Now, let's consider the following example,

which is first order,

linear, non-homogeneous differential equation.

Say, x times y prime + (1 + x)y

is equal to e to_the_negative x times cosine 3x.

First, note that the differential equation is not always standard form.

In other words, the leading coefficients of differential equation,

that is the coefficient of y prime,

is not 1 but x.

So, divide it through by x. Divide it through the equation by x,

then y prime plus (1 + x) over x and y,

that is equal to e to_the_negative x,

over x and times cosine 3 of x.

Now, we can recognize from this expression what is p?

Be careful p is not 1+ x ,

p must be, (1+ x) over x.

What is q?

q(x) is, e to_the_negative x,

over x, times the cosine 3 of x.

What I'm claiming is,

this part is the the little p(x).

And that part is equal to little q(x).

So, what is the integrating factor?

Integrating factor mu of x is equal to,

e to the integral px, dx.

What is p? (1 + x) over x.

So, there we have e to the integral (1 + x) over x, dx.

So, what is the anti-derivative of (1+ x) over x?

That is the same as the 1 over x plus 1.

So you have- actually,

this is equal to e to the log of x and plus x.

By taking the integral constant to be equal to zero,

so what does that mean?

This is equal to, e to_the_x times, e to the log of x,

that is equal to x times e to_the_x.

This is our integrating factor.

This is our integrating factor.

Multiply this integrating factor,

mu of x, on both sides of this equation.

So, what we needed to do is for this whole equation,

multiply it by, x times e to_the_x.

Multiply.

Then what's this left inside?

It should to be, x times e to_the_x times

y and derivative of e. That should be equal to,

x times e to_the_x times e to_the_negative x,

over x times cosine 3x.

This over x canceled out,

e to_the_x, e to_the_negative x canceled out.

So finally you get,

this times that, is equal to cosine 3 of x.

That's the equation we have down here.

Derivative of x times e to_the_x of y, is equal to cosine 3 of x. What does that mean?

x times e to_the_x y is equal to y.

So, from this equation down there,

you have x times e to_the_x,

times y is equal to,

anti-derivative of cosine 3x, dx.

This is equal to one-third.

And the sine of 3 of x plus,

arbitrary constant c. Then finally divided through by,

x times e to_the_x,

you are going to get to our final solution,

y is equal to 1 over x, times e to_the_x,

times one-third sine 3x + c. Now,

let's introduce one another class.

So called the exact differential equation.

First, for any function,

F(x, y) with continuous first partial derivatives,

we call dF.

We call dF. Which is the same as,

dF over dx times dx,

plus dF over dy times dy.

We call this expression,

dF, the total differential of capital F. Then,

along any level curve,

F( x, y ) = constant,

we should have, the total differential,

dF, must be equal to zero.

Because the total differential of any constant is equal to zero.

So, we have such an equation down here.

dF over dx times dx + dF over dy times dy is equal to 0.

Or equivalently, if you wanted to express it,

as y prime is equal to something,

then, you can solve it,

dy over dx is equal to;

negative dF over dx over dF over dy.

Which is the first order differential equation.

Conversely, for a differential equation,

given here, M(x, y) dx + N( x,

y )dy = 0.

This is a first order differential equation.

If this left hand side,

is a total differential of F(x, y ),

then the differential equation can be written as,

because this left hand side is a total differential of capital F,

so you can write is as,

dF(x, y) = 0.

We can solve with this a simple differential equation very easily.

It means simply, F( x,

y) is equal to arbitrary constant c. Arbitrary constant c down there.

So, this is equal to F(x,

y) = c, is a general solution to this first order differential equation.

If this left hand side is known to be the total differential of capital F(x, y).

That motivates us to the following definition.

A force to the differential equation,

M(x,y) dx + N(x,y) dy = 0,

we call it to be exact,

in a plane region R if there is a function capital F(x,

y) such that dF over dx is equal to M(x,

y) and dF over dy is equal to N(x, y).

For all (x, y) in the region R. In other words what?

This left hand side of the differential equation,

is the total differential of capital F(x,

y) in the region R. Then we call,

the given differential equation to be exact.

If it is exactly in the whole plane,

R squared, if R is- the region R,

is the whole plane R_two, R squared.

Then, we simply say that,

the given differential equation, M(x,

y) dx + N(x, y) dy = 0, is exact.