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, (화09:00-10:15)   (목10:30-11:45)

   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: Euler-Cauchy Equations

  4.3 오일러-코시 방정식

  http://www.hanbit.co.kr/EM/sage/1_chap4.html 




Chapter 2.  Second-order 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 second-order equation

              

is a linear ODE.

An other form is nonlinear ODE.


A second-order equation

              

is a homogeneous linear ODE.


If , then a second-order 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 second-order 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

    .

             그림입니다.

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 first-order 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 second-order 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 second-order 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

    .

    그림입니다.

        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

.

      그림입니다.

              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

    .

      그림입니다.

             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 Euler-Cauchy equation


Euler-Cauchy 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 Euler-Cauchy 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

Euler-Cauchy 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 Euler-Cauchy 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 Euler-Cauchy 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 Euler-Cauchy 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

    .

     그림입니다.

              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/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