3.3 Integrating Factor III

Loading...

From the course by Korea Advanced Institute of Science and Technology

Introduction to Ordinary Differential Equations

126 ratings

Explore Related Videos

At Coursera, you will find the best lectures in the world. Here are some of our personalized recommendations for you

Korea Advanced Institute of Science and Technology

Introduction to Ordinary Differential Equations

126 ratings

In this introductory course on Ordinary Differential Equations, we first provide basic terminologies on the theory of differential equations and then proceed to methods of solving various types of ordinary differential equations. We handle first order differential equations and then second order linear differential equations. We also discuss some related concrete mathematical modeling problems, which can be handled by the methods introduced in this course. The lecture is self contained. However, if necessary, you may consult any introductory level text on ordinary differential equations. For example, "Elementary Differential Equations and Boundary Value Problems by W. E. Boyce and R. C. DiPrima from John Wiley & Sons" is a good source for further study on the subject. The course is mainly delivered through video lectures. At the end of each module, there will be a quiz consisting of several problems related to the lecture of the week.

From the lesson

FIRST ORDER DEFERENTIAL EQUATION 2

3.3 Integrating Factor III10:57

3-4 Substitutions, Ex 18:09

Meet the Instructors

Kwon, Kil Hyun
Professor
Department of Mathematical Sciences

Consider nonexact differential equation,

which is M dx plus N dy is equal to zero.

I'm assuming this is nonexact.

If this quantity, say 1 over N times dM over dx,

minus dN over dx, it depends only on x.

Then, let's go back this one.

This equation, right?

If this part, this coefficient part,

1 over N, dM over dy minus dN over dx,

if this is a function of x only,

then you can solve this first order differential equation very easily, right?

Can you recognize it?

Let me remind it here.

Think about the following thing.

Y prime is equal to f(x) times y.

This is of this formula.

The equation is of this point.

What is the solution of this simple first order differential equation?

This is separable differential equation,

and then in fact that you have y is equal to

arbitrary constant times e to the integral of f(x)dx.

This is the general solution. Can you see it?

I skipped to several steps,

from this to that.

But you can fit it up very easily

because this is a separable differential equation, right?

Or you can check it.

Starting from this one,

what is a derivative of y?

From here, derivative of y.

This is the c times derivative of exponential is exponentially safe.

And by the Chain Rule,

you need the derivative of integral f(x)dx.

That is f(x), right?

So this is 1.

C times exponential integral f(x)dx.

This is equal to exactly y.

So that simply you get f(x) times y.

That's the differential equation given.

Okay? So in the first case,

in the equation (8),

when μ(x,y) is equal to μ(x) is a function of only x only,

then how can you find this μ from here,

μ(x) is equal to arbitrary constant times exponential integral of 1 over N,

dM over dy minus dN over dx.

Even though we have arbitrary constant c in front,

but you can take it to be 1 to make a thing this simple because

we do not need whole family of integrating factor.

We simply need any one of them.

So why not take c is equal to 1.

By the same token in the second case,

if there is an integrating factor μ,

which is a function of y only,

so that μ satisfy this simple force to the differential equation.

What is μ? μ is equal to exponential integral of 1 over

M times dN over dx minus dM over dy and the dy, okay?

This so I the summarized in this theorem down there.

So we now consider our nonexact differential equation: M(x,y)dx + N(x,y)dy = 0.

This is nonexact.

If 1 over N times dM over dy minus dN over dx is a function of x only,

is a function of x only,

then the integrating factor we are looking

for can be given by μ(x) is equal to exponential integral,

and 1 over N times dM over dy minus dN over dx and the dx.

This is an integrating factor of this nonexact differential equation.

On the other hand,

if the quantity 1 over capital M times dN over

dx minus dM over dy is a function of a y only,

then we have an integrating factor,

μ(y) which is equal to exponential integral of 1 over M,

dN over dx minus dM over dy, and the dy.

Once we can find such an integrating factors,

the problem of solving the original given equation is easy enough.

Multiply this equation by the integrating factor either this one or that one.

Then apply the rule we found before, right?

I'll check whole process through the examples again.

Here's an example.

Let's try to solve this equation, first order equation.

(x + y)dx + x ln xdy = 0.

What is the dM over dy?

That is 1, right?

How about the dN over dx?

That is equal to ln x + 1, okay?

Subtract these two, that is equal to negative log of x is not identically zero.

That means the given differential equation is not exact.

But let's think about the following quantity.

Say 1 over capital N times d capital M over dy minus d capital N over dx,

and that is equal to negative 1 over x.

It's a function of x only.

What does that mean?

That means there is an integrating factor μ(x) which is

given by exponential of integral minus 1 over x dx.

This is equal to simply 1 over x.

Can you see it?

What does that mean?

Multiply this integrating factor 1 over x on both sides of this equation.

Now you have then 1 over x times x plus y

dx plus x times the log of x over x is simply log of x.

So log of x dy is equal to zero.

And I claim that this is exact. Let me confirm it.

Is that really exact then?

I'm saying that 1 over x,

(x + y) and the dx + ln x dy = 0.

And I claim that this is exact. What does that mean?

This is the μM, this is the μN.

So, you need to check about the dM over dy.

Can you compute the dM over dy from this expression?

And dM over dy is what?

This is the 1 over x and 0 plus 1,

so that is 1 over x, right?

How about the dN over dx?

That's 1 over x again.

dN over dx.

So these two coincide,

and then means this is exact.

Now how to solve that equation then?

This is exact means there is some function capital F(x,y).

Capital F(x,y) whose x of partial is equal to M and whose y

partial is equal to N. Now let me take the first equation.

The x partial of the capital F,

dF over dx is equal to 1 over x times x plus y,

that is equal to 1 plus y over x.

So what is F then?

Through the integration, capital F is equal to integral of 1 + y over x and the d(x).

So that gives you x + y times ln x

+ integral constant which is any differentiable function of a y, okay?

Which I recorded by g(y).

Now that must satisfy another the identity say,

dF over dy must be equal to capital N. The μ capital N down there,

there is a log of x.

So first, dF over dy from this expression,

that is y derivative zero,

y derivative that is log of x,

y derivative that is a g prime of y,

that must be equal to down there log of x.

So, log x canceled from both sides.

Finally, you get g prime of y is equal to zero,

and that means we can take g to be zero.

In other words, finally,

the x + y times ln x is equal to

arbitrary constant c. This is the general solution of given differential equation.

Either this one or the original differential equation is

(x+y)dx + x ln x dy = 0.

This is the equation originally given in this example.

Explore our Catalog

Join for free and get personalized recommendations, updates and offers.

Coursera provides universal access to the world’s best education, partnering with top universities and organizations to offer courses online.

© 2019 Coursera Inc. All rights reserved.

Download on the App Store

Get it on Google Play