0:22

As the third example okay, let's consider the again,

the first order differential equation.

3x squared + 2x- 1 times dx =

sine of y + e to the -y times dy, okay?

The different equation is already given in the seperated from, right?

So simply by taking integration of both sides,

left-hand side, and this right-hand side.

It'll give to you anti-derivative of 3x squared is x

cubed anti-derivative of 2x, that is x squared,

anti-derivative of 1, that is x + some arbitrary

constant c = anti-derivative of sine of y, that is -cosine of y,

anti-derivative of exponential- y, that's -e -y, okay?

So this expression down there, this is implicit form of the general solution for

the given first order nonlinear differentiated equation okay?

The final solution down there as you can see it's impossible

to solve this equation for y in terms of x, right?

So you can leave this form down there and

the claim that this is a general solution to this given,

the first ordered differential equation, okay?

2:01

As one more example, consider the first derivative

differential equation, y prime = to y square- 1, okay?

We can separate the variable easily,

divide the equation through by the y squared- 1 and

multiply dx on both sides then you can separate

the variables as a dx = dy over y squared- 1, okay?

Let's take a partial fraction decomposition for

1 / y squared- 1 which is the same as one-half times 1 / y- 1- 1 / y + 1, okay?

2:49

Can you confirm it right?

So, okay the equation becomes, let's rewrite in this way, right?

The equation is, the separate equation,

multiply 2 on both sides is of that expression.

You're going to get 2 times dx = (1

over y -1,- 1/ y +1) dy, okay?

Taking integration on both sides now then the anti-derivative of 2,

that is = 2 of x, + arbitrary constant c sub 1, okay?

3:26

Anti-derivative of this 4 is to 1/y- 1,

that is the log of absolutely value y- 1, right?

So here you're going to get, log of absolute value

absolute value y- 1 and- anti-derivative

1/ y + 1 is log of absolute value y + 1.

By the log of properties, you are going to get this is the same

as log of absolute value of y- 1/ absolute value y + 1, right?

So we have the expression down there,

2x + 1, 2x + c1 = log absolute

value of y- 1/y + 1, okay?

Let's try to solve this equation,

for absolute variable y- 1/ y +1, okay?

Can you do that?

Then you get, this is the same as 2x + c sub 1, okay?

So absolute variable of y- 1 / y +

1 = e to the 2x + c 1, that is the same

as e to the c1 times e to the 2 / x, right?

Then remove this absolute value sign on the left- hand side,

you are going to get y- 1/ y + 1 = + or -, can you see down there,

+ or- e to the c1 times e to the 2x, right?

So by setting c = + or- e to the c1,

which is nonzero arbitrary number,

we get the y- 1/y + 1 = c times e to the 2x, right?

That's the general solution of this given differential equation, okay?

5:33

This expression, you can solve it for y in terms of x so very easily, okay?

Through the simple algebra, if you solve it for

y, then you're going to get y = 1 + ce to the 2x over 1- ce to the 2x, right?

Remember that until this point, c can be any nonzero real number, okay?

6:02

Let's look at the exceptional case again.

What happen when c = 0?

If c = 0 in this expression, y becomes identically 1, okay?

Is it the solution to the given problem?

If y = identically 1, derivative is 0 and

when y = 1, right hand side= 02.

That means y = identically 1 is also a solution to

the this given differential equation, okay?

So we can say that, this expression down there

y = 1 + ce to the 2x/1- ce to the 2x,

where c is real arbitrary real number is

a general solution to this given differential

equation, are you following me?

Let's go one step further, okay?

Look at this differential equation again.

7:09

Theright hand side becomes the 0 not only when y = 1,

but also for y = to identically to -1, okay?

When y = to identically -1, what happen to this left hand side?

Derivative of any constant function is 0, right?

What does that mean?

y =- 1 is also a solution to this first

order differential equation, okay?

Let's remind the following here,

what do you mean by a singular solution of certain differential equation?

7:48

We have a general solution down there which containing one arbitrary

constant starting from the first order differential equation,

and we have a 1 on other solution y = identically -1, right?

And there is a simple algebra to check that, okay?

You cannot obtain y = -1 via signing any specific

value of c in this expression, okay?

In other words with no choice of c you can get y = -1 from the expression 5 okay?

That will really mean a singular solution, okay?

In other words this is a solution, cannot be obtained from

this general solution 5 for any choice of c, okay?

So, we solve the given first order differential equation,

in the following way.

We first get general solution where c is arbitrary real number and

then we also get another solution, y = -1, which is a singular solution, okay?