Now then, there arise a very nature of question, right? How to find an integrating factor, okay? So, let b(x,y) be an integrating factor of a nonexact to differential equation M(x,y)dx + N(x,y)dy = 0, okay? What does that mean? Then by Theorem 2.2, We should have, right, if you multiply the given equation here by the function view, it must be exact, right? In other words, view M(x,y)dx + N(x,y)dy = 0 must be exact, right? And then missed by Theorem, okay, partial derivative of mu times M with respect to y, must be equal to partial derivative of mu times N with respect to x, right? Take the y partial and the x partial of these two quantities by the product floor and the variety, okay? Then you can get that equation 7 easily, which is M times dmu/dy- N times d mu over dx is equal to dN over dx- dM over dy times mu, okay? In this equation, mu is unknown. It's unknown of two very exact same x and y, okay? This is a positive differential equation for the unknown function mu. Unfortunately, this is a positive differential equation is in general harder to solve. Than the original, ordinary differential equation Mdx + Mdy = 0. Right. It's no good. But think about the following thing. Okay. However, if this unknown integrating factor mu depends only on one variable x or y, then the equation 7, the partial differential equation 7 becomes much simpler okay? Can you see it? Let's go back to the differential equation 7. This is a partial differential equation. The trouble we have for this differential equation is, because mu function of two variables, the x and y, and the equation involves both d Mu and dy and d Mu / dx, so it becomes partial differential equation which is usually very hard to solve. So to make it very simple, now I am assuming that mu is a function of only one variable, either x or y, okay. For example, if mu is a function of only y variable, then dmu/dx is equal to zero. On the other hand, if du is a function form the x variable, then d mu / dy will be zero, and the remaining equation will be an ordinary differential equation for the unknown function mu which is the function of X only or function of Y only. Okay, that's what I mean. Here we have the results. Right. As I said, if mu depends only on one variable, either x or y, okay? For example, if mu is a function of only x variable, then d mu / dy is automatically 0, so that the partial differential equation 7 becomes a much simpler ordinary differential equation, okay? d Mu/ dx = 1/N(dM / dy- dN/dx)mu, okay? On the other hand, if mu(x,y) is a function of only y only, then, the mu over dx will be automatically 0, so that the partial differential equation (7) reduce to ordinary differential equation (9), say dmu/dy = 1/m(dN/dx- dM/dy)mu. Right? But here the note to the following thing, okay? In the equation a, okay? Left hand side must be a function of x only because we assume that mu is a function of x only, okay. And that is unknown, dx is unknown, right? How about this coefficient part? Okay. 1/N(dM / dy)- dN/dx), right? Because both M and N, they are functions of x and y. The quantity down here is usually a function of x and y, okay? If it is really a function of x and x, then the differential equation 8 makes contradiction, right? Because the left hand side is a function of x only, the right hand side is the function of x and y, okay? They cannot be the same thing. By the same token, in the equation 9, when we assume that mu is a function of only y variable, the left hand side is a function of only y, but right side can possibly be a function of both x and y, okay? Unless this part 1/M(dN/dx- dM/dy) is also a function of y only, okay? So, we can make the following observation, okay?
