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Second-Order ODEs

Gyeongsang National University

Dept. of Information & Communication Engineering

2.1 Homogeneous Linear ODEs of

Second Order

Second Order Homogenous Linear ODEs

• Standard form of second order linear ODEs

• Homogenous second order linear ODEs ( 𝑟 𝑥 = 0 )

• Solution of homogeneous linear ODEs

– Superposition principle or linearity principle

• We can obtain further solutions by adding them or by multiplying them with any constant

• Linear combination of two solutions

• Example 1) 𝑦′′ _{+ 𝑦 = 0}

– The functions y = sin 𝑥 and y = cos 𝑥 can be the solutions?

– The function y = 4.7 cos 𝑥 − 2sin 𝑥 can also be the solution?

– Linear combination of sin 𝑥 and cos 𝑥, 𝑐₁ sin 𝑥 + 𝑐₂ cos 𝑥, can also be the solution?

• Proof of superposition principle

– Show that when the solutions of 𝑦′′ _{+ 𝑝𝑦}′ _{+ 𝑞𝑦 = 0 are 𝑦}

1 and 𝑦2, the

linear combination of two solutions, 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, can also be the solution.

Superposition Principle

• Example 2) Verify that y = 1 + sin 𝑥 and y = 1 + cos 𝑥 are solutions of the nonhomogeneous linear ODE

– Sum of them is can be a solution?

• Second order ODEs consist of two initial conditions

• The conditions are used to determine two constant 𝑐₁ and 𝑐₂.

• The particular solution pass through the point (𝑥₀, 𝐾₀) with 𝐾₁ as the

• Example 4) Solve the initial value problem.

Initial Value Problem

 General solution

 Particular solution

 Let 𝑦 = 𝑐₁ cos 𝑥 + 𝑐₂ 𝑘 cos 𝑥 be a general solution. Can we find the particular solution that satisfies initial conditions?

𝑦

+ 𝑦 = 0,

𝑦 0 = 3.0,

𝑦

0 = −0.5

𝑦 = 𝑐

sin 𝑥 + 𝑐

cos 𝑥

𝑦 0 = 𝑐

= 3.0,

𝑦

0 = 𝑐

= −0.5

𝑦 = 3.0 cos 𝑥 − 0.5 sin 𝑥

• For a general solution of second order ODE 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂, 𝑦₁ and 𝑦₂ are not proportional.

• These 𝑦₁, 𝑦₂ are called a basis of solutions.

Basis

 Example 5) For 𝑦′′ + 𝑦 = 0, 𝑦 0 = 3.0, 𝑦′ 0 = −0.5, cos 𝑥 and sin 𝑥 are bases?

Basis

 Example 6) Verify 𝑦₁ = 𝑒𝑥 and 𝑦₂ = 𝑒−𝑥 are solutions of 𝑦′′ − 𝑦 = 0. Then solve initial value problem

• If one solution is known, a second solution can be found by solving a first order ODE.

• For the standard form of second order linear ODE,

If one solution, 𝑦₁, is known, we suppose the second solution as 𝑦₂ = 𝑢𝑦₁.

Substituting them into ODE

• Substituting

Reduction of Order

Since 𝑈 is always lager than 0, u is not a constant. Thus, 𝑦₁, 𝑦₂ are bases.

One solution

Reduction of Order

2.2 Homogeneous Linear ODEs

with Constant Coefficients

• Homogeneous Linear ODEs with Constant Coefficient

• Suppose a solution as 𝑦 = 𝑒𝜆𝑥_{, (𝜆 is a constant)}

Homogeneous Linear ODEs with

Constant Coefficient

• Roots of the characteristic equation

• Types of roots of the characteristic equation

Homogeneous Linear ODEs with

Constant Coefficient

(Case I) Two real roots if 𝑎2 − 4𝑏 > 0

(Case II) A real double root if 𝑎2 − 4𝑏 = 0

• The quotient of 𝑦₁ = 𝑒𝜆1𝑥 and 𝑦

2 = 𝑒𝜆2𝑥 is not constant and thus,

they are bases.

