Okay, as concrete examples, I'd like to solve the following nonhomogeneous differential equations, okay? First, y double prime + y prime + y = -5 cosine 2 x + sine 2 x. Second I'll try, y double prime- 4 y prime + 4 y =- e to the 2 x. And finally, y double prime + 4 y = -4 times sine squared x, right? Let's look at these examples one by one, okay? First example a, what's the equation, let me remind it, okay? It's, Y double prime + y prime + y = -5 cosine 2x + sine of 2x, right? That's the problem. First let's consider it's corresponding homogeneous problem. Y double prime + y prime + y = 0, okay? It's a characteristic equation it's equal to, as you can see easily from this one. r squared + r + 1, right? So from this quadratic equation r squared + r + 1 = 0. We have two distinct roots which are complex conjugates, -1 + or- square root of 3 i over 2, and that implies the complementary solution, okay? The general solution of this homogeneous problem is, which I denoted by y sub c. Which is equal to exponential -x over 2 times c1 cosine square of 3x over 2 + c2 sine square root of 3x over 2, right? On the other hand, here we have a complementary solution. On the other hand I'm looking for the annihilator of this right-hand side. What is the annihilator of this right hand side? Here we have, this is essentially a linear combination of cosine 2x and the sine 2 of x, right? So, in fact they are annihilated by D squared + 4, okay? So, apply this differential polynomial D squared + 4 to the given differential equation then you will get D squared + 4 and from this, you have D squared + (D + 1) of y = D squared + 4 acting on this that should be = 0. D squared + 4 is an annihilator of -5 cosine 2 x + sine 2 of x, right? So we get this homogeneous differential equation of which the general solution will be the complementary solution y sub c of this equation, right? The general solution of this part and plus you will get d1 cosine 2 x + d2 sine 2 x, right? They are coming from this D squared + 4, okay? So, this expression, y = y sub c + d1 cosine 2 x + d2 sine of 2 x, this is a general solution of this differential equation here, okay? Now our general theory says, The tail part, they should have particular solution of this original problem, of the form d1 cosine 2 x + d2 sine 2x, right? So plug in this expression, plug in this yp into the original differential equation, Substituting this into the original differential equation. Let me erase this part. Substituting y sub p into the differential equation then say yp double prime + yp prime + yp, okay? If you simplify this combination, then you are going to get (-3d1 + 2d2) cosine 2x + (-2d1- 3d2) sine of 2x must be equal to this given right inside which is -5 cosine 2x + sine 2x, right? So equating coefficients, coefficient of cosine 2x must coincide, and the coefficient of sine 2x must coincide from both sides of equality. So that you will get simultaneous equation, say -3d1 + 2d2 = -5. Another one is -2d1- 3d2 must be equal to 1, okay? Solving this simultaneous equation, we get d1 = 1 and d2 = -1, okay? Which means our required particular solution is y sub p = cosine 2x- sine 2x, okay? Combining with, Our previous complimentary solution, the general solution of the original problem will be yc + yp which is given by that expression down here, okay? Okay let's consider the second example, example b the problem is y double prime- 4 y-prime + 4y = -e to the 2x, okay? As before first we solve the corresponding homogeneous problem say, consider the y double prime- 4 y-prime + 4y = 0, okay? It's characteristic equation will be r squared- 4r + 4, in other words (r- 2) squared = 0. So that r = 2, which is a double root, right? And that means the general solution of this homogeneous problem, which we denote by yc = e to the 2x times c1 + c2x, where c1 and c2 are arbitrary constant, right? Next, let's look at the annihilate of this right-hand side, right? Right-hand side is constant times the e to the 2x, okay? This is annihilated by D- 2 here, D- 2 acting on -e to the 2x = 0, right? So apply (D- 2) on the original differential equation, okay? And the original differential equation then in fact our original differential equation this is equal to (D- 2) squared y, okay? Apply one more D- 2, so then the (D- 2) cubed y = (D- 2) of right hand side, (-e to the 2x) = 0. You get this high order homogeneous differential equation, and it is easy to find the general solution of this homogeneous problem, okay? This general solution will be y = yc, which is obtained down there, y c, + we have a one more linearly independent solution, say x squared times e to the 2x. So general solution is y = yc + some arbitrary constant d x squared and e to the 2x, right? And what does that mean? That means there must be a particular solution of our original problem of the form, yp = d times x squaed + e to the 2x, right? [INAUDIBLE] constant d, right? Finally to find the constant d, plug this expression yp = dx squared of e to the 2x into this differential equation. So make that expression, yp double prime- 4yp + 4y p-prime +4 yp right here, okay? If you computed this side of then you are going to get 2d times e to the 2x, right? That must be the given right-hand side, -e to the 2x and that implies d must be equal to negative one-half, comparing the coefficients, right? So that the general solution of our original problem will be, as a complementary solution and plus one particular solution which is, negative one-half x squared e 2x. So, this is general solution of our original problem b, okay? As a last example, c, what is the problem c? C is here, the problem c is a solve the problem, a nonhomogeneous problem y double prime + 4y and that is = -4 sine squared x, right? Okay, that's the forgiven problem, okay? As user, we first needed to solve it's corresponding homogeneous problem say, y double prime + 4y = 0. Its characteristic equation is r squared + 4 =0. So that r is equal to + or- 2i, okay? So r = + or- 2i and that means the complementary solution will be y of c, which is the c1 cosine 2x + c2 sine of 2x, right? Now look at the right hand side, okay? -4 times the sine squared x, right? By the trigonometric identity sine squared x = one-half times 1- cosine 2x, okay? So we can find the annihilator of sine squared x, as D times D squared + 4, sine squared x. In other words D times (D squared + 4), (1- cosine 2x) will be equal to 0, okay? So, D times (D squared + 4) is annihilator of this right-hand side, -4 sine squared x, right? And what I mean is D( D squared + 4) acting on -4 sine squared x and that is = 0, right? That an annihilator, okay? We found it, okay? Apply this differential operator to the given one, then you're going to get in fact the following, D and D squared + 4. Now you have here one more D squared + 4 and y and that = 0, okay? So in fact, this is D(D squared + 4) squared and y = 0, right? That's the equation we get in high order homogeneous differential equation. You can write down the general solution of this high order homogeneous constant coefficient differential equation easily, okay? Thus the general solution would be y = first the complementary solution yc down here plus you need three more linear independent solutions. One is coming from D, that is equal to 1, another one is coming from the (D squared + 4) squared so they are x times cosine 2x and x times the sine of 2x, okay? So the general solution of this high order homogeneous differential equation is yc + yc + d1 + d2x cosine 2x + d3x sine 2x. And that means what? That means there must be a particular solution of this given the differential equation in the form of d1 + d2x cosine 2x + d3 x sine 2x, right? To find all the three and determine the coefficients d1, d2 and d3, plug in this expression yp into the given differential equation. So forming yp double prime +4y of p, if you simplify then you are going to get that is = 4d3 cosine 2x- 4d2 sine 2x + 4d1. That must be equal to the given right-hand side that is -4 times sine squared x, okay? That is the same as -4 times one-half times 1- cosine of 2x, okay? Now comparing the coefficients from this and from that you're going to get d1 = negative one-half, d2 = 0, and d3 = one-half, okay? So the required general solution of this given problem is y = c1 cosine 2x + c2 sine 2x- one-half + one-half times x, times the sine of 2x, right? That's how we apply the method of undetermined coefficients to such problems, okay?