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MATH 251 Work sheet / Things to know

Chapter 3

1. Second order linear differential equation

Standard form:

What makes it homogeneous?

We will, for the most part, work with equations with constant coefficients only.

What is the general form of a second order linear equation with constant coefficients?

Ex. 3.1.1 Can you think of any function(s) that satisfy each equation (w/ constant

coefficients) below?

(a) y″ − 25y = 0

(b) y″ + 25y = 0

(c) y″ − 25 y′ = 0

The example (c) above is an instance of a second order linear equation with the y-term

missing. It is essentially a first order linear equation in disguise. All equations of this

type can be solved by changing it into a first order equation with the substitutions u = y′

and u′ = y″, then use the integrating factor method to solve for u, and integrate the result

to find y.


2. The characteristic equation

Given the equation ay″ + by′ + cy = 0, a ≠ 0, what is its characteristic equation?

Any root, r, of the characteristic equation has the property that y = ert always satisfies the

equation above. Therefore, y = ert will be a particular solution for each root r.

Consequently, an important formula to remember for this class is (surprisingly) the

quadratic formula:

a

acbbr

2

42−±−

=

Note that the characteristic equation method does not require the given differential

equation to be put into its standard form first – the quadratic formula simply doesn’t care

whether or not the leading coefficient is 1.

Suppose r1 and r2 are two distinct real roots of the characteristic equation, what is the

general solution of the differential equation?

y =

Ex. 3.2.1 y″ + y′ − 12y = 0

What is its characteristic equation?

What are the roots of the characteristic equation?

Based on the roots, what are 2 particular solutions of the equation?

The general solution is

y =

Ex. 3.2.2 2y″ + 3y′ − 2y = 0

y =


3. Initial Value Problems

What do the initial conditions of a second order differential equation look like?

A second order equation’s general solution will have 2 arbitrary constants / coefficients.

Therefore, an IVP will have 2 initial conditions in order to give 2 (algebraic) equations

needed to solve for them. What must the conditions look like?

Ex. 3.3.1 Take the previous example y″ + y′ − 12y = 0. Find its solution satisfying

(a) y(0) = 0, y′(0) = 2

(b) y(0) = 2, y′(0) = −6

How to (easily) work with initial conditions where t0 ≠ 0?

Ex. 3.3.2 y″ + y′ − 12y = 0, y(45000) = 0, y′(45000) = 2


The Existence and Uniqueness Theorem (for second order linear equations)

It is really the same theorem as the one we saw earlier, except this one is in the context of

second order linear equation.

What does it say?

How to find the largest interval (that is, the interval of validity) on which a particular

solution is guaranteed to exist uniquely?

Ex. 3.3.3 Consider the equation given below. For each set of initial conditions, find the

largest interval on which the particular solution is guaranteed to exist uniquely.

(t2 + 2t – 8) y″ + sin(2t) y′ + (t + 9)

3 y = t

−1e5t

(a) y(3) = 1, y′(3) = −9

(b) y(−1) = 9π, y′(−1) = e−1


4. The general solution of second order linear equations

Structure of a general solution:

homogeneous second order linear equation: y =

A nonhomogeneous equation, however, will have a slightly different form of solution.

Wronskian − know the determinant formula

W(y1, y2)(t) =

*Note that the Wronskian is defined as a function of t.

Fundamental solutions

What are they? Why are they important?

How to determine if any two given solutions are fundamental?

Ex. 3.4.1 Which pair(s) of functions below can be fundamental solutions?

(1) 10, t + 2

(2) e2t, e−2t

(3) et, 0

(4) 2e3t, e3t + 2

(5) cos 6t, −2sin 6t

(6) 5sin t, cos(t + π/2)

(7) cos 2t, sin(2t − 2π)

Ex. 3.4.2 Given that y1 = te−t and y2 = 2te

−t − 7ln(t) are both known solutions of a certain

equation y″ + p(t)y′ + q(t)y = 0. What is the general solution of this equation?


5. The Abel’s Theorem

It gives us a way to determine, up to a constant multiplier, the Wronskian of any pair of

solutions of a given second order homogeneous linear equation, by looking at just the

equation itself only.

What does it say: W(y1, y2)(t) =

(Due to their similarity) do not confuse this formula with another formula we have

seen before. Which formula?

Ex. 3.5.1 (Final Exam, fall 2008) Let y1(t) and y2(t) be any two solutions of the second

order linear equation t2y″ − 6t y′ + cos(3t)y = 0. What is the general form of their

Wronskian, W(y1, y2)(t)?

Ex. 3.5.2 (Exam I, spring 2007) Consider the second order linear differential equation

t y″ − 2y′ + y = 0. Suppose y1(t) and y2(t) are two fundamental solutions of the equation

such that y1(1) = 2, y′1(1) = 0, y2(1) = 2, and y′2(1) = 2. Compute their Wronskian

W(y1, y2)(t).


