Introduction to Differential Equations National Chiao Tung University Chun-Jen Tsai 9/9/2019

Introduction to Differential Equations

National Chiao Tung University Chun-Jen Tsai 9/9/2019

Outline of the Course

 Introduction to differential equations (Chapter 1)

 First-order differential equations (Chapter 2)

 Higher-order differential equations (Chapter 4)

 Modeling with Higher-order differential equations (Chapter 5)

 The Laplace transform (Chapter 7)  midterm around this point!

 Systems of linear 1^st-order differential equations (Chapter 8)

 Power series methods (Chapter 6)

 Fourier series methods (Chapter 11)

 Partial differential equations (Chapter 12)

Textbook and Grading Policy

 Textbook:

 Dennis G. Zill, Differential Equations with Boundary-Value Problems, 9th edition, 2018, Cengage Learning.

(高立圖書代理, 顔俊杰 0921-456030)

 An alternative textbook:

Dennis G. Zill, Differential Equations with Modeling Applications, 11th edition, 2018, Cengage Learning.

(華泰文化, 蕭瑀倢 0933-838337)

 Grading is based on

 Pop Quizzes (25%) – from homework assignments

 Mid-terms exam (35%) – on 11/4/2019

 Final exam (40%) – on 1/6/2020

Before You Move On …

 Homework #0: Check out the following video:

Raffaello D'Andrea’s TED talks

The astounding athletic power of quadcopters.

Jun 2013

I will be asking you questions on this video in our next class!

Differential Equations

 Definition:

An equation containing the derivatives of one or more dependent variables, with respect to one or more

independent variables, is said to be a differential equation (DE).

 Example:

1 2

.

) 0

(t e ^t f

x   xt

dt

dx  0.2

 

dt e

dx _t

2 .

2 0

1 .

0 



In this course, given a blue equation (behavior of a phenomenon),

Why Differential Equations

 For dynamic phenomena, we want to predict their long- term behavior by observing and measuring their short- term behavior

 Long-term behavior of a dynamic system is defined by its underlying rule  hard to measure

 Short-term behavior of a dynamic system is described by its changing characteristics (derivatives)  easier to measure

Classification of DE by Type

 Ordinary differential equation (ODE): an equation contains only ordinary derivatives of one or more dependent variables with respect to a single

independent variable

 Partial differential equation (PDE): an equation involving the partial derivatives of one or more dependent variables of two or more independent variables

ex

dx y

dy  5 

u u

u    

 2

2 2

Classification of DE by Order

 The order of a differential equation is the order of the highest derivative in the equation.

 An n^th-order ODE with one dependent variable can be expressed in the general form:

 ) 0 ,....,

, , ,

(x y y y  y ⁿ  F

a real-valued function of n+2 variables 2nd order 1st order

ex

dx y dy dx

y

d   



 



 5 4

3 2

2

Normal Form of ODE

 F() can be expressed in general in the normal form:

where f is a real-valued function with n+1 variables.

For example, the normal forms of the first order and the 2^nd-order ODEs are:

 ) ,....,

, , ,

(   ^1

 ⁿ

n n

y y

y y x dx f

y d

) , (x y dx f

dy 

) , , (

2

y y x y f

d  

Classification of DE by Linearity

 An nth-order ODE, F, is said to be linear if F is linear in y, y', …., y⁽ⁿ⁾. That is, F can be expressed as:

where a_i(x), i = 0, …, n depend on the independent variable x only

 Example:

 (y – x)dx + 4xdy = 0

 y" – 2y' + y = 0

 y e^x

dx x dy dx

y

d ₃ 3 5 

3

   

 

 

1

1 a x y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n

n  _ ^_    

Nonlinear ODE

 A differential equation with nonlinear functions of the dependent variable or its derivatives.

 Examples: If y is the dependent variable,

 (1 – y)y' + 2y = e^x

 d²y/dx² + sin y = 0

 y⁽⁴⁾ + y² = 0

Solution of an ODE

 Definition: a solution of an ODE is a function y(x), defined on an interval I and possessing at least n derivatives that are continuous on I, which when

substituted into an n^th-order ODE reduces the equation to an identity.

 That is, a solution y(x) of F satisfies:

F(x, y(x), y(x), y(x), …, y⁽ⁿ⁾ (x)) = 0, x  I.

 If an ODE has a solution y(x) = 0, x I, then it is called the trivial solution of the ODE.

All Roads Lead to Rome

 If we have a function y:

Then,

Thus, it doesn’t matter what the constant C is, is a solution of the DE dy/dx = 2xy.

 Often, a differential equation alone has many solutions;

more information is required to resolve ambiguity .

