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Let's think about a very simple example.

Â Say, dy over dx is equal to y squared minus x squared over xy.

Â This equation is homogeneous.

Â Can you read it?

Â The top is a homogeneous

Â of degree two and the bottom is a homogeneous of degree two, right?

Â Or you can rather read it as that way,

Â y squared minus x squared over xy,

Â that is equal to, right, okay?

Â Divided through then, this is the y squared over xy, that is y over x, right?

Â Minus x squared over xy,

Â that is x over y,

Â okay? This is f of y over x, okay?

Â What is f? f of t,

Â that is equal to t minus 1 over t, okay?

Â Makes sense?

Â So that's a homogeneous equation.

Â As I said, now,

Â let's just said u is equal to y over x, right?

Â Then the equation becomes u is equal to y over x.

Â In other words, y is equal to xu.

Â So what is the y prime?

Â By the product where u plus xu prime, okay?

Â Let's write inside, I left inside u plus xu prime, right?

Â How about divide inside?

Â As I said here, down there,

Â this is the u and this is the 1 over u.

Â So you have a new differential equation for the unknown u,

Â u plus xu prime is equal to u minus 1 over u,

Â u and u come in both sides.

Â They canceled out so, finally,

Â you get xu prime is equal to minus 1 over u.

Â That is a separable.

Â Let's separate the variable, then you will get,

Â udu is equal to negative 1 over xdx.

Â Now, really the variables are separated here.

Â Taking integration of both sides of this equation,

Â from the left, you get u squared , right?

Â From the right, you get

Â minus, oh, let me correct it, okay?

Â I made a mistake.

Â Integral of udu, that is the equal to u squared over 2, right?

Â So let me write it here,

Â integral of udu must be equal to integral of negative 1 over xdx, right?

Â So this is equal to u squared over 2.

Â That is equal to negative log of absolute value of x plus C, okay?

Â You get this one.

Â In other words, u squared is equal to

Â negative two times of log absolute value of x and plus 2C, right?

Â You get this one or you get rather,

Â this is equal to log of x squared plus 2C, right?

Â Now, you get u squared equal to the log of x squared plus 2C.

Â C is an arbitrary constant,

Â two times of C is another arbitrary constant so you can

Â simply denote it by another C. That's what I did over there, right?

Â So u squared is equal to which is y over x squared

Â is equal to negative log x squared plus C, okay?

Â Multiply x squared in both side as here,

Â then y squared is equal to, right, okay?

Â Cx squared minus x squared over x squared , right?

Â So, x squared is a common, okay?

Â So you get finally,

Â y squared is equal to x squared C minus log x squared.

Â That is a general solution to this homogeneous fourth order differential equation.

Â Okay. Now, let's consider another one.

Â Equation of the form y prime is equal to f of ax

Â plus by where a and b are constant and the b is a non-zero constant.

Â Then, set u is equal to ax plus by.

Â Then, the equation becomes, right, from this,

Â u prime is equal to a plus by prime.

Â So, let's see here these things and write a computation,

Â y prime is equal to f of ax plus by,

Â where b is a non-zero constant, right?

Â Now, I said that u is equal to ax plus by, then what?

Â Then, u prime is equal to a plus by prime, okay?

Â So, what is the y prime then? So, y prime is equal to u prime minus a over b, right?

Â Plugging these expressions into the equation,

Â then you get, instead y prime,

Â we have a u prime minus a over b.

Â That is equal to f of u, right?

Â So what is the equation for u prime?

Â Then, u prime is equal to bf of u plus a, right?

Â That's the equation down there. I write it, okay?

Â u prime is equal to b over f of u plus a,

Â and that's a separable equation, right?

Â Can you see it?

Â Why this is a separable?

Â That is equal to du over dx, right?

Â So divide the whole thing by bf of u plus a and then multiply dx,

Â then you are going to get 1 over bf of

Â u plus a and du and that is equal to d over x, right?

Â That is really the separable differential equation

Â and we really separate the variables, right?

Â So it is, again, easy to solve, right?

Â Let's a look at the problem.

Â initial value problems, y prime is equal to square root of x plus y

Â minus 1 satisfies the initial condition y of 0 is equal to 1.

Â This is right inside the square root of x plus y minus 1

Â is a function of a single variable say, x plus y.

Â This is of the top of this one, okay?

Â So, we said u is equal to x plus y,

Â so that u prime is equal to 1 plus y prime or y prime is equal to u prime minus 1, right?

Â Plugging that expression into the equation,

Â the differential equation becomes u prime is equal to square root of u.

Â After separating variables, you have u to the negative one f of du is equal to dx, right?

Â Integrating both sides, you will get 2 times square root of u,

Â that is equal to x plus C. What is u?

Â u is equal to x plus y so that our general solution is 2 times square root of x plus y,

Â that is equal to x plus C, right?

Â Finally, using the given initial condition,

Â y of 0 is equal to 1, that gives C is equal to 2, right?

Â y of 0 is equal to 1 when x is equal to zero,

Â y is equal to 1 so C is equal to 2,

Â and that means 2 times square root of x plus y is equal

Â to x plus 2 is the solution of this initial value problem.

Â Here, my claim is,

Â 2 times square root of x plus y,

Â that is the core to x plus 2.

Â Square root both sides, then you get 4 times x of a plus y is

Â equal to x plus 2 squared , right?

Â Solve this equation for y.

Â Solve this is equation for y.

Â Through the simple algebra,

Â you will get the solution in the explicit form,

Â y is equal to one quarter times x squared plus 1, okay?

Â