2019학년도 1학기
반도체 공학과 공학수학1
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 이상구 외 4인, 최신공학수학 I, 1st Edition
실습실: http://www.hanbit.co.kr/EM/sage/ http://matrix.skku.ac.kr/LA/
강의시간: 공학수학1, (화09:0010:15) (목10:3011:45)
담당교수: 김응기 박사
Week 3
3주차
2.1: Homogeneous Linear ODEs of 2nd Order
4.1 2계 선형미분방정식의 개념
2.2: Homogeneous Linear ODEs with Constant Coefficients
4.2 상수 계수 동차선형미분방정식
2.5: EulerCauchy Equations
4.3 오일러코시 방정식
http://www.hanbit.co.kr/EM/sage/1_chap4.html
Chapter 2. Secondorder Linear ODEs
ODEs may be divided into two large classes
1. linear ODEs
2. nonlinear ODEs
Linear ODEs of the second order are the most important ones because of their applications in mechanical and electrical engineering.
Their theory is typical of that of all linear ODEs.
Chapter 2 includes the derivation of general and particular solutions, the latter in connection with initial value problems.
2.1 Homogeneous Linear ODEs of Second Order
We shall now define and discuss linear ODEs of second order
1. mechanical
2. electrical vibrations
a. wave motion b. heat condition c. other parts of physics
A secondorder equation
is a linear ODE.
An other form is nonlinear ODE.
A secondorder equation
is a homogeneous linear ODE.
If , then a secondorder equation
is a nonhomogeneous linear ODE.
is a nonhomogeneous terms.
If and in and are coefficients of the ODE.
For example, a nonhomogeneous linear ODE is
.
Homogeneous linear ODE is
.
Nonlinear ODE is
.
Linear combination
and a function of the form
(, are an arbitrary constants).
is a linear combination of and , where and are an arbitrary constants.
Theorem 1 Fundamental Theorem for the Homogeneous Linear ODE (2)
Let and be solution of on . Then and are arbitrary constants
then is a solution of on .
Proof
Given equation
and be solution of on .
,
is a solution of on .
Theorem 1 is called the linearity principle.
Example 1 Homogeneous Linear ODEs : Superposition of Solutions
The function and are solution of the homogeneous linear ODE
for all .
Solution
Function and are solution of the homogeneous linear ODE .
Linear combination is
where and are an arbitrary constants.
Function is a solution of the homogeneous linear ODE .
This satisfies superposition principle.
Theorem 1 Fundamental Theorem for Homogeneous Linear ODE
For a homogeneous linear ODE , any linear combination of two solutions on an open interval is again of on . In particular, for such an equation, sums and constant multiples of solutions are again solutions.
Proof
Using linearity principle
Given equation
Let be solution of
Hence, is a solution of on .
Example 2 A Nonhomogenous Linear ODE
The function and are solutions of the nonhomogeneous linear ODE
for all .
Solution
The function and are solutions of the nonhomogeneous linear ODE .
Linear combination is
where and are an arbitrary constants.
is not solution of the nonhomogeneous linear ODE
.
Because given equation is not the homogeneous linear ODE.
Example 3 A Nonlinear ODE
The function and are solutions of the nonlinear ODE
for all .
Solution
The function and are solutions of the nonlinear ODE .
Linear combination is
where and are an arbitrary constants.
is not solution of the nonlinear ODE .
Because given equation is not the linear ODE.
Initial Value Problem
For a secondorder homogeneous linear ODE an initial value problem consists of and two initial conditions
(4) , .
To condition and are used to determine the two arbitrary constants and in a general solution
(5)
of the ODE, here and suitable solution of the ODE.
This result in a unique solution, pass through the point with as the tangent direction(the slope) at that point.
That solution is a particular solution of the ODE .
Example 4 Initial Value Problem
Solve the initial value problem
, , .
Solution
The function and are solution of the ODE
, where and are an arbitrary constants.
From the general solution and the initial conditions
Taking , .
Taking ,
The solution of our initial value problem the particular solution is
.
Fig. 29. Particular solution and initial tangent in Example 4
Definition General Solution, Basis, Particular Solution
A general solution of an ODE on an open interval is a solution in which and are solution of on that are not proportional, and and are arbitrary constants. These , are a basis of solutions of on .
A particular solution of on is obtained if we assign specific values to and in .
and are proportional on if for all on
(6) or (, ).
Two function and are linearly independent on an interval where they are define if
(7) , everywhere on implies and
Any and are called linearly independent on if holds for some constants , not both zero. Then if or we can divide and see that and are proportional
or .
Linear independence these functions are not proportional because then we cannot divide in
Define Basis(Reformulated)
A basis of solution on an interval is a pair of linearly independent solution of on .
Singular solutions not obtainable from of a general solution.
Example 5 Basis, General Solution, Particular Solution
and in Example 4 form a basis of solutions of the ODE for all .
().
Hence is general solution.
The solution of the initial value problem is a particular solution.
Example 6 Basis, General Solution, Particular Solution
and are solution of the ODE .