6. Characteristic equation: complex roots case

(Background info) Euler’s formula:

Suppose r = λ ± µi are two complex conjugate roots of the characteristic equation, what is

the general solution of the differential equation?

y =

Ex. 3.6.1 y″ − 4y′ + 40y = 0



y =

What is the particular solution satisfying y(0) = 1 and y′(0) = −6?



7. Characteristic equation: repeated real root case

Suppose r is a repeated real root of the characteristic equation, what is the general

solution of the differential equation?

y =

Ex. 3.7.1 y″ + 12y′ + 36y = 0



y =




8. Reduction of order

The characteristic equation method only works for equations of constant coefficients.

While we have not (and will not, in this course) learned a technique to solve a generic

second order linear equation with non-constant coefficients, we nevertheless have learned

enough to be able to solve such an equation of non-constant coefficients – provided that

we know one nonzero solution of the equation already. This type of problems is called

reduction of order.

Why is it called reduction of order?

There are more than one ways to solve such a problem. What we learn in class is a

“synthetic” method. It is different, and easier, than the method in the textbook.

What are the key steps of our method?

Ex. 3.8.1 Given that y1 = t is a solution, find the general solution of the equation

t2y″ + 2t y′ − 2y = 0.

Find its general solution.


9. Nonhomogeneous second order linear differential equation

Standard form (constant coefficients):

What is its corresponding homogeneous equation?

Structure of its general solution:

y =

What is the complementary / homogeneous solution?

What is the particular / nonhomogeneous solution?

Note that the different context under which the name “particular solution” is used, as

compared to the solution of an IVP.

10. Method of Undetermined Coefficients

What is the idea behind the method?

You should ALWAYS solve for the complementary solution first.

The starting choices:

(1) If g(t) is an exponential function.

Ex. 3.10.1 Given g(t) = 3e−5t, choose Y =


(2) If g(t) is a polynomial, or a power of t.

Ex. 3.10.2 Given g(t) = −2t5 + 11t

3 + t – 4, choose Y =

(3) If g(t) is a sinusoidal function, sine or cosine.

Ex. 3.10.3 Given g(t) = –9sin 3πt, choose Y =

.

If g(t) is a sum or difference of the “basic” function, how do you choose Y?

Ex. 3.10.4 (a) Given g(t) = 3e−5t – 9sin 3πt, choose Y =

(b) Given g(t) = t2 + 2cos 4t, choose Y =

The possible glitch with our starting choices, how to spot it, and how to correct it

When do you need to multiply your starting choice by t (or by t2)? AKA, why you

should always solve for the complementary solution first.

How to spot the problem?

Ex. 3.10.5 Consider the equation y″ + y′ − 12y = g(t)

What is its complementary solution?

(a) Given g(t) = 9e−4t, choose Y =

(b) Given g(t) = 2e3t – 4e

−3t, choose Y =


Ex. 3.10.6 Use the method of undetermined coefficients to solve

y″ + y′ − 12y = 9e−4t.

How to choose Y(t) when g(t) is a product of elementary functions?

What to do when g(t) is a product of any two, or all three, of polynomial/power,

exponential, and sinusoidal functions?

The principles:

(i) If g(t) is a sum of several products, do each part separately. (Review the rule about

handling g(t) being a sum/difference.)

(ii) The starting choice for Y shall be a product of the corresponding starting

choice for each component function in g(t). Every possible term in the resulting

product of the basic choice of functions must be represented.

(iii) Each of the terms in the starting choice shall have its own unique coefficient.

(iv) Lastly, the starting choice must be checked against yc for identical terms. If any

shared term is found, then every term in the starting choice must be multiplied by t.

Repeat until no shared term is found.

Know well how to apply the above principles. Do many exercises!


Ex. 3.10.7 (Exam 1, fall 2007) Determine the most suitable choice of Y(t) for each

equation.

(a) y″ − 4y′ + 8y = 2e2t − 5t

2 + sin 2t

(b) y″ − 4y′ + 8y = − e2tsin 2t + 1

(c) y″ − 4y′ + 8y = t2e−tcos 5t

Ex. 3.10.8 (Exam 2, fall 2001) Solve the initial value problem

y″ + 4y = et, y(0) = 0, y′(0) = 0.


Ex. 3.10.9 Use the method of undetermined coefficients to solve

y″ + 12y′ + 36y = t + 3 − 2e−6t.

Here are a couple exercises to test your familiarity with some of the concepts – how the

different parts are put together to give the solution of a second order linear equation – of

this chapter thus far.

Review Ex 3.1 Given that y = 3e−5t is a known solution of the equation

y″ + 12y′ + 36y = g(t).

What is the general solution of this equation?

Review Ex 3.2 Given that y1 = 2πe−5t + 2e

−4t and y2 = −6e

2t + 2e

−4t are both known

solutions of a certain equation y″ + p(t)y′ + q(t)y = g(t).

(a) What is the general solution of this equation?

(b) What is the equation? (That is, determine a set of functions p(t), q(t), and g(t), such

that y1 and y2 are its solutions.)


11. Mechanical vibrations

Equation at equilibrium: (Use it to find Hooke’s constant.)

