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 2


2주차

1.4: Integrating Factors

  3.4 적분인자

  http://www.hanbit.co.kr/EM/sage/1_chap3.html 


1.5: Linear ODEs, Bernoulli Equation

  3.5 선형미분방정식



1.4 Exact ODEs.


The first-order ODE

(1)                     

is an exact differential equation if the differential form

                         

is exact.


This form is the differential

(2)                       

of some function . Then (1) can be written

                                

By integration we obtain the general solution of (1) in the form

(3)                        ( is an arbitrary constant)

                               implicit solution


Theorem  Suppose , , and are continuous for all within a rectangle in the plane. Then, is exact on in and only if for each in ,

(4)                      .

Proof

If is exact, then there is a potential function and .

Then, for in

    

Conversely, suppose and are continuous on .

Choose any in and define, in ,

(*)              .

From the fundamental theorem of calculus

    ,

since the integral in equation (*) is independent of . Next compute

   

      

      

      

and the proof is complete.


      

               

               


The differential equation is where .

The equation can be written as

   

where is an exact differential. Thus the general solution is or equivalently

   

   (where is an arbitrary constant)

where indicates that the integration is to be performed with respect to keeping constant.



Example 1  An Exact ODE

Solve 

(7)                .

Solution 1

   

   

   

 is exact.

      ( is an arbitrary constant)

      

      

      

      

General solution is . ( is an arbitrary constant)


Solution 2

         

   

Hence .

By integration

   

General solution is . ( is an arbitrary constant)


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

x, y = var('x, y')

M = cos(x + y)

N =  3*y^2 + 2*y + cos(x + y)

bool(diff(M, y) == diff(N, x))

Evaluate

True

x = var('x')

y = function('y')(x)

de = lambda y : ( cos(x + y))*diff(x) + (3*y^2 + 2*y + cos(x+ y))*diff(y) == 0

desolve(de(y), [y, x])

Evaluate

y(x)^3 + y(x)^2 + sin(x + y(x)) == c



Example 2  An Initial Value Problem

Solve the initial value problem

(10)              , .

Solution 1

    

    

    

 is exact.

      ( is an arbitrary constant)

      

      

      

      

General solution is ( is an arbitrary constant)

Initial condition, ,

Particular solution is .


Solution 2

          

    

Hence .

By integration

      

General solution is ( is an arbitrary constant)

Initial condition, ,

Particular solution is .


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

x, y = var('x, y')

M = (cos(y)) * (sinh(x)) + 1

N =  -(sin(y))*(cosh(x))

bool(diff(M, y) == diff(N, x))

Evaluate

True

x = var('x')

y = function('y')(x)

de = lambda y : ( (cos(y))*((1/2)*(e^x - e^-x))+ 1)*diff(x) + (-(sin(y))*((1/2)*(e^x + e^-x)))*diff(y) == 0

desolve(de(y), [y, x], ics=[1, 2])

Evaluate

1/2*((e^(2*x) + 1)*cos(y(x)) + 2*x*e^x)*e^(-x) == 1/2*((e^2 + 1)*cos(2)+ 2*e)*e^(-1)



Integrating Factors


The nonexact differential equation

(1)                         .

Multiplying by function .

(2)                  

is exact.

Function is an integrating factor of .



[a] The following combination are often useful in finding integrating factors

(i)                  (ii)

(iii)                

(iv)          

(v)         

(vi)          



[b] ( is integer) is integrating factors. find



[c] a function of alone, then

                       

is an integrating factor.


If has an integrating factor which depends only on , show that , where .

 is exact. Then

     

Since depends only on , this can written

           

Thus

    

Integration both sides, we obtain

          .

Hence, .



[d] a function of alone, then

    

is an integrating factor.



Example 3,  4  Integrating Factor

Solve .

Solution 1

     multiply by

    

General solution is   ( is an arbitrary constant)


Solution 2

    ,

     

     

Equation is not exact.

    

Integrating factors is

     multiply by

    

     

    

      ( is an arbitrary constant)

     

     

General solution is ( is an arbitrary constant).


