Okay, as an example, let's consider the initial value problem, (x + 2y- 1)dx + (2x + y + 1)dy = 0 satisfying the initial condition y(-1) = 2, okay? This is a differential equation with linear coefficient, right? (x + 2y- 1) and (2x + y + 1), okay? I skipped some necessary computation part which is easy to check, set x = u- 1 and y = v + 1, okay? Then the equation becomes a homogeneous equation, say (u + 2v)du + (2u + v)dv = 0, so w = v over u, okay? Let me remind you the equation we have before. Here we have (u + 2v)du + (2u + v)dv = 0, now I set, w = v over u, so that v = uw, and that means dv = wdu + udw, right? Plug in all of the things into the equation, plug in this expressions into that then the equation becomes (2w + 1) du + (w +2)(udw + wdu) = 0, okay? So collect the terms involving du and the collecting terms involving dw, then you will get w squared + 4w + 1 du + u times the( w + 2)dw = 0. This is a sample of a differential equation, divide the things y u times w squared + 4w + 1, then you separate the variable and we get 1 over u du + w + 2 over w squared + 4w + 1 times dw = 0, right? Take integration from this one you'll get log of absolute value of u, from this, okay? Look at the top and the bottom, derivative of bottom will be 2w + 4, that is a two times of this, w + 2 all right? So, that integral of this expression will be, one-half of log absolute value of w square + 4w + 1 and finally, we need the integration constant to c sub 1, okay? Multiply this by 2 first, so there you have 2 times of log of absolute value of u which is the same as the log of u square and plus log of absolute value of w squared + 4w + 1 = 2 times of c1 way, okay? Removing the log signs, you will get, u squared times w squared, + 4w + 1 = c, which is nonzero constant, right? Here I am skipping the following things, right? For this equation, first multiply 2 on both sides, you get 2 log of absolute value of u + log of w squared + 4w + 1 and that is = 2c1, okay? Through the log formula, this is the same as log of u squared times w squared + 4w + 1, it's absolute value and this is = 2 times of c1. So u squared times the absolute value of w squared + 4w + 1, this is = e to the 2c1, okay? Then removing this absolute value sign, instead you should have + or -, right? Then I equate it by c, new constant c and because of this exponential, it cannot be a 0, it can be arbitrary nonzero constant, right? That's what I'm claiming down there, u squared times w squared + 4w + 1 = c and c is arbitrary nonzero constant, right? Lets use the w = v over u, right? Where is it? w = v over u right? Plugging this into this expression, and to multiply it out then, you will get v squared + 4uv + u squared, that is = arbitrary non-zero constant c, right? What is v? It was y- 1, right? What is the u? That was x + 1, right? Okay, so going back to the original variable x and y, you will get, This is = c and c is a nonzero arbitrary constraint, that is, general solution of original differential equation, okay? Original force to the differential equation right here, right? That's the general solution but we are concerned with the initial value problem with the initial condition of y of -1 = 2. So using this initial condition, finally, it gives c = 1. So that expression down there or expanding everything out you will get y squared + 4 x y + x squared- 2 x + 2 y = 3. This is a solution to our original initial value problem, say this equation.