0:04

As a concrete example,

let's try to solve the following first-order differential equation.

Say, y prime = x + 1 over y squared, right?

Then you can rewrite it, right?

0:20

By multiplying the equation through by the y squared times the dx,

you are going to get y prime dy is equal to (x +1)dx, right.

Now the variables are separated already.

Taking integration on both side is one of the antiderivative of y squared.

That is y cubed over 3, right?

Antiderivative of (x + 1).

That is x squared over 2 plus x, right?

Okay, then we should add one integral constant or we call it c1, okay?

1:17

So, you can solve this equation even for y,

so y is equal tto, okay, cumulative 3x

squared over 2 + 3x + 3c1, okay?

And where the c1 is the arbitrary constant,

then 3c1 is an another arbitrary constant which you may call by, c, okay?

So finally we get, y = cube root of 3x

squared over 2 + 3x + c, okay?

This is a general solution of, given differential equation down there, okay?

So, very simple process, okay?

2:04

Another example.

Now, let us try to solve the Initial Value Problem.

Say, y prime is equal to y-1 over 2-x, and

satisfying the initial condition y over negative one is equal to two, okay?

2:22

And again rewriting this equation into the following form,

okay, 1 over y-1 times dy = 1 over 2-x of dx, right.

Now, the variables are separated here, integrating both side is,

okay, from the left-hand side, antiderivative of 1

over y-1 is log of absolute variable y-1, right?

Likewise, antiderivative of 1 over 2-x is

negative log absolute value of x-2, right?

3:02

Then you needed to add one integral constant, say c sub 1, okay.

Then solving this equation for this absolute value of y-1,

you will get, absolute value of y-1 = e to the,

this expression, right, okay.

This expression down there, okay.

Which is the same as e to the c1 times

e to the negative log absolute value of x-2, right.

3:36

How about this second expression down there?

Can you see that this is the same as, okay, 1 over the absolute value of x- 2?

Because if we have, I think you know that e to the log of x is equal to x, right?

3:54

So, e to the negative log of absolute value of x-2,

is the same as 1 over absolute value of x-2, right.

So together with this part, you will get e to the c1 over absolute value of x-2.

Let's call this e to the c1 to the c2, okay.

The c2 is an arbitrary positive number because exponential

of any real number is just takes to be positive.

Right, okay, so here the constant c2,

this can be arbitrary positive number okay.

So we have this expression, absolute value of y-1

= c2 over absolute value of x-2, right, okay?

Now, let's try to remove the absolute value sign, okay?

4:51

Then we are going to, sorry.

Okay, we are going to get, removing this absolute value sign,

we will get y-1 is equal to plus or

minus c2, 1 over x-2.

By removing this absolute value sign, we have two different choice

of the signs either plus or minus right in front, okay?

Now, lets write to the c to be plus or minus c2.

Since c2 is strictly positive arbitrary number,

plus or minus c2 can be arbitrary non-zero real number, okay.

That equals c, okay.

So finally, we get the expression y-1 is equal to c over

x-2 where c is arbitary non-zero number, okay.

5:49

Let's check the some exceptional case where c is equal to 0.

What happens to this expression when c is equal to 0.

If c is equal to 0, the right-hand side is equal to 0.

So that y must be equal to identically 1, okay.

And y is equal to identically 1 should satisfy the given

differential equation as you can see from here, right.

Let's remind you our original differential equation, y prime is equal to

y-1 over 2-x, right?

So now I claimed that y is equal to identically 1 which you

will obtain by setting c = 0, in this expression, right.

This is a trivial solution to this problem,okay, this differential equation.

Okay, so that means what?

In the expression four,

c can be really arbitrary real number including even 0, right.

Okay, so that one parameter family given by this

expression four, including the zero for c, okay,

with really arbitrary constant c is a general solution

of the given first order differential equation, okay.

Now finally, let's consider the given initial condition

which was the y-1 = 2, right, okay.

So from this expression, right, okay,

you have the general solution y- 1 = c over x- 2, okay.

Then using the initial condition y(-1) = 2, right?

Plugging those values then, when x = -1, y = 2.

So, 2- 1 = c over -1- 2, that is equal to c over -3.

The left-hand side is equal to 1,

so what does that means?

c = -3, right, okay?

7:56

So you can see that, right.

So our solution to this Initial Value Problem will be,

y = -3 over x- 2 + 1, okay.

This is the solution to this given Initial Value Problem, okay.