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

1 |
§1.1 Basic Concepts. Modeling §1.3 Sparable ODEs. Modeling §1.4 Exact ODEs. |

2 |
§1.4 Integrating Factors §1.5 Linear ODEs. Bernoulli Equation. |

3 |
§2.1 Homogeneous Linear ODEs of 2nd Order §2.2 Homogeneous Linear ODEs with constant coefficients. §2.5 Euler-Cauchy Equations. |

4 |
§2.6 Existence and Uniqueness of Solutions Wronskian. §2.7 Nonhomogeneous ODEs |

5 |
§2.10 Solution by Variation of Parameters * §4.1 Systems of ODEs §5.1 Power Series Method |

6 |
§5.2 Legendre's Equation. Legendre Polynomials §5.3 Extended Power Series Method: Frobenius Method |

7 |
§5.4 Bessel’s Equation. Bessel Functions * §5.5 Bessel Functions of the . General Solution |

8 |
중간고사 |

9 |
§6.1 Laplace Transform Inverse Transform. Linearity. s-Shifting §6.2 Transforms of Derivatives and Integrals §6.3 Unit Step Function. t-Shifting |

10 |
§6.4 Short Impulses. Dirac's Delta Function. Partial Fractions §6.5 Convolution. Integral Equations §6.6 Differentiation and Integration of Transforms. ODEs with Variable Coefficients |

11 |
* §6.7 Systems of ODEs §7.1 Matrices, Vectors §7.2 Matrix Multiplication |

12 |
§7.3 Linear Systems of Equations. §7.4 Linear Independence. Rank of Matrix. Vector Space. §7.5 Solutions of Linear System |

13 |
§7.6 For Reference: Second- and Third-Order Determinants §7.7 Determinants. Cramer's Rule §7.8 Inverse of Matrix. |

14 |
§8.1 Eigenvalues, Eigenvectors * §8.2 Some Applications of Eigenvalue Problem |

15 |
§8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrix §8.4 Eigenbases, Diagonalization. Quadric Forms |

16 |
기말고사 |

* 는 시간 상 생략할 수 있음

Week 1

1주차

1.1: Basic Concepts, Modeling

3.1 1계 미분방정식의 개념

1.3: Separable ODEs, Modeling

3.2 변수분리형 미분방정식

1.4: Exact ODEs

3.3 완전미분방정식

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

Chapter 1. First-Order Ordinary Differential Equations

(ODEs)

Ordinary Differential Equations(ODEs)

① Deriving them from physical or other problems (Modeling)

② Solving them by standard mathematical methods

③ Interpreting solutions and their graphs in terms of a given problem

First-Order ODE

The ODEs of the first-order involve only the first derivative of the unknown function and no higher derivatives.

Notation for the unknown function

①

② when the independent variable denotes time .

1.1 Basic Concepts. Modeling

Mathematical modeling

① Mathematical model

If we want to solve an engineering problem, we first have to formulate the problem as a mathematical expression in terms of variables, function, equations. Such an expression is known as a mathematical model of the given problem.

② Mathematical modeling

The process of setting up a model, solving it mathematically and interpreting the result in physicals or other terms is called Mathematical modeling.

Differential equation

An equation containing derivatives of an unknown function are a differential equations.

① Ordinary differential equation

② Partial differential equation

Ordinary differential equation : ODE

An ODE contains one or several derivatives of an unknown function.

For example,

(1) first order

(2) second order

(3) third order

are ordinary differential equations.

, .

Partial differential equation : PDE

A PDEs involve partial derivatives of an unknown function of two or more variables.

For instance, a PDE with unknown function of two variable and is

.

Order

An ODE is said to be of order if the derivative of unknown function is the highest derivative of in the equation.

Degree

The degree of a differential equation is the power to which the highest-order derivative is raised.

The -order ODE

The derivative of unknown function is the highest derivative of in the equation. This is called a -order ODE.

The first-order ODE

An ODE contain only the first derivative and many contain and any given function of .

In general, type of first-order ODEs

(4) Implicit form

Explicit form

For instance. the implicit is ODE , the explicit is ODE .

Concept of solution

A function is a solution of a given ODE on some open interval if is defined and differentiable throughout the interval and is such that the equation becomes an identity if and are replaced with and , respectively.

The curve of is a solution curve.

General Solution

A solution containing an arbitrary constant is a general solution of the ODEs.

Particular Solution

A particular solution of the ODEs does not contain any arbitrary constants.

Singular Solution

A singular solution is a solution which cannot be obtained from the general solution by specifying values of the arbitrary constant.

Integral Curves

A graph of a solution of a first-order ODEs is an integral curve of the equation.

If we know the general solution, we obtain an infinite family of integral curves, one for each choice of the arbitrary constant.

Initial condition

A particular solution is obtained from a general solution by an initial condition with given values and that is used to determine a value of the arbitrary constants .

Initial Value Problem

An ODE together with an initial condition is an initial value problem. Thus, if the ODE is explicit , the initial value problem is

.

initial condition

1.3 Separable ODEs. Modeling

General type of first-order ODE is

: implicit form

or

separable variable equation.

