Another interesting first order equation is equations with linear coefficients, okay? By which I mean, okay, equation of this type, okay? Is a first order differential equation, and the coefficient of the x and the coefficient of dy both are linear functions in x and y, okay. Here, a1 of x + b1 of y + c1, coefficient of dy is a2 of x + b2 of y + c2, right? Okay, that's why I call as an equation with linear coefficients, okay? First to know to that If both c1 and c2 is equal to 0, if there is no c1 and no c2, then coefficient of dx is a1x + b1y. That's a homogeneous polynomial of degree one, can you read it? On the other hand, if c2 is equal to 0, the coefficient of dy is a2x + b2y. It is again a homogeneous of polynomial of degree one. So that the differential equation three becomes a homogeneous differential equation, which we know how to handle. In general, we now set x = u- h and y = v- k, where u and v are new variables, and h and k are constants to be determined later. Through this simple substitution, the differential equation three becomes, a1u + b1v- a1h- b1k + c1 and du. On the other hand, a2u + b2v- a2h- b2k + c2 and dv, and that is equal to 0, okay? Seems okay? If we can choose two undetermined coefficients, h and k, so that this part will be equal to 0, and that part is equal to 0 too, what do I mean? If there's no this part and there's no part, then the equation number four becomes a homogeneous equation in u and v. Right, can you read it, because in that case, the coefficient of du becomes only a1u + b1v. The coefficient of dv is a2u + b2v, right, okay, that is our goal, okay. In other words, I'd like to choose two constants h and k, so that -a1h- b1k + c1 = 0. And -a2h- b2k + c2 = 0, okay, and we summarize it here, okay? We'd like to choose two constants h and k, so that a1h + b1k = c1, and a2h + b2k = c2. We have two unknowns, h and k, and we have two equations. This is a so-called simultaneous equation for two unknowns, h and k, very simple one. So if we choose h and k so that equation five holds, then the differential equation becomes homogeneous. Look at this simultaneous equation, if the quantity a1b2- a2b1 = 0, okay, can you recognize this quantity on the left hand side? Okay, can you recognize it, this is so called the determinant of the coefficient matrix of this system of equation. Okay, a1, b1 and a2, b2, right, determinant of this two-by-two matrix is a1b2- a2b1. Can you recognize it, if this is equal to 0, then the equation three, what is equation three? This one, okay, if the coefficients, four numbers, satisfy this is equal to zero, right, okay. Then this differential equation, my claim can be put into the form of y prime = f(ax + by), okay? So we assume that, in this case, we know how to handle it, right? We've seen this type of first order differential equation, okay, some time ago, right? So we assume that, okay, this quantity is never equal to zero, okay, it's not equal to zero. Then this is simultaneous first order to the equation has a unique solution for h and k, okay. Do you know how to solve this simultaneous system of an equation? Using the simple factors from the linear algebra, it's best to say that. By the Cramer rule, the first unknown h is equal to determinant of this two-by-two matrix, over determinant of this coefficient matrix, and which is given by that. And this second unknown k is given, similarly, by this, Quantity, with these two choices of h and k in the equation six, right, the equation four becomes, as I said before, right? The equation four, okay, I choose h and k so that this will be equal to 0, and that will be equal to 0. So that equation four becomes a1u + b1v du + a2u + b2v, dv is equal to 0, okay? This what I'm writing down there, okay, there. And we know that this is a homogeneous first order differential equation. Which if you know how to solve it, for example, we said w is equal to d over u, right? That is simple substitution, make this a homogenous equation into a for our differential equation.