Then solve the initial value problem , , .
Solution
Function and are solution of the ODE .
Linear combination is
where and are an arbitrary constants.
Function is solution of the ODE
Solution of the ODE is
From the general solution and the initial conditions
, ,
Particular solution is .
Find a Basis if One Solution Is Known. Reduction of Order
Reduction of Oder
Then a second linearly independent solution can be obtained by solving a first order ODE.
We applied reduction of order to homogeneous linear ODE
is a solution of given ODE.
We need a second linearly independent solution of on .
To get , we substitute,
into .
This gives
(8)
Hence
We divide this remaining ODE by and set and
is firstorder ODE.
Separation of variables and integration gives
By taking exponents we finally obtain
(9)
Here , so that .
Hence the desired second solution is
.
The quotient cannot be constant (since ), so that and form a basis of solution.
Example 7 Reduction of Order if a Solution is Known Basis
Find a basis of solution of the ODE
.
Solution
is a solution
.
Let .
Then
.
This ODE is of first order in , namely,
.
Separation of variable and integration gives
By integrating both sides with respect to , we obtain
Hence
Since and are linearly independent,
We have obtained a basis of solution, valid for all positive .
The general solution is , where and are an arbitrary constants.
2.2 Homogeneous Linear ODEs with Constant Coefficients
A secondorder homogeneous linear ODEs whose coefficients and are constant,
(1) .
An operator is a transformation that transforms a function into another function
is differential operator
Linearity principle
If and are differentiable, then exists for any constants and and, .
is differential operator
is
by left sides of
is polynomial operator.
is the characteristic equation or auxiliary equation of .
The characteristic equation of solution is the characteristic roots.
A secondorder linear differential equation with constant coefficients
(1)
where and are constant.
The characteristic equation is roots , .
Case I. Two real roots if .
Case II. A real double roots if .
Case III. Complex conjugate roots if .
Distinct real roots
If and are distinct real roots (), then the general solution of is
where , are an arbitrary constants.
Proof
Let .
, by integration both sides
is linear equation, by integration factor
.
:
Example 1 General Solution in the Case of Distance Real Roots
Solve .
Solution
Characteristic equation is
.
Characteristic roots are and .
General solution is
,
where and are an arbitrary constants.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve(m^2  1 == 0, m) 
Evaluate
[m == 1, m == 1]
var('x') y = function('y')(x) de = desolve(diff(y, x, 2)  y == 0, [y, x]) print de 
Evaluate
k1*e^x + k2*e^(x)
Example 2 Initial value Problem in the Case of Distance Real Roots
Solve the initial value problem , , .
Solution
Characteristic equation is
.
Characteristic roots are and .
General solution is
,
where and are an arbitrary constants.
We obtain from the general solution and the initial condition.
Taking , we get .
Taking , we get .
Hence , .
Particular solution is
.
Fig. 30. Solution in Example 2
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') # 2 solve(m^2 + m  2 == 0, m) 
Evaluate
[m == 1, m == 2]
var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + diff(y, x)  2*y == 0, [y, x], ics=[0, 4, 5]) print de 
Evaluate
3*e^(2*x) + e^x
Double roots
If and are equal roots, say (), then the auxiliary yields the general solution of is
where , are an arbitrary constants.
Proof
The general solution of .
or
,
By integration both sides
,
By integration both sides
.
where and are an arbitrary constants.
Example 3 General Solution in the Case of a Double Real Roots
Solve .
Solution
The characteristic equation is
.
The characteristic roots are .
The general solution is
,
where and are an arbitrary constants.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve(m^2 + 6*m + 9 == 0, m) 
Evaluate
[m == 3]
var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + 6*diff(y, x) + 9*y == 0, [y, x]) print de 
Evaluate
(k2*x + k1)*e^(3*x)
Example 4 Initial value Problem in the Case of a Double Real Roots
Solve the initial problem , , .
Solution
The characteristic equation is
.
The characteristic roots are .
The general solution is
,
where and are an arbitrary constants.
We obtain from the general solution and the initial condition.
Taking , we get .
Taking , we get .
Hence and .
The particular solution is
.
Fig. 31. Solution in Example 4
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve(m^2 + m + 0.25 == 0, m) 
Evaluate
[m == (1/2)]
var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + diff(y, x) + 0.25*y == 0, [y, x], ics=[0, 3.0, 3.5]) print de 
Evaluate
(2*x  3)*e^(1/2*x)
Complex conjugate roots
In this case the roots and of the auxiliary equation are complex numbers. We can write
where and are real numbers. In fact, , .
The solution of the differential equation is
Using Euler’s equation
where , .
This gives all solutions (real or complex)of the differential equation.
The solutions are real when the constants and are real.
If the roots of the auxiliary equation are the complex numbers , , then the general solution of is
Example 5 Complex Roots. Initial Value Problem
Solve the initial problem , , .
Solution
The characteristic equation is
.
The characteristic roots are .
The general solution is
where and are an arbitrary constants.
We obtain from the general solution and the initial condition.
