Introduction to First Order Equations - Sections 2.1-2...Introduction to First Order Equations...

A B I L E N E C H R I S T I A N U N I V E R S I T Y

Department of Mathematics

Introduction to First Order EquationsSections 2.1-2.3

Dr. John EhrkeDepartment of Mathematics

Fall 2012

A B I L E N E C H R I S T I A N U N I V E R S I T Y D E P A R T M E N T O F M A T H E M A T I C S

Course GoalsThe study of differential equations has three principal goals:

1 To discover the underlying differential equation that describes aspecific physical situation.

2 To find, either exactly or approximately the appropriate solution of thatequation.

3 To interpret the solution that is found in the context of the model beingstudied.

There are three different techniques that we will employ this semester toaddress these goals:

Analytic Solution techniques like separation of variables, integratingfactor method, and variation of parameters.

Geometric Slope fields, phase portraits, integral curves, . . .Numerical Euler’s method, Runge-Kutta (approximation schemes)

/24 — Dr. John Ehrke — Lecture 1 — Fall 2012

Geometric Slope fields, phase portraits, integral curves, . . .

Numerical Euler’s method, Runge-Kutta (approximation schemes)

Motion of a Falling Body

Example (Physical Description of the Problem)An object falls through the air toward the Earth. Assuming that the onlyforces acting on the object are gravity and air resistance, determine thevelocity of the object as a function of time.

Solution: The physical law governing the motion of objects is Newton’ssecond law which states the net forces exerted on an object are equal to theobject’s mass times its acceleration, given by

F = ma (1)

which we can rewrite as using a = dv/dt as

F = m(dv/dt) (2)

At this point, we have the beginnings of a real differential equation, butwhere do we go from here?

Motion of a Falling Body

Example (Physical Description of the Problem)An object falls through the air toward the Earth. Assuming that the onlyforces acting on the object are gravity and air resistance, determine thevelocity of the object as a function of time.

Solution: The physical law governing the motion of objects is Newton’ssecond law which states the net forces exerted on an object are equal to theobject’s mass times its acceleration, given by

F = ma (1)

which we can rewrite as using a = dv/dt as

F = m(dv/dt) (2)

At this point, we have the beginnings of a real differential equation, butwhere do we go from here?

So, we have a basic model

where m represents the mass of the object, dv/dt represents its acceleration,and F represents the total force on the object. To develop a better model, weneed to explore the net forces exerted on the mass during free fall:

• The force due to gravity is expressed by mg where g is the accelerationdue to gravity. It is convenient to define v as positive when it is directeddownward.

• The force exerted by air (i.e. the air resistance or drag) on the body actsin opposition to the force exerted by gravity, so we can reasonablyrepresent this force by −γv where γ is the positive drag constantdependent on the shape of the object and the air density. We use thenegative sign because air resistance opposes the motion of the fallingbody.

Together, these two quantities define F more completely, and so we canupdate our model as follows:

= mg− γv (3)

where m represents the mass of the object, dv/dt represents its acceleration,and F represents the total force on the object. To develop a better model, weneed to explore the net forces exerted on the mass during free fall:• The force due to gravity is expressed by mg where g is the acceleration

due to gravity. It is convenient to define v as positive when it is directeddownward.

= mg− γv (3)

Solving the ModelIn order to solve our model, (3), we will employ a technique calledseparation of variables, and requires us to treat dv and dt as differentials.Rewriting, (3), we obtain

dvmg− γv

Integrating the separated equation, we obtain∫dv

mg− γv=

∫dtm

=⇒ −1γ

ln | mg− γv |= tm

+ c (5)

Solving for v, we obtain

v =mgγ− Aγ

e−γ(t/m) (6)

where A = e−γc has the same sign as (mg− γv).

In the context of the problem, what would you call the constant mg/γ?

dvmg− γv

mg− γv=

∫dtm

=⇒ −1γ

+ c (5)

v =mgγ− Aγ

e−γ(t/m) (6)

dvmg− γv

mg− γv=

∫dtm

=⇒ −1γ

+ c (5)

v =mgγ− Aγ

e−γ(t/m) (6)

dvmg− γv

mg− γv=

∫dtm

=⇒ −1γ

+ c (5)

v =mgγ− Aγ

e−γ(t/m) (6)

TerminologyLooking at the falling body problem, if this were a specific case, we wouldbe given the values of m, g, and γ. The determination of the integrationconstant c and therefore A requires we also know the initial velocity, v0 ofthe object. We can now formally define the problem we’ve been working onsuitable for use in a differential equations class as

= mg− γv, v(0) = v0 (7)

Initial Value Problem (IVP) The equation, (7) interprets the physicalproblem as a differential equation dependent on an initialcondition. Such problems are called initial value problems orIVPs.

Boundary Value Problem (BVP) If instead of specifying the initial velocitywe specified both a starting velocity and ending velocity onsome discrete time interval, we would obtain a boundaryvalue problem or BVP.

General Solution The solution in (6) is called the general solution of thedifferential equation since it contains all possible solutionsrepresenting an infinite family of curves.

Particular Solution If we solve the IVP, then we obtain a specificrepresentative of this infinite family called a particularsolution. The particular solution depends on the nature of theinitial condition, v0.

