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First-Order ODEs

Gyeongsang National University

Dept. of Information & Communication Engineering

Differential Equations?

The amount of salt after 𝑡 minutes? Inflow of salt water 10 l/min

100g of salt/1 Liter

10 l/min outflow Water 1000 l

𝑦(𝑡) : The amount of salt after 𝑡 minutes

𝑦′(𝑡) : Increase in salt at time 𝑡 (rate of change) 𝑦′ 𝑡 = 1000 − 0.01𝑦 𝑡

Basic Concept

• Differential equations

– An equation containing a function that differentiates an unknown function more than once

• Ordinary differential equations, ODE

– An equation consisting of functions that differentiate an unknown function 𝑦(𝑥) more than once

– Differential equations can include 𝑦(𝑥), a known function of 𝑥, and a constant

• 𝑛

_𝑡ℎ

-order ODE

– When the 𝑛_𝑡ℎ-order derivative of 𝑦(𝑥) is the derivative of the highest

order in the equation

– The order of the following differential equation?

• First-order ODE always includes 𝑦

′

_,

and can include 𝑦, 𝑥

• Implicit form of first-order ODE

• Explicit form of first-order ODE

• A funiction 𝑦 = ℎ 𝑥 is called of a given ODE:

– If ℎ 𝑥 is defined and differentiable on the open interval 𝑎 < 𝑥 < 𝑏 – If the equation becomes identity if 𝑦 and 𝑦′ are replaced with ℎ and ℎ′

• Example 1) Find the solution of 𝑦

′

_{= cos 𝑥}

Concept of Solution

• Example 2) Find the solution of 𝑦′ = 0.2𝑦

Concept of Solution

• A solution containing an arbitrary constant 𝑐 is called a general

solution of the ODE, e.g., 𝑦 = 𝑐𝑒

0.2𝑡

.

• If we choose a specific 𝑐 (e.g., 𝑐 = 1), the solution is called a

particular solution of the ODE.

– A particular solution does not contain any arbitrary constant.

• A particular solution is obtained from a general solution by an initial

condition 𝑦 𝑥

₀

= 𝑦

₀

with given value 𝑥

₀

and 𝑦

₀

.

– The solution curve should pass through the point (𝑥₀, 𝑦₀).

• An ODE, together with an initial condition, is called initial value

problem.

• Example 4)

• Step 1. The transition from the physical situation(physical system) to

is mathematical formulation(mathematical model)

• Step 2. The solution by a mathematical method

• Step 3. The physical interpretation of the result

Example 5) Given an amount of a radioactive substance 0.5𝑔, find the

amount present at time 𝑡?

Physical information) Experiments show that at each instant a

radioactive substance decomposes – and is thus decaying in time –

proportional to the amount of substance present.

• Step 1. Setting up a mathematical model of the physical process

– 𝑦(𝑡) : the amount of substance present at any time 𝑡 – 𝑦′(𝑡) : the time rate of change of substance

– 𝑑𝑦

𝑑𝑡 = −𝑘𝑦 𝑘 > 0 , 𝑦 0 = 0.5

• Step 2. Mathematical solution

– 𝑦(𝑡) = 𝑐𝑒−𝑘𝑡 : general solution

– 𝑦(𝑡) = 0.5𝑒−𝑘𝑡 : particular solution

• Step 3. Interpretation of result

• Separable ODE: ODEs that can be solved by separating variables.

 First-order ODE

 Separable equation

 Separating variables

 Replace

1 𝐺(𝑦)

to 𝑔(𝑦)

 Integrating both sides

• Example 1) Find a general solution of 𝑦

′

_{= 1 + 𝑦}

2

Separable ODEs

Separating variables, Integrating, 1 𝑎2 _{+ 𝑦}2 𝑑𝑦 → 1 𝑎tan −1 𝑥 𝑎 + 𝑐

Separable ODEs

Separable ODEs

• Solving previous example by separating variables

– 𝑦′ = 0.2𝑦 – 𝑦′ _{= −0.2𝑦}

• Example 4) Radiocarbon dating. A mummy was founded in ice. The ration of carbon 14₆

_C

_to

6

12_{C carbon} 146C

6

12_C in this mummy is 52.5%. When did this

mummy approximately live and die?

– Physical information) In living organisms, the ratio of radioactive carbon 14_{6C to}

ordinary carbon 12_{6C is constant. When an organism dies, the amount of}

6

14_C

decreases by half every 5715 years. radioactive substance decomposes proportional to the amount of substance present.

Differential Equations?

