﻿ 2019학년도 1학기

2019학년도 1학기

반도체 공학과 공학수학1

주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition

부교재: 이상구 외 4인, 최신공학수학 I,  1st Edition

강의시간:  공학수학1, (화09:00-10:15)   (목10:30-11:45)

담당교수:  김응기 박사

Week 4

4주차

2.6: Existence and Uniqueness of Solutions. Wronskian

2.7: Nonhomogeneous ODEs

4.4 비동차 선형미분방정식

3.6 매개변수변화법(1계)

2.6 Existence and Unique, Wronskian

The general theory of homogeneous linear ODEs

(1)

with continuous, but otherwise arbitrary variable coefficients and .

This will concern the existence and form of a general solution of as well as the uniqueness of the solution of initial value problems consisting of such an ODE and two initial conditions

(2)       ,

with given , and .

The two main results will be Theorem 1, stating that such an initial value problem always has a solution which is unique, and Theorem 4, stating that a general solution

(3)          ( and are arbitrary constants)

includes all solutions.

Hence linear ODEs with continuous coefficients have no “singular solutions.”

Theorem 1  Existence and Uniqueness Theorem for Initial Value Problems

If and are continuous function on some open interval and , then the initial value problem consisting of and has a unique solution on the interval .

Linear Independence of Solutions

A general solution on an interval is made up from a basis , on , that is, from a pair of linearly independent solution on . Here we call , linearly independent on if the equation

(4)       on    implies   ,

We call , linearly dependent on if this equation also holds for constants , not both . and are proportional , that is

(5)       (a)   or   (b) (for all on )

Theorem 2  Linear Dependence and independence of solutions

Let the ODE have continuous coefficients and on some open interval . Then two solution and of on are linearly dependent on .

Wronskian

The Wronskian or Wronski determinant is differentiable function and of

(6)       .

If then and are linearly dependent on .

If , then and are linearly independent on .

Example 1 Illustration of Theorem 2

The function and are solution .

Their Wronskian is

These solutions are linearly independent        .

.

For we have , which implies linear dependent.

Sage Coding

 var('x, w') print wronskian(cos(w*x), sin(w*x), x) print wronskian(cos(w*x), sin(w*x), x).simplify_full()

Evaluate

w*sin(w*x)^2 + w*cos(w*x)^2

w

Example 2 Illustration of Theorem 2 for a Double Root

A general solution of on any interval is .

.

.      and are linear independence on any interval.

Sage Coding

 print wronskian(exp(x), x*exp(x), x) print expand(wronskian(exp(x), x*exp(x), x))

Evaluate

(x*e^x + e^x)*e^x - x*e^(2*x)

e^(2*x)

A General Solution of Includes All Solution

Theorem 3  Existence of a General Solution

If and are continuous on an open interval , then has a general solution on.

Theorem 4  A General Solution Includes All Solution

If the ODE has continuous coefficients and on some open interval , then every solution of on is of the form

(8)

where is any basis of solution of on and , are suitable constants.

Hence does not have singular solution(that is solution not obtainable from a general solution).

2.7 Nonhomogeneous ODEs

The second-order nonhomogeneous linear ODE is

(1)

where .

A “general solution” of is the sum of a general solution of the corresponding homogeneous ODE

(2)

and a “particular solution” of .

The “general solution of ” and “particular solution of ” are defined as following

Definition  General solution, Particular solution

A general solution of the nonhomogeneous ODE on an open interval is

(3)

here is a general solution of the homogeneous ODE on and is any solution of on containing no arbitrary constants.

A particular solution of is a solution obtained from by assigning specific values to the arbitrary constants and in .

General solution of is

.

Theorem 1 Relations of solution of to Those of

(a) The sum of a solution of on some open interval and a solution of on is a solution of on . In particular, is a solution of on .

(b) The different of two solution of on is a solution of on .

Theorem 2 A General Solution of a Nonhomogeneous ODE Includes All Solutions

If the coefficients , , and the function in are continuous on some open interval , then every solution of on is obtained by assigning suitable values to the arbitrary constants and in a general solution of on .

Method of Undetermined Coefficients

Method of find a solution of .

Method of Undetermined Coefficients is suitable for linear ODEs with constant coefficients and

(4)

Choice Rules for the Method of Undetermined Coefficients

(a) Basic Rule

If in is one of the function is the first column in Table , choose in the same line and determine its undetermined coefficients by substituting and its derivatives into .

(b) Modification Rule

If a term in your choice for happens to be a solution of the homogeneous ODE corresponding to , multiply your choice of by (or by if this solution corresponds to a double root of the characteristic equation of the homogeneous ODE).

(c) Sum Rule

If is a sum of functions in the first column of Table , choose for the sum of the function in the corresponding lines of the second column.

Table Method of Undetermined Coefficients

 Term in Choice for

Example 1 Application of the Basic Rule

Solve the initial value problem

, , .

Solution 1

Step 1. General solution of the homogeneous ODE.

