 So let's see how much further we can take our method of using the differential operator to solve differential equations. So remember the term homogeneous is used in a couple of different ways in mathematics, and just to make things more complicated, it's actually used in a couple of different ways in differential equations. So in this particular case, if we have a differential equation of the form this horrible mess, equal to zero, where our ci's are constant, we say that we have a homogeneous equation with constant coefficients. It's homogeneous because all of our non-zero terms are either y or a derivative of y, and constant coefficients because our coefficients are, well, constant. Since this is also a linear differential equation, rewriting this using our linear differential operator l gives us a characteristic polynomial, and then the roots of the characteristic polynomial can then be used to find the general solution of the differential equation. What happens if we don't have a homogeneous equation with constant coefficients? There are two strategies. First, we can make an ansatz, an inspired guess of what the solution might be, or we can make the equation homogeneous. So an ansatz is an initial solution that works. It's basically an inspired guess. And here's everything you can be taught about coming up with an ansatz. Yeah, it's not much. Coming up with a good ansatz requires quite a bit of mathematical intuition, and by definition, you can't be taught intuition. On the other hand, a more algorithmic approach uses an annihilator, which annihilates the non-homogeneous part of the differential equation. Now, to some extent, finding the annihilator is like finding an ansatz. It takes a little bit of intuition to come up with a good one. However, we can be a little bit more systematic about this, and so we might make the following observations. First, it doesn't do any good to add, subtract, or multiply, because that won't eliminate a non-homogeneous term. It'll just move it around. But, well, we are talking differential equations. We might be able to differentiate it away. So, for example, let's say I have the non-homogeneous first-order linear differential equation, d y over dx minus y equals 8x minus 5. And the thing to recognize here is if I differentiate both sides, we get rid of the x term. Of course, we still don't have a homogeneous equation, because we still have this non-zero constant left, but a useful idea in life and in mathematics is that anything you can do once, you can do as many times as you need to. I differentiate it once, and let's differentiate again. And now we have a homogeneous third-order linear differential equation. And we know how to solve those using our linear differential operator. So, rewriting our differential equation in operator notation and that corresponds to characteristic polynomial d cubed minus d squared. We find the roots, which will be 0 twice and 1. And this gives us our general solution, y equals c1 plus c2x plus c3e to power x. And that seems great until you remember a very important idea. A kth-order differential equation will have k undetermined constants. And we're starting with a first-order linear differential equation. We should only have one undetermined constant. We have too many constants. So what can we do? Now let's think about this for a moment. Note that if we consider the homogeneous equation, dy over dx minus y equals 0, the linear differential operator will be d minus 1, giving us root 1 and solution y equals c3e to power x. And that's this part of our general solution. Using the annihilator, introduce two additional roots, 0 and 0, which corresponds to the solution y equals c1 plus c2x. That's this part of the general solution. These extra roots must correspond to definite but unknown constants, which we need to find. Since the problem reduces to finding the value of undetermined coefficients, we might call this, in a fit of inspired creativity, the method of undetermined coefficients. Yes, mathematicians are very bad at coming up with original names for things. Just as a side note, usually the method of undetermined coefficients is reserved for the part of the solution after finding the ansatz, either by an inspired guess or by annihilating the non-homogeneous term. In other words, somehow you get this general solution and then apply the method of undetermined coefficients. So here's an important simplification. Since c3e to power x corresponds to the root 1 for the characteristic polynomial for the homogeneous equation dy dx minus y equals 0, we know that when we substitute this into our equation, this part will be 0, and the linearity of the derivative will allow us to ignore this part of the general solution. And what that means is we don't need to worry about this part of our general solution because it will satisfy the homogeneous equation, and in particular, it doesn't matter what c3 is, it's never going to give us 8x minus 5. So we only need to find c1 and c2, where c1 plus c2x solves our original differential equation. So let's go ahead and find y and dy dx. So for y, we can take the part of the general solution that comes from the annihilator, and so our derivative will be, then dy minus y is equals means replaceable, solve for place dy dx minus y with 8x minus 5, and let's rearrange things on the right-hand side. And now I have this equation, 8x minus 5 equals minus c2x plus c2 minus c1. Since this must be true for all x, we need the coefficients of corresponding terms to be the same, and so we need minus c2, the coefficient of x, to be equal to 8, the coefficient of x. Likewise, we need c2 minus c1, the constant term on the right, to be equal to minus 5, the constant term on the left. And solving this system gives us c2 equal to minus 8, and c1 equal to minus 3. And so c1 and c2 are no longer undetermined coefficients, we've determined what they are, and so now we have our general solution to the original differential equation, y equals minus 3 minus 8x plus c3 e to the power x.