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

x, y = var('x, y')

M = -y

N = x

bool(diff(M, y) == diff(N, x))

Evaluate

False

x, y = var('x, y')

M = -y

N = x

u = (diff(N, x)-diff(M, y))/N

print u

print exp(-integral(u, x))

Evaluate

2/x

e^(-2*log(x))

x = var('x')

y = function('y')(x)

de = lambda y : (-y/x^2)*diff(x) +(1/x)*diff(y) == 0

desolve(de(y), [y, x])

Evaluate

c*x 


Example 5  Integrating Factor, Initial Value Problem

Find an integrating factor and solve the initial value problem

.

Solution

     ,

     

     

Equation is not exact.

    

Integrating factor is

    , multiply by

    

    

    

    

Equation is exact.

      ( is an arbitrary constant)

      

      

      

      

General solution is ( is an arbitrary constant)

Initial condition, ,

Particular solution is .


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

x, y = var('x, y')

M = e^(x+ y) + y*e^y

N = x*e^y - 1

bool(diff(M, y) == diff(N, x))

Evaluate

False

x, y = var('x, y')

M = e^(x+ y) + y*e^y

N = x*e^y - 1

u = (diff(N, x)-diff(M, y))/M

print u

print exp(integral(u, y))

Evaluate

-1

e^(-y)

x = var('x')

y = function('y')(x)

de = lambda y : (e^(x+ y) + y*e^y)*diff(x) +(x*e^y - 1)*diff(y) == 0

desolve(de(y), [y, x], ics=[0, -1])

Evaluate

((x*y(x) + e^x)*e^y(x) + 1)*e^(-y(x)) == e + 1





1.5 Linear ODEs


    


A first-order ODE is said to be linear equation

               .   Standard from

    Then the ODE is homogeneous linear equation.

    Then the ODE is nonhomogeneous linear equation.


The function on the right may be a force, and the solution a displacement in a motion or an electrical current or some other physical quantify. In engineering, is frequently called the input, is the output or the response to the input(given initial condition).



Homogeneous Linear ODE

The ODE becomes

                   

is homogeneous.

By separating variables, 

    .

By integrating on both sides with respect to

    

or 

    

Taking exponents on both sides, 

     

or 

      (where is an arbitrary constant)

General solution of the homogeneous ODE ,

      (where is an arbitrary constant)

here we choose and obtain the trivial solution for all .



Nonhomogeneous Linear ODE

Write the equation as

            

then .

    

    

Equation is not exact.

    

is a function of alone, then

    

is an integrating factor.

Multiplying on both sides by integrating factor , the equation becomes

    

or

    

Integrating on both sides, we obtain

     ( is an arbitrary constant)

Diving this equation by

           ( is an arbitrary constant)

              

We see the following

Total Output Response to the Input + Response to the Initial Data.



Example 1  First-order ODE, General Solution

Solve the linear ODE .

Solution

    

    

    

    

This equation is not exact.

     is a function of alone, then

is an integrating factor.

Multiplying on both sides the differential equation by to get

    

     

Integrating on both sides with respect to , we obtain

     ( is an arbitrary constant)

Multiplying on both sides this equation by , we obtain the general solution 

    .  ( is an arbitrary constant)



Example 2  First-order ODE, General Solution. Initial Value Problem

Solve the initial value problem .

Solution 1

   

    

    

    

Equation is not exact.

 is a function of alone, then

      

is an integrating factor.

Multiply the differential equation by to get

     

     

Integrating on both sides with respect to , we obtain

 ( is an arbitrary constant)

General solution is .  ( is an arbitrary constant)

Initial condition, ,      

Particular solution is .


Solution 2

    , , ,

    

      

      

      

      

      

      

General solution is .  ( is an arbitrary constant)

Initial condition, ,      

Particular solution is .


