 So let's introduce a powerful way to represent and solve differential equations known as a linear differential operator. So this goes back to one of the many ways we have of introducing the derivative. One way to indicate the process of differentiation is to use what's known as the differential operator, which we write as d over dx. And so the first derivative of y with respect to x is d over dx applied to y, which gives us dy over dx. The second derivative of y with respect to x is then going to be the differential operator, d over dx, applied to dy over dx. And that gives us d2y over dx squared. And if you've ever wondered why the second derivative is written in this fashion where we have a d2y over dx squared, it's because of this operator notation. Differential notation and the differential operator are very helpful when trying to solve differential equations. So we'll be using this differential operator a lot, and so for simplicity we'll let capital D be the differential operator, d over dx. Since D represents the derivative, it has the following useful properties. D applied to a sum, f plus g, is df plus dg, which is to say the derivative of a sum of functions is the sum of the derivative of the individual functions. Likewise, d applied to cf for any constant c is c times df. Again, the derivative of a constant times a function is the constant times the derivative of the function. So except for the fact that we're using d to represent the differentiation operation, these are just the familiar rules of derivative from calculus. There is one new thing, any function that acts this way, where if we apply it to a sum, we get the sum of the function applied to the individual terms. And where we apply it to a constant times something, we get the constant times the function applied to the something. It's what we call a linear operator. And so we say that d is a linear operator. The differential operator allows us to rewrite differential equations in operator form. For example, if we have the differential equation, second derivative of y plus first derivative of y minus 2y equals zero. So the second derivative is the derivative of the derivative of y. The derivative is just the derivative of y. And minus 2y is just 2y. Or using our capital D notation for the differential operator, this is d squared y plus dy minus 2y equals zero. So it seems that all of these terms have a common factor of y that we could remove. But they don't because this does not represent a multiplication of d and y. It's the derivative of y. But if you remember your function notation, I can add two functions f plus g of x. And what that means is f of x plus g of x. f of f of x, well that's really f2 of x. And a constant times f of x, we can write that as cf of x. And so while it's important to understand that this does not say we're multiplying y by d squared or y by d, function notation allows us to rewrite it as if it were a product. So we can rewrite this as d squared plus d minus 2 applied to y. And this looks like an awful lot like a polynomial. And so we call d squared plus d minus 2 the characteristic polynomial for the differential equation. Or because it consists only of derivatives and constants, it's also a linear operator. And so we also call it the linear differential operator. So let's try to identify the linear differential operator if possible for the two differential equations. Since our first equation is linear, it has no powers of y or any of its derivatives. It has a linear differential operator. So first we can rewrite this in operator notation. So we start with our original differential equation. This is the fourth derivative of y. This is the second derivative of y. And this is just 8 times y. So the fourth derivative, that's really just the derivative operator applied four times. So we can write that as d to the fourth. The second derivative is our operator applied twice, d squared. And 8 is just a constant. And again, it looks like we can remove a common factor of y, but we can't. But we can rewrite this in function notation as if we were removing a common factor of y. In any case, our linear differential operator is this function d to the fourth minus d squared plus 8. How about the second equation? So the first thing to recognize here is that this is not a linear differential equation because it has this product of a derivative and the function itself. And since this equation is not a linear differential equation, it has no linear differential operator.