The amount of salt after 𝑡 minutes? Inflow of salt water 10 l/min

100g of salt/1 Liter

10 l/min outflow Water 1000 l

• Example 6) Heating an office building. Suppose that in winter the daytime temperature in an office building, is maintained at 70°F. Heating is shup off at 10 PM and turned on again 6 AM. On a certain day the temperature inside the building at 2 AM was founded to be 65°F. The outside temperature was 45°F at 10 PM to 6 AM. What was the temperature inside the building when the heat was turned on at 6 AM?

– Physical information) Experiments show that the time rate of change of the temperature T of the object is proportional to the difference between T and the temperature of the surrounding medium (Newton’s law of cooling). Therefore, the rate of temperature change inside building is proportional to the temperature between inside and outside of building.

Extended Method:

Reduction to Separable Form

• Extended method: Certain non-separable ODEs can be separable

by transformations.

• Substitution to

𝑦

Extended Method:

Reduction to Separable Form

Extended Method:

• Partial derivatives

From 𝜕𝑢 𝜕𝑥, 𝜕𝑢 𝜕𝑦, find 𝑢(𝑥, 𝑦)

• From

𝜕𝑢 𝜕𝑥

= 2𝑥𝑦

3

_{+ 1,find}

𝜕2𝑢 𝜕𝑥𝜕𝑦

• From

𝜕𝑢 𝜕𝑦

= 𝑥

2

_3𝑦

2

_{+ 1, find}

𝜕2𝑢 𝜕𝑥𝜕𝑦

Partial Derivatives

(By regarding y as a constant, differentiate u by x) (By regarding x as a constant, differentiate u by y)

• Total differential

Exact Differential Equation

Total differential of 𝑢 𝑥, 𝑦 = 𝑥 + 𝑥2𝑦3 = 𝑐:

General solution of the differential equation: 𝑢 𝑥, 𝑦 = 𝑥 + 𝑦2𝑦3 = 𝑐 𝑦′ = 𝑑𝑦

𝑑𝑥 = −

1 + 2𝑥𝑦3 3𝑥2_𝑦2

• Definition of exact differential equation

– Transform first-order ODE 𝑀 𝑥, 𝑦 + 𝑁 𝑥, 𝑦 𝑦′ = 0 as following form:

– If 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 is in the form of 𝑑𝑢 = 𝜕𝑢

𝜕𝑥𝑑𝑥 + 𝜕𝑢

𝜕𝑦𝑑𝑦, we call this

equation as exact differential equation.

Exact Differential Equation

•

Determine whether it is an exact differential equation or not

 If first-order ODE 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0 become an exact DE,

we have 𝜕𝑢 𝜕𝑥 = 𝑀, 𝜕𝑢 𝜕𝑦 = 𝑁, equivalently we have 𝜕2𝑢 𝜕𝑥𝜕𝑦 = 𝜕𝑀 𝜕𝑦, 𝜕2𝑢 𝜕𝑥𝜕𝑦 = 𝜕𝑁 𝜕𝑥.

• General solution of the exact DE

– From 𝜕𝑢

𝜕𝑥 = 𝑀, obtain 𝑢 = 𝑀𝑑𝑥 + 𝑘(𝑦) by integration

– From 𝑢 = 𝑀𝑑𝑥 + 𝑘(𝑦) , we obtain 𝜕𝑢_𝜕𝑦 and then, obtain 𝑘(𝑦) by comparing with 𝑁(𝑥, 𝑦).

– From 𝑢 = 𝑁𝑑𝑥 + 𝑙(𝑥), we obtain 𝜕𝑢

𝜕𝑥 and then, obtain 𝑙(𝑥) by comparing

with 𝑀(𝑥, 𝑦).

 General solution has an implicit form as 𝑢 𝑥, 𝑦 = 𝑐.

• Example 1) find a solution of the following equation

cos 𝑥 + 𝑦 𝑑𝑥 + 3𝑦

2

+ 2𝑦 + cos 𝑥 + 𝑦 𝑑𝑦 = 0

Exact Differential Equation

 Step 1: determining whether the exact DE or not

• Example 2) Find the solution of 𝑥 − 𝑦 𝑑𝑥 − 𝑑𝑦 = 0.

• Example 3) Find the solution of −𝑦𝑑𝑥 + 𝑥𝑑𝑦 = 0

Exact Differential Equation

 Step 1: determining whether the exact DE or not

 This equation is not an exact DE but, if we multiply 1

𝑥2 to both sides,

it becomes the exact DE.