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

x, y = var('x, y')

M = y*tan(x) - sin(2*x)

N = 1

bool(diff(M, y) == diff(N, x))

Evaluate

False

x, y = var('x, y')

M = y*tan(x) - sin(2*x)

N = 1

u = (diff(N, x)-diff(M, y))/N

print u

print exp(integral(u, x))

Evaluate

-tan(x)

e^(-log(sec(x)))

x = var('x')

y = function('y')(x)

de = lambda y : diff(y,x) + y*tan(x) == sin(2*x)

desolve(de(y), [y, x], ics=[0, 1])

Evaluate 

-(2*cos(x) - 3)/sec(x)



Example 3  Electric Circuit

Model the RL-circuit in Fig and solve the resulting ODE for the current (amperes), where is time. Assume that the circuit contains as an EMF (electromotive force) a battery of (volts), which is constant, a resistor of (ohms), and an inductor of (henrys), and that the current is initially zero.

           그림입니다.

                    Circuit                 Current 

Solution

The model of the RL-circuit is

          .

Find the general solution.

                

Integrating on both sides

          

                               

                                

                                  (where, )

                                (where, )

The general solution is . ( is an arbitrary constant)

    , ,       ,

                                  .

Initial value gives ,

Particular solution

          .


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

var('t')

R = 11;E = 48; L = 0.1

I = function('I')(t)

de = L*diff(I, t) + R*I == E

expand(desolve(de, [I, t], ics=[0, 0]))

Evaluate

-48/11*e^(-110*t) + 48/11




Bernoulli Equation

A Bernoulli equation is a first-order equation

    

in which is a real number.

, then the equation is separation of variables.

, then the equation is linear equation.


Equation is a nonlinear equation.

Dividing on both sides by

    

Let , then differential both sides with respect to , we obtain

          

By

          

Formula is a linear equation by

    

    

    

is a function of alone, then

    

is an integrating factor.

Multiply on both sides the differential equation by to get

    

    

Integrating on both sides, we obtain

    

General solution of is

      ( is an arbitrary constant)

General solution of is

    

   ( is an arbitrary constant)



Example 4  Logistic Equation (Verhulst equation)

So we the following Bernoulli equation, known as the logistic equation

(8)                  .

Solution 1

     or

Dividing on both side with respect to

    

Let . Differentiating on both side with respect to

        or   

Hence we have obtained the linear ODE.

    

     when

    

     (where,   is an arbitrary constant)


Solution 2

     or

Dividing on both side with respect to

    

Let . Differentiating on both side with respect to

        or   

Hence we have obtained the linear ODE.

       or   

    

    

    

Equation is not exact.

     is a function of alone, then

    

is an integrating factor.

Multiply on both sides the differential equation by to get

    

         

By integrating on both sides with respect to , we obtain

    

Dividing this equation by , we obtain the general solution

     thus

Since , we obtain the general solution of

(9)          (where, is an arbitrary constant)

       그림입니다.

   Logistic population model. Curves (9) in Example 4 with .


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

var('t, A, B')

y = function('y')(t)

de = diff(y, t) == A*y - B*y^2

desolve(de, [y, t])

Evaluate

-(log(B*y(t) - A) - log(y(t)))/A == c + t



Example 

Solve .

Solution

    

Dividing on both side with respect to , we obtain

     .

Dividing on both side with respect to , we obtain

    .

Let . Differentiating both side with respect to , we obtain

       or  .

Hence we have obtained the linear equation

     or

Formula is a linear equation by , we obtain

    

    

    

    

This equation is not exact.

      is a function of alone, then

is an integrating factor.

Multiplying on both sides the differential equation by to get

    .

Let .

By integrating both sides with respect to , we obtain

    .

Multiplying on both sides the differential equation by , we obtain

    .

Since .

The general solution of the Bernoulli equation is

     (namely, is an arbitrary constant).


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

var('x')

y = function('y')(x)

de = x*diff(y, x) + y == x^2*y^2

desolve(de, [y, x])

Evaluate

1/((c - x)*x)




Example

Solve .

Solution

    .

Dividing on both side with respect to , we obtain

    .

Let . Differentiating both side with respect to , we obtain

      or  .

