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One more example.

Now, let's think about the initial value problem,

differential equation is a 6xydx + (4y + 9x squared) dy = 0.

And initial condition is y(0) = 1.

Again, let's check the differential equation,

to be exact or not.

dM over dy - dN over dx.

From this one, dN over dy that is a 6x.

From this expression, dN over dx is 18x.

Subtract it, you get negative 12 times x which is not identical as 0.

So, the given equation is not exact.

Let's compute the following quantity.

1 over M (dM over dx minus dM over dy).

Then that is equal to 2 over y

because this quantity is negative 12x divided by 6xy that is equal to 2 over y.

So, this is a function of y only.

Then there is an integrating factor on view which is also a function of a y only.

And in fact, v over y is a given by

exponential integral of 2 over y_dy that is equal to y squared.

So this is an integrating factor.

Multiply the given equation by this integrating factor.

So you get 6xy cubed dx +

(4y cube + 9x squared y squared) dy is equal to 0.

And this must be exact.

This must be exact. What does that mean?

Again, there must be a function capital F(x,y)

whose x partial is equal to 6xy cubed and

whose y partial derivatives is 4y cubed + 9x squared y squared.

Let me use the first equation.

Say, dF over dx is equal to 6xy cubed.

Then capital F(x, y) through the x integration it'll be

3x squared y cube plus arbitrary integral constant which is arbitrary function of,

arbitrary differentiable function of y, g(y).

Now take y partial of this capital F. You have a 9x squared y squared

+ g prime of (y) must be equal to, dN.

dN, right here.

dN, right here.

So, compare these two equations then.

9x squared y squared cancelled out,

and then this g prime of (y) is equal to to 4y cubed.

So then this g(y) must be equal to y to_the_four.

Therefore, 3x squared y cubed + y

to_the_four is equal to arbitrary constant c. That's a general solution.

Finally, let's use the initial condition y(0) = 1.

When x is equal to zero,

y is equal to one,

that gives immediately c is equal to one.

So the solution of the initial value of problem is

3x squared times y cubed + y to_the_four.

And that is equal to 1.

So far, I introduced

a subclass of first order differential equations which are rather easy to solve.

For example, I introduced the linear first order differential equation,

and the separable first order differential equation,

and the exact differential equation,

for which we have a simple analytic tools to solve those differential equations.

But there are many other- the first order differential equation,

which is neither linear,

separable or nor exact.

In this section, I would like to show

some other subclass of first order differential equations which may be

transformed into another form by a suitable substitution,

which is easy to solve.

As a first of such example,

let's consider this as specific differential equations.

Say, xe to_the_y y prime + e to_the_y = cos x.

Just look at this differential equation carefully.

And then, try the following.

I will set u = e to_the_y.

Then, u prime is equal to by the chain rule,

derivative of exponential function is itself.

And because of y is the function of x,

so derivative of e to_the_y will be e to_the_y times y prime.

With this a simple substitution,

the given differential equation becomes now e to_the_y y prime,

there is a u prime.

So xu prime + e to_the_y, that is u,

and the xu prime + u is the same as derivative of x times u.

Then, it must be equal to cosine of x.

Now, look at the differential equation we obtained now for the due unknown function u.

This is the first order differential equation,

which is very easy to solve.

What is x times u of x?

It's sine to derivative of cos x.

In other words, that is sin x + c. So now, we have this one.

x times u, that is x times e to_the_y because u is equal to e to_the_y,

that is equal to sin x + c. That's a general solution of the given differential equation.

We can solve this first to the differential equation through a very simple substitution.

Say, u is equal to exponential y.