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.
2-4 Integrating Factor
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
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Korea Advanced Institute of Science and Technology
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First Order Differential Equation 1
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Kwon, Kil HyunProfessor
Department of Mathematical Sciences
As a second class of dispersion first order, differential equation,
let me introduce to you the following.
A first order differential equation is linear if it is of the form
a sub 1 of x times the y prime plus a naught of x times y is equal to g of x.
So really linear in the variables of y and this derivative way.
So we call it as a linear first order differential equation.
Furthermore, if this right hand side,
the g of x is identical to zero,
then we call it as homogeneous first order linear differential equation.
Otherwise, in other words,
if g of x is not identically equal to zero then we record it as
nonhomogeneous first order linear differential equation.
If right hand side is identical to zero then homogeneous.
Otherwise, we record it as a nonhomogeneous.
Let us say, a naught
over a 1 to the p of x and g over a sub 1 to be q of x.
Then starting from the original equation
which I reminded here, differential equation one,
it was a sub 1 of x y prime
plus a naught of x y and that is equal to g of x, right.
Where people call this a sub 1 of x has a leading coefficients.
This is the coefficient of the highest derivative of y appearing in this expression way.
So divided through a 1 of x divided through the equation through the a 1 of x,
then you are going to get y prime plus a naught over
a 1 times y that is equal to g over a sub 1.
We call this a second equation as an domain form of the original equation.
I introduce just another notation for this part,
I will call it to be letter p. This part I will call it to be letter q.
So we have expressions to y prime plus p of x times the y is equal to q of x.
This is a standard form of the linear first order differential equation.
Very special case if p x is equal to identical zero
then the equation reduce to y prime is equal to q of x.
This is the simplest the differential equation
whose general solution is very simply given by general solution y is equal
to any anti derivative of q which I denoted by integral q of x of
the x plus arbitrary integral constant to c. This is a general solution,
as we know quite well from the integral calculus.
In general or in other words the if p of x is not identically zero,
what can we do then?
In that case what we are going to do is the following thing.
Okay. Here we have a standard form of the first order linear differential equation.
Let us say y prime plus p of x times the y and that is equal to q of x.
This is a so called standard form of the first order linear differential equation.
I am looking for some unknown function µ of x,
µ of x such that if I multiply this equation by
µ then the right hand side
will be µ times the q a.
And the other hand,
of the left inside will be µ y prime plus µ p times y. I like to choose µ,
this unknown function µ.
So that, this right hand side is the same as
µ times y and this derivative way.
I would like to choose a µ so that this equality holds.
What does that mean?
Just divide a product for the derivative of µ times y is the same
as µ prime y plus µ y prime by the product through, right.
Again, let us set up the equality between this and that expression.
Then here you have a µ y prime.
Here you have the same term down there so that we can cancel them out.
Then finally you are going to get µ prime of y is equal to µ over p times of y.
In which y is common again so that
you may require that µ prime is equal to p times of µ.
So what I am looking for is,
I am looking for some unknown function µ,
such that if I multiply the given first order differential equation by this µ.
Then µ times the y prime plus
µ times the p of y can be written as a derivative of µ times of y.
Which is the same as the µ prime is equal to p times of µ.
So we get another first order differential equation.
But now, this differential equation is a separable, right.
Can you see it?
We have two variables,
one independent variable x,
and the unknown function µ,
this is an other dependent variable.
So let us separate the variable then you are going to get one over µ
and d µ that is equal to p of x and the d of x.
We really separate the variables here.
How to solve µ? Take intuition of both sides.
From this side you are going to get log of absolute value of a µ.
From this side anti derivative of letter p of x, d of x.
Any one of the anti derivative of it plus arbitrary constant of c,
c sub 1 for example.
You are going to get this expression.
So, for the absolute value of the µ will be
equal to e to the c 1 into letter p of x and the d of x.
So, this is my mistake because I am taking the exponential.
So, it should be e to the anti derivative into letter p x and d of x.
We did it this type of thing the couple times already. What is a µ?
µ is equal to plus or minus into this c 1,
e to the p of x and the d of x to get this work.
So what does that mean?
If I choose µ of x unknown function µ of x to be,
this expression down there.
If I multiply the given differential equation by any one of such functions in
µ then the problem becomes µ times the q.
Now µ is given by this one that is equal to derivative of µ of y.
That is a very one of the simplest differential equation that we can solve.
So here, you have some arbitrary choice of c 1,
but that you may call it as c.
Now the way the c is equal to plus or minus e to the c
1 which is the arbitrary one non zero real number.
So, why not choose it?
Because we do not need all sorts of possibilities for µ.
We need the only one possibility.
So, why not choose the c is equal to one?
That is my choice down there.
Why not choose c is equal to one?
By choosing c is equal to one,
you will get µ of x is equal to exponential integral p x and d x.
That is the equation for down here.
So, multiply the given standard form of
the first order linear differential equation by this function µ of x.
Then the problem becomes a very simple one which is easy to solve.
As you can see now.
So, multiply with this µ with choice of c is equal to
one so multiply this whole equation by this µ then.
Then left hand side will be µ times
the q and the right hand side will be derivative of µ times y.
So you are going to get this equation.
And that means what?
You can not solve this equation very easily.
What is the µ x of y?
µ x of y must be equal to
anti derivative of µ x times the q of x.
Any one of this said derivative plus arbitrary constant c. Now,
from this expression divided the whole thing by the µ of x,
you are going to get y is equal to 1 over
µ x times the integral µ x times the q x and d x plus
arbitrary constant c. This is the general solution
of our given first order linear differential equation down there.
So, once we find the µ of x which is given by this expression then we can
find the general solution of this linear
first order differential equation in the form of the equation five.
We call this a specific choice of µ,
µ is equal to exponential integral p x of d x
as an integrating factor for the linear differential equation, right.
For this linear differential equation we call
this the µ of the x to be on integrating factor, okay?
