Now, let's introduce another one,

which is the so-called Bernoulli equation,

a first order differential equation of the form,

y prime plus pxy is equal to qx times y to the n,

n is a constant, okay?

Any real constant.

It's called the Bernoulli equation.

Here, I have shown n is not equal to 0 or 1,

and qx is not identical to 0.

Why? If otherwise, if n is equal to 0,

then it is like a y prime plus pxy is equal to q of x.

That's a linear equation.

If n is equal to 1, y prime plus pxy is equal to qxy.

That's, again, a linear equation, okay.

If qx is identical is 0, then it's simpler,

y prime plus pxy is equal to 0, right?

In all those cases,

the problem is a linear problem,

which we can handle easily so that we assume that n is not equal

to 0 or 1 or qx is not identically 0, okay?

Note here that before giving the general solution of this Bernoulli equation,

okay, also have the following effect.

If n is any positive constant,

if n is any positive constant,

then y is equal to identically zero,

it's the trivial solution.

Can you see that? If this power is a strictly positive,

and if y is equal to identical is zero,

then right hand side is zero and left hand side is the zero, right?

So, the differential equation is trivially satisfied.

So, y is equal to 0, constant function.

This is the trivial solution, okay?

Now, let's look at this equation again, Bernoulli equation,

divide this equation, divide the both sides by y to the n,

then you are going to get,

y to the negative n y prime plus p

of x y to the 1 minus n and that is equal to q of x, right?

Divide through the divide of the given Bernoulli equation by y to the n,

then you are going to get this one, okay?

Now set, u is equal to y to the 1 minus n, okay? My claim is that this is a substitution.

Transform this Bernoulli equation into a linear equation say,

u prime plus 1 minus n times pxu is equal to 1 minus n q of x,

which is a really linear force

to the differential equation for the unknown function u.

Why is it that? It's simple, right?

So, from this equation I set, u is equal to y to the 1 minus n, right?

What happened then? What is u prime?

u prime is equal to 1 minus n times Y to the negative n times y prime, okay?

Look at this part.

Look at this equation,

y to the negative ny prime. That is, 1 over 1 minus nu prime , right?

So, the left hand side,

it becomes 1 over 1 minus nu prime plus px,

times what is y to the n minus n?

That is u, and that is equal to q, right?

So multiply through by 1 minus n,

you are going to get

this linear force to the differential equation for the unknown u, right?

That is my claim, okay?

And be careful, here,

I'm dividing the equation through by 1 minus n, right?

Since I'm assuming that n is not equal to one,

1 minus n is never 0 so that we can safely divide this the question by 1 minus n, right?

That's another reason why I'm assuming that n is not equal to one, okay?

Anyway, this last first order linear differential equation,

we know how to handle it, right?

That means we know how to handle this Bernoulli equation, okay?

I will check this thing through the exam first.

The problem is so y prime plus 2 over xy is equal to 2x times y to the 1/2, right?

What is n in this example?

n is equal to one half, right?

Which is not equal to zero,

not equal to one, okay,

divide the equation through y to the 1/2, right?

Then, you are going to get by y to

the negative 1/2y prime plus 2 of xy to the 1/2, and that is equal to 2 of x, right?

Set y is equal to y to the 1/2, right?

Set u is equal to y to the 1/2, right?

Then u prime is equal to 1/2 times y to the negative 1/2 y prime.

So y to the negative 1/2y prime,

it's the same as the two times over u prime and this is equal to 2 over xn times u,

that is equal to 2x, right?

Divide it through by 2 then,

you are going to get u prime plus x to the negative 1 times u, that is x.

Multiply through by x then,

you will get xu prime plus u,

which is the same as derivative of x times u,

that must be equal to x squared, right?