Equation of motion:

Parameters for the motion equation: mass, damping constant, Hooke’s constant,

applied forcing function, initial displacement and initial velocity.

Undamped free vibration

Equation becomes:

Solution: u(t) =

The displacement is a simple harmonic motion, oscillating with constant amplitude at the

system’s natural frequency.

Know natural frequency and natural period. (What they are and how to find them.)

The amplitude and phase-angle form of the displacement

u(t) =

Know how to find the amplitude and phase angle.

Do not confuse the amplitude of the simple harmonic motion, which is constant,

with amplitude of an underdamped system or a system undergoing resonance – in

both latter cases the amplitude is a function of time! The above formula does not

apply to those latter cases.


Damped free vibration

With nonzero damping constant γ, and no forcing function, there are 3 possible types of

displacement function, depending on the roots of the characteristic equation.

Equation:

Know the 3 types of damped system, which can be classified according to the roots of the

characteristic equation. Be sure to understand the differences in their behavior.

Overdamped:

Critically damped:

Underdamped:

(1) Which case(s) do not produce oscillation?

(2) How often could the equilibrium position be crossed in each case?

(3) Know how to find quasi frequency and quasi period.

(4) As t → ∞, what happens to the displacement?

(5) The amplitude of an underdamped system is not constant (as in an undamped

system), but decreasing with time. How to find the maximum/peak displacement?


Ex. 3.11.1 Classify the mass-spring system described by each of the equations below as

undamped, underdamped, critically damped, or overdamped.

(a) u″ + 2u′ + 3u = 0

(b) 2u″ + 8u′ + 8u = 0

(c) 3u″ + 300u = 0

(d) 4u″ + 12u′ + 8u = 0

Ex. 3.11.2 (Exam 2, summer 2003) A mass of 1 kg stretches a spring 50 cm. The mass-

spring system has damping of 4 N sec/m. At t = 0 the mass is set in motion from its

equilibrium position with downward velocity of 2 m/sec. Assume that the gravitational

constant,g = 10 m/sec2. (a) Set up an initial value problem that describes this situation.

(b) Solve the initial value problem. (c) What is the quasi-frequency of the system?


Undamped forced vibration

We will only study the case of a periodic forcing function, for example F(t) = F0 cos ωt

or F(t) = F0 sin ωt.

The displacement function behaves differently depending on whether or not the forcing

function’s frequency is equal to the undamped system’s natural frequency.

What is beat?

When ω = ω0, resonance occurs. See next section.


12. Resonance

A system undergoing resonance will oscillate with progressively larger amplitude that

grows unbounded, increasing linearly with time (it does not grow exponentially, contrary

to a popular belief).

What are the precise conditions, mathematically, necessary for resonance to occur?

Ex. 3.12.1 A mass-spring system is described by the equation

u″ + 64u = 2cos(ωt) – 3sin(2ωt)

For what value(s) of ω will resonance occur?

Ex. 3.12.2 Which equation below describes a system undergoing resonance?

(a) u″ + 2u′ + u = sin(t)

(b) u″ − 4u = 4cos(2t)

(c) −u″ − 9u = 15sin(3t)

(d) 4u″ + 36u = −6cos(3t)


MATH 251 Work sheet / Things to know

Chapter 4

1. Higher order linear equations

Linear equations of higher ( ≥ 3) order, with constant coefficients, can be solved in the

same fashion as those of the second order. For homogeneous linear equations with

constant coefficients, the characteristic equation method solves them all.

Know the four rules of the characteristic equation method. For an n-th order equation the

general solution has exactly n terms. The characteristic equation also has exactly n roots

(counting duplicates individually). Each root gives one of the n fundamental solutions,

which together form the general solution, according to the four rules:

i. For distinct real roots:

Ex. 4.1.1 If r = 2, 4, −6, −8 are distinct roots, what are the fundamental solutions?

ii. For repeated real roots:

Ex. 4.1.2 If r = 6, 6, −5, −5, −5 are real roots (counting repetitions), what are the

fundamental solutions?

iii. For distinct complex conjugate roots:

Ex. 4.1.3 If r = −2 ± 2i, 5 ± i, 1 ± 7i are distinct roots, what are the fundamental

solutions?

iv. For repeated complex conjugate roots:


Ex. 4.1.4 If r = 3 ± 5i, 3 ± 5i, 3 ± 5i are complex roots (counting repetitions), what are

the fundamental solutions?

The above 4 rules are cumulative and can be applied together.

Ex. 4.1.5 Suppose an 8th order linear equation has a characteristic with the following

roots: r = 3, 3, 3, 5, −4 ± i, −4 ± i. Write down its general solution.

Ex. 4.1.6 Find the general solution of

y(5) − 16y″ = 0.

Question: How many initial conditions must an n-th order linear equation IVP have?

For nonhomogeneous linear equations of constant coefficients, the approach remains the

same: use the characteristic equation to find yc, then the method of undetermined

coefficients can be used to find the particular solution Y. The general solution is their

sum, y = yc + Y.

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