, )

(x Ce ² C R

y  ^x 

   

dx C

dy _{x x}







x2

Ce y 

Solution is not Guaranteed

 Expressing a phenomenon as a differential equation does not guarantee that it has a solution. Obviously,

(y)² + y² = –1 has no (real-valued) solution.

Interval of Definition

 A solution of an ODE includes a function y(x) and the interval of definition, I.

 I is usually referred to as the interval of definition, the interval of existence, the interval of validity, or the

domain of the solution.

 I can be an open interval (a, b), a closed interval [a, b], an infinite interval (a, ), and so on.

Solution Curve

 The graph of a solution y(x) of an ODE is called a

solution curve. Since y(x) is a differentiable function, it is continuous on its interval of definition.

 There maybe a difference between the graph of y(x) and the graph of the solution of the ODE.

y = 1/x, (0,∞)

y

1 x 1

y

1 x 1

Explicit and Implicit Solutions

 Definition: A solution in which the dependent variable is expressed solely in terms of the independent

variable and constants is called an explicit solution.

 Definition: An equation G(x, y) = 0 is said to be an implicit solution of an ODE on an interval I provided

that there exists at least one function y that satisfies the relation as well as the differential equation on I.

Verification of an Implicit Solution

 Example:

The relation x²＋y² = 25 is the implicit solution of the differential equation dy/dx = –x/y on the interval

–5 < x < 5 Verification:



 ^{ }

dx y d

dx x

d   2  2  0

dx y dy x

y dx x

dy   /

Solving for Explicit Solution

 One can solve an implicit solution for explicit solutions.

In the previous example,

Implicit solution Explicit solution 1 Explicit solution 2 x²＋y² = 25 _{y1  25 x}²_{, 5  x  5 25} ²_{, 5 5}

2    x   x 

y

- 5 5

x y

5

5 x y

5

5 x y

5

- 5 - 5

- 5 - 5

Families of Solutions

 A solution to a 1^st-order DE containing an arbitrary constant represents a set G(x, y, c) = 0 of solutions is called a one-parameter family of solutions.

 For n^th-order DE, an n-parameter family of solutions can be represented as

G(x, y, c₁, c₂, …, c_n) = 0.

If the parameters c₁, c₂, …, c_n are resolved, then it’s called a particular solution of the DE.

 Example:

y – cx = 0 is a family of solutions of xy′ – y = 0.

y

x c < 1 c = 1 c > 1

Singular Solutions

 Definition: A singular solution is a solution that cannot be obtained by specializing any of the parameters in the family of solutions.

 Example:

Both y = x⁴/16 and y = 0 are solutions of dy/dx = xy^1/2 on the interval (–, ). The ODE possesses the family of solutions y = (x²/4 + c)². However, y = 0 is not in the

family of solutions.

General Solutions

 Definition: If every solution of an n^th-order ODE

F(x, y, y, y, …, y⁽ⁿ⁾) = 0 on an interval I can be obtained from an n-parameter family of equations

G(x, y, c₁, c₂, …, c_n) = 0 by appropriate choices of the parameters c_i, i = 1, 2, …, n, we then say that the n-

parameter family of equation is the general solution of the D.E.

Example: Two-Parameter Family

 The functions x = c₁cos4t and x = c₂sin4t, where c₁ and c₂ are arbitrary constants, are solutions of x" + 16x = 0.

For x = c₁cos4t, the first two derivatives w.r.t. t are x' = –4c₁ sin 4t and x" = –16c₁cos4t.

Substituting x" and x' into the DE gives

x" + 16x = –16c₁cos4t + 16(c₁cos4t) = 0.

Similarly, for x = c₂sin4t, we have

x" + 16x = –16c₂sin4t + 16(c₂sin4t) = 0.

Their linear combinations are a family of solutions.

Example: Piecewise Solutions

 One can verify that y = cx⁴ is a solution of xy' – 4y = 0 on the interval (–, ). The following piecewise defined solution is a particular solution of the ODE:

 This particular solution cannot be obtained by a single choice of c.









 

0 ,

0 ,

4 4

x x

x y x

x c = 1, x0 y

c =–1, x < 0 c =1

c =–1 x y

Initial Value Problem

 Definition:

On some interval I containing x₀, the problem:

Solve:

Subject to: y(x₀) = y₀, y(x₀) = y₁, …, y^(n–1)(x₀) = y_n–1, where y₀, y₁, …, y_n–1, are arbitrarily specified real constants, is called an initial value problem (IVP).

The values of y(x) and its first n – 1 derivatives at x₀ are called initial conditions.