Hence we have obtained the linear equation

     or

Formula is a linear equation by , we obtain

    .

    

    

    

This equation is not exact.

      is a function of alone, then

    

is an integrating factor.

Multiplying on both sides the differential equation by to get

    .

Let .

By integrating both sides with respect to , we obtain

    .

Multiplying on both sides the differential equation by , we obtain

    .

Since  .

The general solution of the Bernoulli equation is

     (namely, is an arbitrary constant).


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

var('x')

y = function('y')(x)

de = diff(y, x) + y/x == 3*x^2*y^3

desolve(de, [y, x])

Evaluate

1/(sqrt(c - 6*x)*x)



Population Dynamics

The logistic equation plays an important role in population dynamics, a field that models the evolution of populations of plant, animals or humans over time .


Malthus’s law

If , then is . This solution is and gives exponential growth, as for a small population in a large country. The term in is a “braking term” that prevents the population from growing without bounded.


    

If , then , so that an initially small population keeps growing as long as  . If , then and the population is decreasing as long as .

    .


An ODE in which does not occur explicitly is of the form

(10)          

and is called an autonomous ODE. Thus the is autonomous. Equation has constant solution, called equilibrium solutions or equilibrium points. These are determined by the zeros of , because gives by hence . These zeros are known as critical points of . An equilibrium solution is called stable if solutions close to it for some remain close to it for all further . It is called unstable if solutions initially close to it do not remain close to it as increases. is an unstable equilibrium solution, and is a stable one.



Example 5  Stable and Unstable Equilibrium Solutions. “Phase Line Plot”

The ODE has the stable equilibrium solution and the unstable , as the direction field in figure suggests. The values and are the zeros of the parabola in the figure.

ODE is autonomous, we can “ondense”the direction field to a “phase line plot” giving and and the direction of the arrows in the field.

Giving information about the stability or instability of the equilibrium solutions.

       그림입니다.

          Example 5. (A) Direction field. (B) “Phase line”. (C) Parabola



Riccati Equation (Pro 30)

A differential equation of the form

    

is a  Riccati equation.

 is one particular solution.

    

General solution is .

Differentiating both side with respect to , we obtain

Substituting

    

     

     

     

which is a Bernoulli equation . Unfortunately, one finds by guessing.

The substitution that is needed to solve this Bernoulli equation.

Dividing this equation by

    

Let .

Differentiating on both side with respect to

     or

Substitution into

    

or

    

linear equation by

    

    

    

    

Integrating factor is

    

    

    

    

General solution of is

    

     

     ( is an arbitrary constant)


Example

Consider the Riccati equation .

Solution

 is one particular solution.

Let . Then

    

    

Dividing on both side with respect to

     

Let . Differentiating both side with respect to

      or

     ,   is linear equation by

     

     

     

     

integrating factor is

     

     

 By integrating both sides with respect to , we obtain

     

     

      or

      (where, is an arbitrary constant)


Sage Coding

http://math3.skku.ac.kr/   http://sage.skku.edu/   http://mathlab.knou.ac.kr:8080/

var('x')

y = function('y')(x)

de = diff(y, x) + 2/x - y/x - y^2/x == 0

desolve(de, [y, x])

Evaluate

1/3*log(y(x) - 1) - 1/3*log(y(x) + 2) == c + log(x)



Clairaut Equation (Pro 30)

A differential equation of the form

    

is called a Clairaut equation.

Proof

    

By differentiating both sides with respect to , We obtain

    so

From ②

Let , then ( is an arbitrary constant)

Substitution ③ into ①

The general solution is (  is an arbitrary constant)

     

     

To eliminate the given differential equation but cannot be obtained from the general

solution for any choice of and so is a singular solution.


    .



Example

Solve where .

Solution

The equation can be solved explicitly for . Differentiation with respect to yields

     from which

    so

case 1. . In this case and so the general solution is

         ( is an  arbitrary constant)

case 2. . In this case or

        ,

To eliminate , note that .

The equation satisfies the given differential equation but cannot be obtained

from the general solution for any choice of and so is a singular solution.



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