Okay, let's introduce homogeneous equation. We call a function f(x, y) to be homogeneous of degree alpha, where the alpha is a real number, okay? If f(tx, ty) = t to the alpha times f(x, y), okay? For example, x squared- 2xy + 3y squared is a homogeneous of degree 2. Can you see it? Replace x by t(x), y by t over y,then it'll be t squared times x squared minus 2xy + 3y squared, right? So this is a homogeneous of degree 2. On the other hand, exponential y over x is a homogeneous of degree 0, right? Okay, e to the ty over tx, this is the same as e to the y of x, right? So, okay, the function e to the y over x is a homogeneous of degree 0, okay? Now, we call a first order differential equation. Say Mdx plus Ndy = 0 is a homogeneous if both M and N are homogeneous of the same degree, okay? Then recall the differential equation is simply homogeneous, okay? Assume now that both capital M and the N are homogeneous of degree alpha, okay? And then set u = y over x. In other words, y = x times u, right? Then by the product for dy = udx + xdu. So that the equation (1) becomes, okay, M(x, y)dx, y becomes xu, and + N(x, xu). And instead of dy, we can write udx + xdu, right? Since I'm assuming that the differential equation is a homogeneous, which means both the capital M and N are homogeneous of the same degree, okay? Let the same degree to be alpha, okay? Let the same degree to be alpha then. From this expression, because M is homogeneous, M is a homogeneous of degree alpha. M(x, xu) is the same as x to the alpha times [M(1, u), right? On the other hand, capital N is also homogeneous to the degree alpha, so N(x, xu) is the same as x to the alpha N(1, u), right? So from this two, x to the alpha is common and times [M(1,u)dx + N(1,u)(udx+xdu) = 0]. Divide it through by the x to the alpha, we get M(1, u )dx + N(1, u)(udx + xdu) = 0. Collect the terms involving dx, and collect the terms involving DU, okay? Then you will get, okay, collecting times involving the x. First, you have the m1 comma u, right? Another one you have down there is a u times N(1,u) and dx, right? So coefficient of dx will be M(1, u) + uN(1, u) right? On the other hand, what is the terms involving du, okay? What is the term involving du? We have only one such a case, right? N(1,u) times x, right? So down the xN(1, u), okay, du = 0. This is the new differential equation for the unknown function, u, right? I claim this is separable, right? Divide it through by, okay, x times, M(1, u) + uN(1, u), right, divide it through by that, then you will get 1 over x dx, okay? + N(1, u) over M (1, u) + uN(1, u) and du = 0, right? That's a separable way, okay? And which is easier to solve, right? Just through the integration, right? Likewise, we may take v = x over y instead of u = y over x, right? Okay, this substitution, v = x over y also leads to a original given equation into a separable equation for the u unknown v, okay? Note that if the given differential equation (1), okay, can you remind what the equation (1) is? Then we remind it here for you. Differential equation (1) is M(x, y)dx + N(x, y)dy = 0, right? Okay, it can be written in the following form, right? Okay, dy over dx is equal to, right? That is equal to, right, -M(xy) over N(xy), okay? The differential equation is homogeneous means what? Both the coefficient, capital M and capital N, they are homogeneous of the same degree, say alpha, okay? Then, okay, This is -x to the alpha M(1, y over x), right? Can you read it? Right? And the capital N is equal to x to the alpha N(1, y over x), right? We can write it in that way, okay? So that, this x to the alpha and that x to the alpha canceled out. Finally, you'll get, okay, Down there, dy over dx is equal to initially negative M over N, now using the homogeneity that is equal to negative M(1, y over x over N(1, y over x), right? What is that expression? It's a function of the x and y, but you can read it also as a function of only a single variable, the ratio y over x), right? Say, you may call it as this one as capital F(y over x), right? Okay? That's the characteristic property of homogeneous differential equation, okay? And the converse is also true, right? If we have a force differential equation dy over dx is equal to some function F(y over x). This differential equation is homogeneous, okay?