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are and .

The general solution is

.

Step 2. Solution of the nonhomogeneous ODE.

Let

Equating the coefficients of , , constant on both sides, we have

, ,       , ,

The particular solution is

Step 3. General solution of the nonhomogeneous ODE.

The general solution is

Step 4. Solution of the initial value problem.

We obtain from the general solution and initial condition.

Taking , we get

Taking , we get

The particular solution is .

as wall as the quadratic parabola about which is oscillating, like a sine curve

since the cosine term is smaller by a factor of about .

Sage Coding

 var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + y == 0.001*x^2, [y, x], ics=[0, 0, 1.5]) print de

Evaluate

1/1000*x^2 + 3/2*sin(x) + 1/500*cos(x) - 1/500

Solution 2

Step 1. General solution of the homogeneous ODE.

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are and .

The general solution is

.

Step 2. Solution of the nonhomogeneous ODE.

By .

The particular solution is .

Step 3. General solution of the nonhomogeneous ODE.

The general solution is

Step 4. Solution of the initial value problem.

We obtain from the general solution and initial condition.

Taking , we get

Taking , we get

The particular solution is .

as wall as the quadratic parabola about which is oscillating, like a sine curve

since the cosine term is smaller by a factor of about .

, where is Multiple Integrals of

(1)

(2)

(3)

Example 2  Application of the Modification Rule

Solve the initial value problem

, , .

Solution 1

Step 1. General solution of the homogeneous ODE.

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are .

The general solution is

Step 2. Solution of the nonhomogeneous ODE.

Let

Equating the coefficients of , , constant on both sides, we have

The particular solution is

Step 3. General solution of the nonhomogeneous ODE.

The general solution is

Step 4. Solution of the initial value problem.

We obtain from the general solution and initial condition.

Taking , we get

Taking , we get

The particular solution is

The curve begins with a horizontal tangent, crosses the -axis at and approaches the axis from below as increases.

Sage Coding

 var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + 3*diff(y, x) + 2.25*y == -10*exp(-1.5*x), [y, x], ics=[0, 1, 0]) print de

Evaluate

-5*x^2*e^(-3/2*x) + 1/2*(3*x + 2)*e^(-3/2*x)

Solution 2

Step 1. General solution of the homogeneous ODE.

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are .

The general solution is

Step 2. Solution of the nonhomogeneous ODE.

By .

.

The particular solution is .

Step 3. General solution of the nonhomogeneous ODE.

The general solution is

.

Step 4. Solution of the initial value problem.

We obtain from the general solution and initial condition.

Taking , we get

Taking , we get

The particular solution is

The curve begins with a horizontal tangent, crosses the -axis at and approaches the axis from below as increases.

where, means real part

where, means imaginary part

Example 3 Application of the Sum Rule

Solve the initial value problem

, , .

Solution 1

Step 1. General solution of the homogeneous ODE.

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are and .

The general solution is .

Step 2. Solution of the nonhomogeneous ODE.

Let

Equating the coefficients of , constant on both sides, we have

, , ,

, , ,

The particular solution is

Step 3. General solution of the nonhomogeneous ODE.

The general solution is

.

Step 4. Solution of the initial value problem.

We obtain from the general solution and initial condition

Taking , we get

Taking , we get

Hence and

The particular solution is .

Sage Coding

 var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + 2*diff(y, x) + 0.75*y == 2*cos(x) - 0.25*sin(x) + 0.09*x, [y, x], ics=[0, 2.78, -0.43]) print de

Evaluate

3/25*x + 31/10*e^(-1/2*x) + sin(x) - 8/25

Solution 2

Step 1. General solution of the homogeneous ODE.

The ODE has the general solution .

The characteristic equation is .

The characteristic roots are and .

The general solution is .

Step 2. Solution of the nonhomogeneous ODE.

By .

The particular solution is

The particular solution is

Step 3. General solution of the nonhomogeneous ODE.

The general solution is

.

Step 4. Solution of the initial value problem.

We obtain from the general solution and initial condition

Taking , we get

Taking , we get

Hence and

The particular solution is .

Stability

A the roots of characteristic equation of the homogeneous ODE in have negative or negative real part, then a general solution of this ODE goes to as , so that the “transient solution of approaches the “steady- state solution.

① In this case the nonhomogeneous ODE and the physical or other system modeled

by the ODE are called stable.

② Otherwise they are called unstable.

[한빛 아카데미] Engineering Mathematics with Sage:

[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화

Contents

A. 공학수학 1 – 선형대수, 상미분방정식+ Lab

Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html

Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html

Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html

Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html

Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html

Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html

Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html

Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html

B. 공학수학 2 - 벡터미적분, 복소해석 + Lab

Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html

Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html

Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html

Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html

Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html

Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html

Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html

Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html

Made by Prof. Sang-Gu LEE  sglee at skku.edu

http://matrix.skku.ac.kr/sglee/   with Dr. Jae Hwa LEE