(1)

Integrate on the both sides with respect to by ,

(2) ( is an arbitrary constant)

or

(3) ( is an arbitrary constant)

and are continuous function, the integrals in exist, and by evaluating them we obtain a general solution of .

This method of solving ODEs is the method of separating variable.

is called separable equation, because in the variables an new separable ; appears only on the right and only on the left.

Example 1 Separable ODE

Find the ODE is separable.

Solution

is separable equation.

By integration on both sides with respect to

or . ( is arbitrary constant)

General solution is . ( is 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) == 1 + y^2 desolve(de, [y, x]) |

Evaluate

arctan(y(x)) == c + x

Example 2 A Separable ODE

Find the ODE is separable.

Solution

is separable equation.

By integration on both sides with respect to

or . ( is arbitrary constant)

General solution is . ( is 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) == (x + 1)*e^(-x)*y^2 desolve(de, [y, x]) |

Evaluate

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

Example 3 Initial Value Problem. Bell-shaped Curve

Solve the ODE , .

Solution

is separable equation.

By integration on both sides with respect to

or . ( is arbitrary constant)

Given initial condition is .

Initial value problem has the solution .

Particular solution is .

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 desolve(de, [y, x], ics=[0, 1.8]) |

Evaluate

9/5*e^(-x^2)

Example 4 Radiocarbon Dating

When did Oetzi approximately live and die if the ratio of carbon to carbon in this mummy is of that of a living organism?

Solution

Radioactive decay is governed by the ODE,

By separation and integration where is time and is the initial ratio to

The general solution is . ( an arbitrary constant)

We use the half-life to determine .

When , half of the original substance is still present.

Finally, we use the radio for determining the times when Oetzi died.

About years ago

Sage Coding

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

var('k, t') y = function('y')(t) de = diff(y, t) == k*y desolve(de,[y, t]) |

Evaluate

c*e^(k*t)

var('H') solve(e^(k*H)==0.5, k) |

Evaluate

[k == -log(2)/H]

(-log(2)/5715).n(digits=5) |

Evaluate

-0.00012129

solve(e^(-log(2)/5715*t) == 0.525, t) |

Evaluate

[t == 5715*log(40/21)/log(2)]

(5715*log(40/21)/log(2)).n(digits=5) |

Evaluate

5312.7

Example 5 Mixing Problem

Mixing problem occur quite frequently in chemical industry. We explain here how to solve the basic model involving a single tank. The tank contains gal of water in which initially lb of salt is dissolved. Brine runs in at a rate of , and each gallon contains lb of dissolved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at . Find the amount of salt in the tank at any time .

Solution

Let denote the amount of salt in the tank at time .

Salt’s time rate of change is

Salt inflow rate - Salt outflow rate Balance law

Salt inflow rate

Salt outflow rate

This is of the total brine content in the tank,

hence of the salt content , this is, .

Thus the model is the ODE

.

By integrating on the both sides with respect to

The general solution is ( is an arbitrary constant)

The model of initial value problem

.

Initially the tank contains of salt.

Hence is the initial condition, that will give the unique solution.

Substituting and in the last equation gives

Hence the amount of salt in the tank at time is

.

Tank Salt content

Sage Coding

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

var('t') y = function('y')(t) de = diff(y, t) == 50 - 0.01*y f= desolve(de, [y, t], ics=[0, 100]) print f plot(f, (t,0,500)) |

Evaluate

100*(50*e^(1/100*t) - 49)*e^(-1/100*t)

Example 6 Heating an Office Building (Newton’s Law of Cooling)

Suppose that in Winter the daytime temperature in a certain office building is maintained at . The heating is shut off at and turned on again at . On a certain day the temperature inside the building at was found to be . The outside temperature was at and had dropped to by . What was the temperature inside the building when the heat was turned on at ?

Solution

is the temperature inside the building

is the temperature outside the building

By Newton’s Law of Cooling

varied between and .

The unknown function replaced with the average of the two known values, or

We will give us a reasonable approximate value of in the building at .

For constant

By separation and integrating on the both sides with respect to

.

General solution is

( is an arbitrary constant)

Initial value problem

(we choose to be )

Given initial condition is (we choose to be )

Particular solution is

We use , where is

is (namely hours ) and

The temperature is the building dropped , a result that looks reasonable.

Sage Coding

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

var('t, k') T = function('T')(t) de = diff(T, t) == k*(T- 45) desolve(de, [T, t], ics=[0, 70]) |

Evaluate

25*e^(k*t) + 45

Homogeneous Differential Equation (I)

A first-order differential equation is homogeneous if it has the form

A homogeneous equation is always transformed into a separable one by the transformation

. Then

By integrating on both sides with respect to

Thus,

( is an arbitrary constant)

Example 8 Reduction to Separable Form

Solve .

Solution

divide the given equation by

.

Let and by product differentiation

New substitute and from and then simply by subtracting on both sides

By integration on both sides with respect to

.

Multiply the last equation by to obtain.

General solution is or . ( is an arbitrary constant)

General solution represents a family of circles passing through the origin with centers on the -axis.

Sage Coding

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

var('x') y = function('y')(x) de = 2*x*y*diff(y, x) == y^2 - x^2 desolve(de, [y, x]) |

Evaluate

-x/(x^2 + y(x)^2) == c

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

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