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.
7-1 Differential polynomials – Ex. 1
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
126 個評分
Korea Advanced Institute of Science and Technology
126 個評分
從本節課中
Linear Second order equations 3
與講師見面
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
In the last section,
we have studied how to solve
homogeneous constant coefficients linear differential equation.
We now consider a nonhomogeneous constant coefficients linear differential equation.
Say, differential equation one,
which is a_sub_n times nth derivative
of y plus a_sub_(n-1) times (n-1) derivative of y,
plus a_sub_1 times y prime plus a_sub_0 times y,
which is equal to g(x),
where the all the coefficient a_sub_i are real constants,
and we assume that the leading coefficient a_sub_n is not equal to zero.
Then we know that
the general solution of such differential equation,
such a nonhomogeneous differential equation,
can be expressed as a sum of y_sub_c plus y_sub_p,
where the complementary solution y_sub_c is
a general solution of corresponding homogeneous differential equation.
Say a_sub_n times the nth derivative of y,
plus a_sub_(n-1) times (n-1) derivative of y,
plus a_sub_1 times y prime,
plus a_sub_0 times y which is equal to zero.
And y_sub_p is any particular solution of
original nonhomogeneous differential equation one.
In section 4.3, we have studied how to find the complementary solution y_sub_c.
So, we now consider the problem of finding
a particular solution y_sub_p of the given nonhomogeneous differential equation.
The original differential equation we are considering is
differential equation
one which is a_sub_n times nth derivative of y, a_sub_(n-1) times (n-1) derivative of y,
and plus a_sub_1 times y prime,
plus a_sub_0 times y,
and that is equal to g(x).
So with a notation capital D
is equal to d over dx and the family of constant a_sub_i,
we call any expression of
the form P of capital D which is equal to a_sub_n times D to the n,
plus a_sub_(n-1) times D to the n-1,
plus a_sub_1 times D plus a_sub_0 are differential polynomial.
Any such expression with
the constant coefficient is a differential polynomial.
Then it's easy to see
that any two differential polynomials P(D) and the Q(D), they commute.
In other words, P(D) times Q(D)
acting on y is equal to Q(D) times P(D) acting on y.
We can switch the order of operation to differential operators P(D) and Q(D).
We can exchange them.
From P(D)Q(D) into Q(D)P(D).
As a very simple example,
let's consider the (D-2) times (D+1) of y.
What is it? (D-2) times and the (D+1) of y,
forced acting (D+1) on y,
then you will get (D-2),
D y means y prime and plus one times y that is y.
Apply (D-2) again, then apply D on this expression.
You will get y double prime plus
y prime minus 2 times of this is minus 2 y prime, minus 2 y.
So, finally, we'll get y double prime minus y prime minus 2 y,
which is the same as this D squared minus D minus 2 times of y.
On the other hand,
if you exchange the order of operations for these two differential polynomials,
then you will get d plus one times d minus two.
On the other hand,
(D+1) times (D-2) of y which is equal to
(D+1) and D acting on y is the y prime and minus two times of y,
that is 2 y.
Now apply D again,
then you will get y double prime minus 2
y prime plus one times of this will be y prime minus 2 y,
which is equal to y double prime minus y prime minus 2 y,
which is exactly equal to this expression.
So that we confirm that these two.
(D-2) times (D+1) is the same as (D+1) times (D-2).
In fact, they are equal to this D squared - D - 2.
But be careful, such a commutativity then not be true in general.
For example, if the differential operator has
a variable coefficient then you
cannot exchange the order of those two differential operators.
As a very simple example,
consider the x D and D acting on y,
you will get x of D and y prime that is equal to x y double prime.
However, if you exchanged the order of these two,
then you will get D times x D of y.
So that is equal to D of x y prime.
Then, this is equal to the by the product rule,
x y double prime and plus y prime.
So that these two things are not the same thing.
In other words, you cannot exchange the order of these two operations,
x D times D is not equal to D times x of D.
It's because the one of the operator x D has variable coefficient.
