This method of undetermined coefficients cannot be used for a linear differential equation with constant coefficients P(D)(y) = g, unless g has a differential polynomial annihilator, okay? This is a crucial part, this right hand side must have differential polynomial annihilator for the method of undetermined coefficients to be applied, okay? For example 1 over x, or for example the secant x has no differential polynomial annihilator, okay? Can you see why? The class of functions g(x), the right-hand side to g(x) which allow the method of undetermined coefficients include polynomials in x. Or the exponential functions e to the alpha of x, or the trigonometry functions cosine beta of x, or sine beta of x, or finite linear combinations. And the products of all these functions, say polynomials in x or the exponential functions and trigonometric functions, okay? As our last example, let's consider the following problem. Determine the form of a particular solution for y triple prime- 2y prime + y prime = -3x squared + 2x- xe to the x + 2e to the 2x. The given differential equation, you can write it as D cubed- 2D squared + D y = g(x), right? g(x) is equal to right here,- 3 x squared + 2x- xe to the x + 2 times e to the 2x, right? That's the given problem, okay? Let's try to find the annihilator of this g of x, right? Let's work it term by term then. First is the polynomial -3x squared + 2x is annihilated by trivial d cubed. Then the next term, xe to the x is annihilated by (D- 1) squared, right? Finally, e to the 2x is annihilated by (D- 2), okay? So apply D cubed times (D- 1) square times (D- 2) to the given differential equation here, right? So what I mean is apply the D cube and the (D- 1) square and then D- 2 to the both sides of that equation, that is D cubed- 2D squared + D y, okay? That must be = 0, so you get that equation. If you simplify this left-hand side then, that is this one D cubed times (D- 1) squared, times (D- 2), times D cubed- 2D squared + D, okay? Simplify, that is the same as D to the 4, times (D- 1) to the 4, times (D- 2), and y = 0, okay? So now we have a higher order homogeneous differential equation, okay? It's general solution is it's easy to write, okay? That's c1 + c2x + c3x squared + c4x cubed. They're all coming to this part, okay? Then, from the (D- 1) to the 4 part you're going to get c5e to the 2x + c6 x times e to the 2x + c7x squared times e to the 2x, okay? Next c8 x cubed times e to the 2x, finally from D- 2 part, you're going to get c9e to the 2x, right? So this long expression down here, okay? This is, general solution of this high order homogeneous differential equation, right? Among these nine linearly independent solutions say 1, x, x squared, x cubed, e to the x, xe to the x, x squared e to the x, x cubed e to the x and e to the 2x. Let us separate out, which form fundamental set of solution of corresponding homogeneous equation, okay? Say y triple prime- 2y double prime + y-prime = D cubed- 2D squared + D and y. In other words, this is equal to D(D- 1) squared and y = 0, okay? This is the corresponding homogeneous problem of the original equation. Among this long linear combinations I would like to separate out, okay? I would like to separate out the general solution of this corresponding homogeneous problem, okay? Can you see the fundamental set of solutions to this problem? I think you can see it. First unit 1, next unit e to the x, and because the r is equal to 1 is double root of characteristic equation, you have xe to the x, right? D to three functions are former fundamental set of solutions to this problem, okay? So let's separate out, okay? The general solution of this problem then, where do we have a 1? 1 is a c1, then where we have e to the x? Second part you have, no you have c5 e to the x, right? Where do we have x times e to the x? That is right here, okay? So it's x times e to the x, right? So what I mean is my c1, right here and the e to the x, that is c5e to the x, next, The x times e to the x, that is a c6 and x times e to the x, right? The linear combination, yc, that is the c1 + C5 e to the x plus c6 times x times e to the x is a complementary solution, okay? In other words, the solution to this homogeneous problem, okay? Then the remaining part, they gives us the form of a particular solution, okay? What's the remaining part? c 2 x, and the c 3 x squared, c 4 x cubed + c 7 x squared e to the x, c 8 x cubed e to the x, plus c 9 e to the 2 x. They determine the form of the particular solution of this original problem, okay? In other words, we have particular solution of the form, y p, that is the d 1 of x + d 2 x squared, next missing term is that this d 3 x cubed, then d 4 x squared e to the x, and + d 5 x cubed e to the x, finally d 6 e to the 2x, right? This last line is the form of the particular solution of this given problem, okay? Those undetermined coefficients from d 1 through d 6 can be obtained by plugging the last expression into the original differential equation, okay? That's how the method of the undetermined coefficients works, right?