[MUSIC] Now, let's introduce the notion of the solution, right? Differential equation contains some unknown function, so we are expected to find it, okay? Or in other words, we are expected to solve it, okay, for the solution. Any function phi(x) is a solution, okay, of differential equation for, right, this ordinary differential equation of order n, right?. On the interval I, if phi(x) is well defined and has nth derivative on the interval I and satisfy the equation for all x in the interval I when phi(x) is substituted for y in this equation. In other words capital F(x, phi(x), phi prime of x, nth derivative of phi) is identical to 0 on the interval I, okay? In this case, we say that phi(x) satisfies the differential equation 4 on the interval I, okay? If the interval I is rather clear from the context, then we usually only refer to the interval I, okay? For example, I claim, for instance, phi(x) = x squared- x to the -1 is the solution of y double prime-2x to the -2 times y = 0 on the interval from negative infinity to 0 and the interval from 0 to infinity, okay? Let's check it through this simple computation, right? So say we have here, phi(x) = x squared- x to the -1, okay? And the given differential equation is linear second-order, so we need to know the second derivative of phi. So first, phi prime of x = 2x, right, and + x to the -2, right? So phi double prime of x is equal to 2- 2 times x to the -3, right? Let's plug these expressions into the differential equation then, okay? So therefore, phi double prime -2x to -2 times phi, this is equal to 2- 2x to the -3 and -2x to the -2 times phi. Phi is equal to x squared -x to the -1, right? And we expect it to be equal to, identical to 0, okay? How much is the product of these two? That is -2, right, including this negative sign in front, so 2, and this cancelled out, right? Now you have a -2 times x to the -3 and -2 times x to the -2 times -x to the -1. The product of these two will be 2 times over x to the -3, right, okay? And because of two negatives down there, you have a plus. So, okay, what I'm saying is that you have this, and this times that makes +2 times x to the -3, that is identical to 0, right, okay? On the interval from minus infinity to 0 and 0 to infinity, okay? I'm excluding the point 0 because of this term, x to the -1 down there, right, which is 1 over x. This is well defined, okay, for non-zero values of x only, right, okay, because you cannot divide anything by 0, okay? Now let's look at the second example. That's again a, First-order linear differential equation. My claim is phi(x) = 2x cubed is a solution to this first-order linear ordinary differential equation on the whole real line, okay? So let's compute phi and phi prime. Phi is now given by the 2x to the cubed, right? What is phi prime? That is a 6x squared, right? Plugging these two expressions into the differential equation, then, how much is x phi prime? x times phi prime, that is equal to 6x cubed, okay? That is equal to 3 times phi, right, okay, very simple, okay? And the solution is well defined and satisfy the differential equation on this whole interval.