Taking , we get .
Taking , we get .
Hence and .
The particular solution is
.
Fig. 32. Solution in Example 5
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve(m^2 + 0.4*m + 9.04 == 0, m) 
Evaluate
[m == (3*I  1/5), m == (3*I  1/5)]
var('x') y = function('y')(x) de = desolve(diff(y, x, 2) + 0.4*diff(y, x) + 9.04*y == 0, [y, x], ics=[0, 0, 3]) print de 
Evaluate
e^(1/5*x)*sin(3*x)
Example 6 Complex Roots
A general solution of the ODE .
Solution
The characteristic equation is
.
The characteristic roots are .
The general solution is
where and are an arbitrary constants.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
var('m, w') solve(m^2 + w^2 == 0, m) 
Evaluate
[m == I*w, m == I*w]
var('x, w') assume(w>0) y = function('y')(x) de = desolve(diff(y, x, 2) + w^2*y == 0, [y, x]) print de 
Evaluate
k1*sin(w*x) + k2*cos(w*x)
Case 
Roots of 
Basis of 
General Solution 
I 
Distinct real



II 
Real double root



III 
complex conjugate



is called the Euler formula.
Example Distinct Real Roots
Solve the ODE .
Solution
Let .
The characteristic equation is
The roots are
General solution is
where , and are an arbitrary constants.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve(m^3  2*m^2  m + 2 == 0, m) 
Evaluate
[m == 2, m == 1, m == 1]
2.5 EulerCauchy equation
EulerCauchy equation are ODEs of the form
(1)
with given constants and and unknown .
We substitute
(2)
(3)
Hence is a solution of is a root of .
The roots of are
(4) , .
Case I. Real different roots are gives two real solutions
and .
The corresponding general solution for all is
where and are an arbitrary constants.
Example 1 General Solution in the Case of Different Real Roots
The EulerCauchy equation .
Solution
Let .
Characteristic equation is
.
Characteristic roots are and .
Hence a basis of solution for all positive is and .
,
General solution is
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve( m^2 + 0.5*m  0.5 == 0, m) 
Evaluate
[m == (1/2), m == 1]
var('x') y = function('y')(x) de = desolve(x^2*diff(y, x, 2) + 1.5*x*diff(y, x)  0.5*y == 0, [y, x]) print de 
Evaluate
k2*e^(1/2*log(x))  1/3*k1/x
Case II. If and are equal roots, say (), then the general solution of is
where, and are an arbitrary constants.
Proof
and is one solution.
Find
EulerCauchy equation is
and using reduction of order, we get
The general solution is
where and are an arbitrary constants.
Example 2 General Solution in the Case of a Double Roots
The EulerCauchy equation .
Solution
Let .
The characteristic equation is
.
The double root is .
The general solution for all positive is
where and are an arbitrary constants.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve( m^2  6*m + 9 == 0, m) 
Evaluate
[m == 3]
var('x') y = function('y')(x) de = desolve(x^2*diff(y, x, 2)  5*x*diff(y, x) + 9*y == 0, [y, x]) print de 
Evaluate
(k2*log(x) + k1)*x^3
Case III. Complex Roots
If and are complex roots, say , then the general solution of is
where and are an arbitrary constants.
Proof
where .
Example 3 Real General Solution in the Case of Complex Roots
The EulerCauchy equation .
Solution
Let .
Characteristic equation is
The roots are complex conjugate and
General solution for all positive is
where and are an arbitrary constants.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
m = var('m') solve( m^2  0.4*m + 16.04 == 0, m) 
Evaluate
[m == (4*I + 1/5), m == (4*I + 1/5)]
var('x') y = function('y')(x) de = desolve(x^2*diff(y, x, 2) + 0.6*x*diff(y, x) + 16.04*y == 0, [y, x]) print de 
Evaluate
(k1*sin(4*log(x)) + k2*cos(4*log(x)))*x^(1/5)
Example 4 Boundary Value Problem. Electric Potential Field Between Two Concentric Sphere
Find the electrostatic between two concentric spheres of radius and kept at potentials and , respectively.
Solution
is a solution of the EulerCauchy equation , where .
The auxiliary equation is .
The roots are and .
General solution is
where and are an arbitrary constants.
Taking , we obtain
Taking , we obtain
Hence, , .
Particular solution is
.
Fig. 49. Potential v(r) in Example 4
http://matrix.skku.ac.kr/sglee/
[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/EMSage/EMathChapter1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EMSage/EMathChapter2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EMSage/EMathChapter7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EMSage/EMathChapter8.html
B. 공학수학 2  벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EMSage/EMathChapter9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EMSage/EMathChapter10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EMSage/EMathChapter11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EMSage/EMathChapter12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EMSage/EMathChapter13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EMSage/EMathChapter14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EMSage/EMathChapter15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EMSage/EMathChapter16.html
Made by Prof. SangGu LEE sglee at skku.edu
http://matrix.skku.ac.kr/sglee/ with Dr. Jae Hwa LEE