일반해: y

• Integrating factor

– For a non-exact DE:

– if a DE multiplied by function F(x, y) is an exact DE,

– We call F(x, y) as an integrating factor.

• Equation in example 3 has multiple integrating factors.

Integrating Factors

• Since the condition that 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0 become an exact ODE is 𝜕𝑀

𝜕𝑦 = 𝜕𝑁

𝜕𝑥,

• the condition that 𝐹𝑃𝑑𝑥 + 𝐹𝑄𝑑𝑦 = 0 become an exact ODE is

• By the product rule, we have (𝐹_𝑦 = 𝜕𝐹

𝜕𝑦, 𝐹𝑥 = 𝜕𝐹 𝜕𝑥)

• However, it is difficult to find the integrating factor to satisfy this condition.

Integrating Factors

• Finding the integrating factor that has a single variable (𝑥 or 𝑦). • Let 𝐹 be 𝐹 = 𝐹(𝑥), we have 𝐹_𝑦 = 0, 𝐹_𝑥 = 𝐹′ = 𝑑𝐹

𝑑𝑥, and then

– Dividing both sides by 𝐹𝑄,

– Theorem 1. if 𝑅 is a function of x only, the integrating factor is obtained by.

Integrating Factors

• Example 5) Solve the following differential equation:

• 단계 1) Determining whether the exact DE or not

• 단계 2) Finding the integrating factor and general solution

– R is not a function of x only, we cannot obtain integrating factor from Theorem 1.

– Thus, we obtain the integrating factor as 𝐹∗ _{𝑦 = 𝑒}−𝑦 and the equation

becomes an exact DE as

– The general solution is obtained as

– By the initial condition, the particular solution is obtained as

• Example 6) Find the general solution of 𝑥

2

_{+ 𝑦}

2

_{𝑑𝑥 − 2𝑥𝑦𝑑𝑦 = 0}

1.5 Linear ODEs. Bernoulii

• If the first-order ODE can be represented by following form, we call it

first-order

linear ODE

and the following form is called as

standard

form

• For 𝑓 𝑥 𝑦

′

_{+ 𝑝 𝑥 𝑦 = 𝑟(𝑥), it can be represented by the standard}

form as

• Represent 𝑦

′

_{cos 𝑥 + 𝑦 sin 𝑥 = 𝑥 as the standard form}

• Determine whether the following DEs are linear or not.

• For the linear ODE with 𝑟 𝑥 = 0, we call it

homogeneous

linear

ODE.

• By separating variables,

• For the linear ODE with 𝑟(𝑥) ≠ 0, we call it

nonhomogeneous

linear

ODE.

• Transforming it to exact DE by multiplying the integrating factor

• Or we can obtaining the solution by following steps:

1. Multiplying an arbitrary 𝐹(𝑥) to the standard form equation:

2. Finding 𝐹(𝑥) that makes the left side becomes 𝐹𝑦 ′ _{= 𝐹}′_{𝑦 + 𝐹𝑦′:}

3. Substituting 𝐹 = 𝑒ℎ and ℎ′ _{= 𝑝 to the equation:}

4. Integrating the equation:

5. Obtaining a general solution as

 General solution of the equation:

Non-homogeneous Linear ODEs

• Example 1) Find a particular solution of 𝑦′ _{+ 𝑦 tan 𝑥 = sin 2𝑥 , 𝑦 0 = 1}

1. Transforming to the standard form

2. Finding ℎ

3. Finding 𝑒

ℎ

Non-homogeneous Linear ODEs

Non-homogeneous Linear ODEs

• Example 3) When initial current is 0 A, find 𝐼(𝑡) at time 𝑡.

Voltage drop: 𝑅𝐼(𝑡)

Bernoulli Equation

• Some of non-linear ODE can be transformed into linear ODE.

• Standard form of Bernoulli equation

• Example 4) Find a general solution of 𝑦

′

_{= 𝐴𝑦 − 𝐵𝑦}

2

Bernoulli Equation

From 𝑢 𝑥 = 𝑦 𝑥 −1,

• Example 5) Find a general solution of 𝑦

′

_{+ 𝑥𝑦 = 𝑥𝑦}

−1

• 1계 상미분방정식(first order ordinary differential equation, ODE)

– 상미분 방정식(ODE) – 하나의 미지 함수 𝑦(𝑥) (또는 𝑦(𝑡))를 미분한 함수들로 이루어진 방정식 – 1차 미분이 방정식에서 가장 높은 차수인 미분방정식 – 일반해(general solution) – 임의 상수 𝑐를 포함하는 해 (부정적분을 통해서 해를 얻음) – 특수해(particular solution) – 초기 조건 𝑦₀ = 𝑦 𝑥₀ 로 부터 𝑐값을 결정 – 모델화(modelling) • 1) 자연 현상을 수학적 문제로 만들기 • 2) 수학의 방법을 이용해서 풀기

