that our next computations to be easy, okay. Differentiating two twist this one, okay, this is the equation two, right. Differentiating this yp you will get the derivative of yp will be u1y1 prime plus u2y2 prime plus u1 prime y1 and plus u2 prime y2, right. It's easy by the product rule of differentiation, right. From this expression, right, we require, okay, we require the second part of to be equal to 0, say u1 prime y1 plus u2 prime y2 to be equal to 0, okay. We simply require it without the proper reasoning at this moment. Then, okay, yp prime will be such that this forced apart, right. In other words u1y1 prime plus u2y2 prime, right. So now we have our y, right from yp prime, this is equal to u1y1 prime plus u2y2 prime, this is the first part plus u1 prime y1 plus u2y2, okay, we require this is equal to 0. We simply require this is equal to 0. And what does that mean then, that means yp is equal to only this part. So it's the second derivative will be u1y1 double prime u2y2 double prime plus u1 prime y1 prime plus the u2 prime y2 prime by the product group again. So using this yp and the yp prime and the yp double prime, okay, make the left hand side of the differential equation, say yp double prime plus p(x) times yp prime plus q(x) y sub p, right, okay. Here's the computation, right, yp double prime plus py y prime plus qy p. If we reorganize this some of these three times then you are going to get u1 times y1 double prime plus py1 prime plus qy1 plus u2 times y2 double prime plus py2 prime qyq plus u1 prime y1 prime plus u2 prime and the y2 prime, okay. Let's pay attention to the quantities in this bracket and the quantities in this bracket, okay. What can you say about these two things? What I mean is the y1 double prime plus py1 prime plus qy1 and y2 double prime plus py2 prime plus qy2 okay, what can you say about this too, okay? Can you remind, what are those y1 and y2, right? I said okay, the y1 and the y2. This is a two linearly independent solution of this is linearly independent solutions to corresponding homogeneous problem, right. Y double prime plus px y prime plus qxyz equal to 0, right? In other words, y1 double prime plus py1 prime plus qy1 or y2 double prime plus py2 prime plus qy2 both of them must be equal to 0, right. Because they are the y1 and y2 are solutions of this responding homogeneous problem, okay. What does that mean? The first bracket is equal to 0. The second bracket that is equal to 0. So that means right, in fact this part and that part they disappeared. So we will get only u1 prime times of y1 prime plus u2 prime times y2 prime, okay, because of yp. This is a particular solution of our original problem, okay. Some of these three terms must be equal to g, right? Can you see it right? So, okay, we have one more condition, right here unknowns the u1 and the u2, right, okay. We have one another such a condition, say u1 prime y1 plus u2 prime y2 is equal to 0, okay. So let me summarize those two equations here, okay.