 





 ⁿ

n n

y y

y x dx f

y d

First Order IVP

 A first order IVP tries to solve dy/dx = f(x, y), subject to y(x₀) = y₀. In geometric term, we are seeking a solution so that the solution curve passes through the

prescribed point (x₀, y₀).

y

(x₀, y₀)

solutions of the DE

Second Order IVP

 A second order IVP tries to solve d²y/dx² = f(x, y, y), subject to y(x₀) = y₀, y(x₀) = y₁. In geometric term, we are seeking a solution so that the solution curve not only passes through the prescribed point (x₀, y₀), but also with a slope y₁ at this point.

y solutions of the DE

m = y₁

(x₀, y₀)

Example: 1st-Order IVPs

 It is easy to verify that y = ce^x is a one-parameter family of solutions of the simple first-order equation y' = y on the interval (–, ). If y(0) = 3, we have

3 = ce⁰ = c

 y = 3e^x is a solution of IVP:

y' = y, y(0) = 3. ^x

y (0, 3)

(1, –2)

Existence of Unique Solution

 Two key questions of solving an IVP are:

 Do solutions exist for the differential equation?

 Given an initial condition, is the solution unique?

 Examples:

 The IVP y' = 1/x, y(0) = 0 has no solution. By integration, we have y(x) = ln |x| + c; but ln |x| is not defined at 0!

 The IVP dy/dx = xy^1/2, y(0) = 0 has at least two solutions: y = 0 and y = x⁴/16.

Example: Multiple IVP Solutions (1/2)

 Consider the IVP dy/dx = xy^1/2, y(0) = 0:

The DE has a constant solution y = 0 and a family of solution

The IVP has infinite solutions:

For any a  0,

4 .

2 2



 



 

 x c

y

 









 

a x

a x

a y x

, 16 / ,

0

2 2 2

y

(0, 0) x

a = 0 a > 0

Example: Multiple IVP Solutions (2/2)

 Consider only the case c  0,

let c = –b, b  0: ^{2 2}

4 

 



 

 x c

y

b a

a x

b x

b y x

2 ,

16 / ) (

4 ) 2 (

4 4 4

2 2 2

2 2 2

2 2













 





 



 



-10 -8 -6 -4 -2 0 2 4 6 8 10

0 50 100 150 200 250 300 350 400

c = –4 c = 0 c = 4

Existence and Uniqueness Theorem

 Theorem: Let R be a rectangular region in the xy-plane defined by a  x  b, c  y  d, that contains the point

(x₀, y₀) in its interior. If f(x, y) and f/y are continuous on R, there exist some interval I₀: x₀–h < x < x₀+h, h > 0, contained in a  x  b, and a unique function y(x)

defined on I₀ that is a solution of the first-order initial- value problem:

. )

( ),

,

(x y y x₀ y₀ dx f

dy  

y d

c

R

(x₀, y₀)

Example:

 Again, let’s revisit the IVP: dy/dx = xy^1/2, y(0)=0.

Since

f(x, y) = xy^1/2, and f/y = x/(2y^1/2),

they are continuous in the upper half-plane defined by y > 0. Therefore, for any (x₀, y₀), y₀ > 0, there is an

interval centered at x₀ on which the given DE has a unique solution.

However, There is no unique solution for the IVP since

f/y is undefined at (0, 0).

DE as Mathematical Models

Hypotheses Mathematical

formulation in DE Turn hypotheses

into equations

Explicit solution (model) Compare model

to observations

Solving DE

Predict outputs from the model Real-world

phenomenon observationsand

Correct hypotheses and models

Natural Growth and Decay Models

 The differential equation

is a widely used model for natural phenomena whose rate of change over time is proportional to its current population  what is the solution?

 If a population has birth and death rates

b

d

respectively. The differential change in size P(t) of the population changes is

Falling Bodies

 Newton’s second law of motion: F = ma

 Question: what is the position s(t) of the rock relative to the ground at time t?

Acceleration of the rock: d²s/dt²

 

Model: d²s/dt² = –g, s(0) = s₀, s'(0) = v₀. Solution: s(t) = –gt²/2 + v₀t + s₀

dt mg s

m d ²₂   g

dt s

d ²₂  

building s₀

s(t) rock

v₀

ground

Torricelli’s Model of a Draining Tank

 Torricelli’s Law of draining tank:

Derivation: Torricelli assumes that a drop of water from the surface escapes the hole at the speed

hole area a

V(t) y(t)

Leaking water

 If i(t) = dq/dt is the electric current across the circuit, the voltage drops across different electric components are:

 Inductor:

 Resister:

 Capacitor:

 Kirchhoff’s second law of circuits:

Voltage drop = Impressed Voltage, that is:

Series Circuit

) 1 (

2

t E dq q

q R

L d   

dt L di v 

Ri v 

C q v  1

L

R

C E(t)