Let's think about a very simple example. Say, dy over dx is equal to y squared minus x squared over xy. This equation is homogeneous. Can you read it? The top is a homogeneous of degree two and the bottom is a homogeneous of degree two, right? Or you can rather read it as that way, y squared minus x squared over xy, that is equal to, right, okay? Divided through then, this is the y squared over xy, that is y over x, right? Minus x squared over xy, that is x over y, okay? This is f of y over x, okay? What is f? f of t, that is equal to t minus 1 over t, okay? Makes sense? So that's a homogeneous equation. As I said, now, let's just said u is equal to y over x, right? Then the equation becomes u is equal to y over x. In other words, y is equal to xu. So what is the y prime? By the product where u plus xu prime, okay? Let's write inside, I left inside u plus xu prime, right? How about divide inside? As I said here, down there, this is the u and this is the 1 over u. So you have a new differential equation for the unknown u, u plus xu prime is equal to u minus 1 over u, u and u come in both sides. They canceled out so, finally, you get xu prime is equal to minus 1 over u. That is a separable. Let's separate the variable, then you will get, udu is equal to negative 1 over xdx. Now, really the variables are separated here. Taking integration of both sides of this equation, from the left, you get u squared , right? From the right, you get minus, oh, let me correct it, okay? I made a mistake. Integral of udu, that is the equal to u squared over 2, right? So let me write it here, integral of udu must be equal to integral of negative 1 over xdx, right? So this is equal to u squared over 2. That is equal to negative log of absolute value of x plus C, okay? You get this one. In other words, u squared is equal to negative two times of log absolute value of x and plus 2C, right? You get this one or you get rather, this is equal to log of x squared plus 2C, right? Now, you get u squared equal to the log of x squared plus 2C. C is an arbitrary constant, two times of C is another arbitrary constant so you can simply denote it by another C. That's what I did over there, right? So u squared is equal to which is y over x squared is equal to negative log x squared plus C, okay? Multiply x squared in both side as here, then y squared is equal to, right, okay? Cx squared minus x squared over x squared , right? So, x squared is a common, okay? So you get finally, y squared is equal to x squared C minus log x squared. That is a general solution to this homogeneous fourth order differential equation. Okay. Now, let's consider another one. Equation of the form y prime is equal to f of ax plus by where a and b are constant and the b is a non-zero constant. Then, set u is equal to ax plus by. Then, the equation becomes, right, from this, u prime is equal to a plus by prime. So, let's see here these things and write a computation, y prime is equal to f of ax plus by, where b is a non-zero constant, right? Now, I said that u is equal to ax plus by, then what? Then, u prime is equal to a plus by prime, okay? So, what is the y prime then? So, y prime is equal to u prime minus a over b, right? Plugging these expressions into the equation, then you get, instead y prime, we have a u prime minus a over b. That is equal to f of u, right? So what is the equation for u prime? Then, u prime is equal to bf of u plus a, right? That's the equation down there. I write it, okay? u prime is equal to b over f of u plus a, and that's a separable equation, right? Can you see it? Why this is a separable? That is equal to du over dx, right? So divide the whole thing by bf of u plus a and then multiply dx, then you are going to get 1 over bf of u plus a and du and that is equal to d over x, right? That is really the separable differential equation and we really separate the variables, right? So it is, again, easy to solve, right? Let's a look at the problem. initial value problems, y prime is equal to square root of x plus y minus 1 satisfies the initial condition y of 0 is equal to 1. This is right inside the square root of x plus y minus 1 is a function of a single variable say, x plus y. This is of the top of this one, okay? So, we said u is equal to x plus y, so that u prime is equal to 1 plus y prime or y prime is equal to u prime minus 1, right? Plugging that expression into the equation, the differential equation becomes u prime is equal to square root of u. After separating variables, you have u to the negative one f of du is equal to dx, right? Integrating both sides, you will get 2 times square root of u, that is equal to x plus C. What is u? u is equal to x plus y so that our general solution is 2 times square root of x plus y, that is equal to x plus C, right? Finally, using the given initial condition, y of 0 is equal to 1, that gives C is equal to 2, right? y of 0 is equal to 1 when x is equal to zero, y is equal to 1 so C is equal to 2, and that means 2 times square root of x plus y is equal to x plus 2 is the solution of this initial value problem. Here, my claim is, 2 times square root of x plus y, that is the core to x plus 2. Square root both sides, then you get 4 times x of a plus y is equal to x plus 2 squared , right? Solve this equation for y. Solve this is equation for y. Through the simple algebra, you will get the solution in the explicit form, y is equal to one quarter times x squared plus 1, okay?