Summary

음함수 형태 양함수 형태

• 변수 분리형 상미분방정식(separable ODE)

– 변수를 분리하여 해를 얻을 수 있는 형태의 미분방정식 – 양변을 적분해서 일반해를 구함

Summary

𝑔 𝑦 = 1 𝐺(𝑦)

• 완전 상미분방정식(exact ODE)

– 미분 방정식의 형태가 어떤 함수의 전미분(total differential) 형태와 같으면 완전 미분방정식 – 완전 미분방정식의 일반해 구하기 • 𝑢 = 𝑀𝑑𝑥 + 𝑘(𝑦)로 부터 𝜕𝑢_𝜕𝑦를 구한 후, 𝑁(𝑥, 𝑦)와 비교하여 𝑘(𝑦)를 구한다.

Summary

𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0

𝜕𝑀

𝜕𝑦

=

𝜕𝑁

𝜕𝑥

𝑑𝑢 =

𝜕𝑢

𝜕𝑥

𝑑𝑥 +

𝜕𝑢

𝜕𝑦

𝑑𝑦 = 0

=

• Integrating factor

– 불완전 미분방정식에 함수 F를 곱해서 완전 미분방정식이 되면 함수 F를 integrating factor라고 함 – Integrating factor를 이용해 완전 미분방정식으로 바꾼 후 해를 구함 • 1) 완전 미분방정식인지 아닌지 판별

Summary

𝑃 𝑥, 𝑦 𝑑𝑥 + 𝑄 𝑥, 𝑦 𝑑𝑦 = 0

𝐹𝑃 𝑥, 𝑦 𝑑𝑥 + 𝐹𝑄 𝑥, 𝑦 𝑑𝑦 = 0

불완전 미분방정식 완전 미분방정식

• Integrating factor찾기

– 하나의 독립변수(𝑥 또는 𝑦)를 갖는 integrating factor를 찾는다. • 𝐹(𝑥) 찾기 • 𝐹∗_{(𝑦) 찾기}

Summary

𝑅 𝑥 = 1 𝑄 𝜕𝑃 𝜕𝑦 − 𝜕𝑄 𝜕𝑥 가 𝑥만의 함수이면 𝐹(𝑥)가 존재 𝐹 𝑥 = exp 𝑅 𝑥 𝑑𝑥 𝑅∗ 𝑦 = 1 𝑃 𝜕𝑄 𝜕𝑥 − 𝜕𝑃 𝜕𝑦 가 𝑦만의 함수이면 𝐹 ∗_{(𝑦)가 존재} 𝐹∗(𝑦) = exp 𝑅∗ 𝑦 𝑑𝑦

𝑃 𝑥, 𝑦 𝑑𝑥 + 𝑄 𝑥, 𝑦 𝑑𝑦 = 0

• 선형 상미분방정식(linear ODE)

– 선형 상미분방정식의 표준형 – 𝑦′와 𝑦는 1차 • homogenous 선형 미분방정식 • non-homogenous 선형 미분방정식

Summary

𝑦′ + 𝑝 𝑥 𝑦 = 0

𝑦

′

+ 𝑝 𝑥 𝑦 = 𝑟 𝑥

일반해 𝑦 = 𝑐𝑒− 𝑝 𝑥 𝑑𝑥 _{= 𝑒}−ℎ_, 𝑦′ + 𝑝 𝑥 𝑦 = 𝑟 𝑥 일반해 𝑦 = 𝑒−ℎ 𝑒ℎ𝑟 𝑥 𝑑𝑥 + 𝑐 , ℎ = 𝑝 𝑥 𝑑𝑥 ℎ = 𝑝 𝑥 𝑑𝑥

• Bernoulli 방정식

– 일부 비선형 미분방정식을 선형 미분방식으로 바꿀 수 있다. – 선형 미분방정식을 풀어서 𝑢를 구한 후, 𝑢를 이용해 𝑦를 구한다.

Summary

𝑦

′

+ 𝑝 𝑥 𝑦 = 𝑔 𝑥 𝑦

𝑎 비선형 미분방정식

𝑢

′

+ 1 − 𝑢 𝑝 𝑥 𝑦 = 1 − 𝑎 𝑔 𝑥

선형 미분방정식 𝑢 𝑥 = 𝑦 𝑥 1−𝑎