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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?