0:23

This is a crucial part, this right hand side must

Â have differential polynomial annihilator for

Â the method of undetermined coefficients to be applied, okay?

Â For example 1 over x, or for example the secant x has

Â no differential polynomial annihilator, okay?

Â Can you see why?

Â 0:52

The class of functions g(x), the right-hand side to g(x) which

Â allow the method of undetermined coefficients include polynomials in x.

Â Or the exponential functions e to the alpha of x, or

Â the trigonometry functions cosine beta of x, or

Â sine beta of x, or finite linear combinations.

Â And the products of all these functions, say polynomials in x or

Â the exponential functions and trigonometric functions, okay?

Â 1:32

As our last example, let's consider the following problem.

Â Determine the form of a particular solution for

Â y triple prime- 2y prime + y prime

Â = -3x squared + 2x- xe to the x + 2e to the 2x.

Â 2:00

The given differential equation,

Â you can write it as D cubed-

Â 2D squared + D y = g(x), right?

Â g(x) is equal to right here,- 3 x squared + 2x-

Â xe to the x + 2 times e to the 2x, right?

Â That's the given problem, okay?

Â 2:37

Let's try to find the annihilator of this g of x, right?

Â Let's work it term by term then.

Â First is the polynomial -3x squared +

Â 2x is annihilated by trivial d cubed.

Â Then the next term, xe to the x is annihilated by (D- 1) squared, right?

Â Finally, e to the 2x is annihilated by (D- 2), okay?

Â So apply D cubed times (D- 1) square times

Â (D- 2) to the given differential equation here, right?

Â So what I mean is apply the D cube and

Â the (D- 1) square and

Â then D- 2 to the both sides of that equation,

Â that is D cubed- 2D squared + D y, okay?

Â That must be = 0, so you get that equation.

Â 3:58

If you simplify this left-hand side then,

Â that is this one D cubed times (D- 1) squared,

Â times (D- 2), times D cubed- 2D squared + D, okay?

Â Simplify, that is the same as D to the 4,

Â times (D- 1) to the 4, times (D- 2), and y = 0, okay?

Â So now we have a higher order homogeneous differential equation, okay?

Â It's general solution is it's easy to write, okay?

Â That's c1 + c2x + c3x

Â squared + c4x cubed.

Â They're all coming to this part, okay?

Â Then, from the (D- 1) to the 4 part you're

Â going to get c5e to the 2x + c6 x times e to

Â the 2x + c7x squared times e to the 2x, okay?

Â Next c8 x cubed times e to the 2x, finally from D- 2 part,

Â you're going to get c9e to the 2x, right?

Â So this long expression down here, okay?

Â This is, general solution of this high order

Â homogeneous differential equation, right?

Â 5:34

Among these nine linearly independent

Â solutions say 1, x, x squared, x cubed,

Â e to the x, xe to the x, x squared e to the x,

Â x cubed e to the x and e to the 2x.

Â Let us separate out, which form fundamental set of

Â solution of corresponding homogeneous equation, okay?

Â Say y triple prime- 2y

Â double prime + y-prime = D

Â cubed- 2D squared + D and y.

Â In other words, this is equal to

Â D(D- 1) squared and y = 0, okay?

Â This is the corresponding homogeneous problem of the original equation.

Â Among this long linear combinations I would like to separate out, okay?

Â I would like to separate out the general solution of

Â this corresponding homogeneous problem, okay?

Â Can you see the fundamental set of solutions to this problem?

Â I think you can see it.

Â First unit 1, next unit e to the x, and because the r is equal

Â to 1 is double root of characteristic equation, you have xe to the x, right?

Â D to three functions are former fundamental set of

Â solutions to this problem, okay?

Â So let's separate out, okay?

Â 7:22

The general solution of this problem then, where do we have a 1?

Â 1 is a c1, then where we have e to the x?

Â Second part you have, no you have c5 e to the x, right?

Â Where do we have x times e to the x?

Â That is right here, okay?

Â So it's x times e to the x, right?

Â So what I mean is my c1,

Â right here and the e to the x,

Â that is c5e to the x, next,

Â 8:00

The x times e to the x, that is a c6 and

Â x times e to the x, right?

Â The linear combination, yc, that is the c1 + C5 e to the x

Â plus c6 times x times e to the x is a complementary solution, okay?

Â In other words, the solution to this homogeneous problem, okay?

Â Then the remaining part, they gives us the form of a particular solution, okay?

Â What's the remaining part?

Â c 2 x, and the c 3 x squared,

Â c 4 x cubed + c 7 x squared e to the x,

Â c 8 x cubed e to the x, plus c 9 e to the 2 x.

Â They determine the form of the particular solution of this original problem, okay?

Â In other words, we have particular solution of the form,

Â y p, that is the d 1 of x + d 2 x squared,

Â next missing term is that this d 3 x cubed,

Â then d 4 x squared e to the x, and + d 5 x cubed e to the x,

Â finally d 6 e to the 2x, right?

Â This last line is the form of the particular

Â solution of this given problem, okay?

Â Those undetermined coefficients from d 1 through d

Â 6 can be obtained by plugging the last expression

Â into the original differential equation, okay?

Â That's how the method of the undetermined coefficients works, right?

Â