• General Solution:

• Example 1)

Case 1. Two Distinct Real-Roots 𝜆

₁

and 𝜆

₂

• Characteristic equation: 𝜆2 − 1 = 0 Roots: 𝜆

• Example 2) Initial value problem

• From 𝑎2 _{− 4𝑏 = 0, we directly obtain 𝜆}

1 = 𝜆2 = −𝑎/2

• Only one solution: 𝑦₁ = 𝑒− 𝑎/2 𝑥 • Reduction of order

In 𝑦′′ _{+ 𝑎𝑦}′ _{+ 𝑏𝑦 = 0 (𝑎, 𝑏 are constant), since we have 𝑝 𝑥 = 𝑎,}

• Example 3) 𝑦′′ _{+ 6𝑦}′ _{+ 9𝑦 = 0}

• Example 4) 𝑦′′ _{+ 𝑦}′ _{+ 0.25𝑦 = 0, 𝑦 0 = 3.0, 𝑦}′ _{0 = −3.5}

• For 𝑎2 _{− 4𝑏 < 0 , we define 𝜔}2 _{= 𝑏 −} 1 4𝑎

2→ 𝜔 = 𝑏 − 1 4𝑎

2

• By using superposition principle, we can obtain a basis of real solutions

• Real general solution

• Example 5) Initial value problem

Case. 3 Complex Roots

General solution : 𝑦 = 𝑒−0.2𝑥 Acos 3𝑥 + 𝐵 sin 3𝑥 Particular solution : 𝑦 = 𝑒−0.2𝑥 sin 3𝑥

2.4 Modeling of Free Oscillations

of a Mass-Spring System

• Attach an iron ball with mass 𝑚 at lower end of coil spring

• The length of the spring is increased by 𝑠₀ to downward direction (positive direction)

• We choose the downward direction as positive

• Weight(force of gravity pulling down)

– 𝑔 is a constant of gravity (Earth 𝑔 = 9.8𝑚/𝑠𝑒𝑐2 )

• Restoring force (force of the spring to return to original position)

– 𝑘(𝑘 > 0) is a spring constant

Static Equilibrium

 Two forces are generated when a body with mass 𝑚 is suspended.

𝐹

₀

= −𝑘𝑠

₀

Static equilibrium – state in which 𝑊 and 𝐹 are blanced

𝑊 = 𝑚𝑔

• By Hooke’s law, the restoring force 𝐹₁ is generated

• Location of the ball 𝑦(𝑡) is determined according to the force generated.

Modeling of Free Oscillations

 At static equilibrium (𝑦 = 0), we pull the ball down by an amount

𝑦 𝑦 > 0

• Undamped system – the force that determines the motion is 𝐹₁ = −𝑘𝑦 only

Undamped System

 The motion of system is determined by Newton’s second law

𝑚𝑦

′′

= 𝐹

₁

= −𝑘𝑦

𝑚𝑦

′′

+ 𝑘𝑦 = 0

ODE of The Undamped System

 General solution

𝜆

+

𝑘

𝑚

= 0

𝜔 =

𝑘/𝑚

𝑦 = 𝐴 cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡 = 𝐶 cos(𝜔𝑡 − δ)

tan 𝛿 = 𝐵 𝐴 (a and b are constant)

• If a mass-spring system with an iron ball of weight 𝑊 = 98[𝑛𝑡, 𝑘𝑔 x 𝑚/𝑠𝑒𝑐2] can be regarded as undamped, and the spring is such that the ball stretches it 1.09𝑚, how many cycles per minute will the system execute? What will its motion be if we pull the ball down from rest by 16𝑐𝑚 and let it start with zero initial velocity?

– 1[𝑛𝑡] is the force to move an object with mass 1𝑘𝑔 to 1𝑚/𝑠𝑒𝑐2

– damping force – force to suppress oscillation

– 𝑐(𝑐 > 0): damping constant

Damped System

𝐹

= −𝑐𝑦′

𝑚𝑦

= 𝐹

+ 𝐹

= −𝑘𝑦 − 𝑐𝑦

 ODE of damped system

𝑦

+

𝑐

𝑚

𝑦

′

₊

𝑘

𝑚

𝑦 = 0

 Roots of characteristic equation (𝜆2 + 𝑐 𝜆 + 𝑘 = 0)

Damped System

• 𝑐2 _{> 4𝑚𝑘 일 때 – 큰 마찰 계수, 특성 방정식이 두 개의 실근을 가짐}

• 𝛼 − 𝛽 > 0, 𝛼 + 𝛽 > 0 이므로, 𝑡가 증가할수록 𝑦(𝑡) 가 빠르게 0 에 수렴함

• 𝑐2 _{= 4𝑚𝑘 일 때 – 특성 방정식이 중근을 가짐}

• 𝛽 = 1

2𝑚 𝑐

2 _{− 4𝑚𝑘 = 0, 𝜆 = −𝛼.}

• 𝑐2 _{< 4𝑚𝑘 일 때 - 작은 마찰 계수, 특성 방정식이 허근을 가짐} • 𝛽 = 1 2𝑚 𝑐 2 _{− 4𝑚𝑘 = 𝑗} 1 2𝑚 4𝑚𝑘 − 𝑐 2 _{= 𝑗𝜔}∗_{, α =} 𝑐 2𝑚 • 𝜆₁ = −𝛼 + 𝑗𝜔∗

Case 3. Underdamping (저감쇠)

• 𝑘 = 90, 𝑚 = 10𝑘𝑔인 물체의 감쇠 진동에서 감쇠상수가 아래와 같은 경우, 초기 값 𝑦 0 = 0.16, 𝑦′ 0 = 0 일 때 물체의 운동을 구하시오. a) 𝑐 = 100𝑘𝑔/𝑠𝑒𝑐 b) 𝑐 = 60𝑘𝑔/𝑠𝑒𝑐 c) 𝑐 = 10𝑘𝑔/𝑠𝑒𝑐

예제 2

• 상수계수를 갖지 않는 2계 제차 미분방정식의 일반해를 구하기는 어려움 • 그러나 특별한 형태의 오일러-코시 방정식의 해는 구할 수 있음 • 해의 형태를 𝑦 = 𝑥𝑚_{, (𝑚은 상수)로 가정.}

Euler-Cauchy Equations

• 특성방정식의 해 구하기 • 특성방정식의 근의 유형

Euler-Cauchy Equations

𝑦

= 𝑥

𝑚 =

−(𝑎 − 1) ±

𝑎 − 1

_{− 4𝑏}

2

𝑦

= 𝑥

𝑎 − 1

− 4𝑏 > 0 이면, 서로 다른 두 실근 𝑚

, 𝑚

𝑎 − 1

− 4𝑏 = 0 이면, 중근 𝑚 = 𝑚

= 𝑚

• 𝑎 − 1 2 _{− 4𝑏 > 0 이면, 𝑦} 1 = 𝑥𝑚1 와 𝑦2 = 𝑥𝑚2 의 비 가 상 수 가 아니므로 아래 미분방정식의 기저가 된다. • 일반해는 𝑦 = 𝑐₁𝑦₁ + 𝑐₂𝑦₂ = 𝑐₁𝑥𝑚1 + 𝑐 2𝑥𝑚2 • 예제 1)

• 𝑎 − 1 2 _{− 4𝑏 = 0 이면, 중근 𝑚} 1 = 𝑚2= 1−𝑎 2 • 유일해: 𝑦₁ = 𝑥1−𝑎2 • 차수의 줄임을 이용한다. 𝑥2𝑦′′ + 𝑎𝑥𝑦′ + 𝑏𝑦 = 0를 표준형으로 바꿔준다. 𝑦′′+ 𝑎 𝑥 𝑦 ′₊ 𝑏 𝑥2 𝑦 = 0이므로 𝑝 𝑥 = 𝑎 𝑥가 된다. →

• 예제 2)

• 𝑎 − 1 2 _{− 4𝑏 < 0 이면 공액복소근을 가짐}

• 중첩의 원리를 이용해 아래 두 개의 실수 기저를 갖는다.

• 아래의 일반해를 갖는다.

Case. 3 Complex Roots

𝑚 = − 𝑎 − 1

2 ± 𝑗 𝑏 −

𝑎 − 1 2

• 예제 3)