Equilibrium Solution The particular solution corresponding tov(0) = mg/γ is a solution that does not change over time. Inthe physical context this represents the terminal velocity ofthe falling body. We call such solutions equilibrium solutionsbecause regardless of the values of v0, the velocity v(t)approaches this solution as t→∞.

Parameters The values of m, γ are chosen based on the specifics of thephysical problem and would be known prior to solving theequation. Such values are called parameters. Parameters aredifferent from the independent variable t, and dependentvariable v, and the constant g, since its value is determined apriori for any problem chosen.

Falling Body Problem Solutions

Figure: *

Graph of v(t) for nine different initial velocities v0. (g = 9.8m/sec2, m/gamma = 5 sec)

First Order Differential EquationsThe first few weeks of class will deal primarily with differential equations offirst order

= f (t, x) (8)

where the unknown function, f (t, x) is related to the first derivative of thedependent variable in some way. Even though first order differentialequations represent the simplest class of differential equations to solve, theyvary wildly in not only the techniques required to solve them, but thedifficulty they pose.

Rank the following equations in terms of difficulty.

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

Hardest, impossible to obtain an analytic solution, mustbe handled numerically.

Linear equation, we will handle this soon. Not verydifficult.

Easiest, separable equation, really more of a Calculus IIproblem.

= f (t, x) (8)

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

= f (t, x) (8)

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

= f (t, x) (8)

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

= f (t, x) (8)

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

= f (t, x) (8)

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

= f (t, x) (8)

(a) y′(x) = x− y2

(b) y′(x) = y− x2

(c) y′(x) = −x/y

Creation and Annihilation OperatorsConsider the two first order equations(

y = 0 (9)

dx− x)

y = 0 (10)

The equations (9) and (10) are called the annihilation and creation operators,respectively. These operators are useful in the study of quantum harmonicoscillators. In particular, the annihilation operator decreases the number ofparticles in a given state by one, hence its name. You will explore thecreation operator in your first homework set.

ExampleObtain an analytic solution of the annihilator equation, (3), and completelydescribe the family of solutions.

Solution(d

dx+ x)

=⇒ dydx

= −xy

=⇒ dyy

= −x dx

=⇒∫

∫−x dx

=⇒ ln | y |= −x2

=⇒ y = e−x2/2 + c

=⇒ y = ec · e−x2/2

=⇒ y(x) = Ae−x2/2

Remarks

What values is A allowed to take? Well,clearly A > 0, since A = ec. Technicallyalso A < 0 works, since really we shouldhave written,

| y |= Ae−x2/2 ⇐⇒ y = ±Ae−x2/2.

What about A = 0? A = 0 should beincluded also since this would give thesolution y = 0, which is indeed a solutionto the ODE. (It satisfies, equation (9).)This is sometime referred to as the lostsolution because it is lost in the process ofseparating the variables when we dividedby y. Note that if A = 1/

√2π, then the

solution is the standard normal pdf.

Polling Question #1Solve the separable IVP

y(x) = xdydx− y = 2x2y, y(1) = 1.

A. y(x) = xex

B. y(x) = xe ex2

C. y(x) = x2

D. y(x) = 2x− 1x

Geometry of SolutionsLast time, we looked at an analytic method: separation of variables, and someof the issues that arise in solving differential equations analytically. In thislecture, we will consider the geometric view when solving differentialequations.

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

direction field (also called a slope field)

integral curve

slope of the direction field at (x0, y1)

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

integral curve

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

integral curve

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

integral curve

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

integral curve

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

integral curve

Analytic:

y′(x) = f (x, y)

y(x), solution

y1(x0) = f (x0, y1)

Geometric:

integral curve

Drawing a Slope FieldThe natural question is: how might one draw a direction field for an associatedODE?

Computer Way:

1 Pick (x, y) with somesort of spacingcondition.

2 Calculate f (x, y) at thatpoint.

3 Draw in a line element.

Human Way:

1 Pick a slope c.

2 Set f (x, y) = c, this is called an isoclineor level curve from Calculus III.

3 All points (x, y) that lie on f (x, y) = c,should have line elements with slopec.

Computer Way:

Human Way:

1 Pick a slope c.

Computer Way:

Human Way:

1 Pick a slope c.

Computer Way:

Human Way:

1 Pick a slope c.

Computer Way:

Human Way:

1 Pick a slope c.

Computer Way:

Human Way:

1 Pick a slope c.

Method of IsoclinesWe will work out a few examples to get used to this method, called themethod of isoclines for drawing a direction field.

ExampleDraw the isoclines and a typical solution curve for f (x, y) = −x/y. This wasthe easy problem identified earlier.

Analytic Solution:

dy/dx = −x/y =⇒ y dy = −x dx =⇒∫

y dy = −∫

=⇒ y2/2 = −x2/2 + c

=⇒ x2 + y2 = 2c

=⇒ x2 + y2 = r2 (let 2c = r2)

So the solutions are circles of arbitrary radius.

Method of IsoclinesWe will work out a few examples to get used to this method, called themethod of isoclines for drawing a direction field.

ExampleDraw the isoclines and a typical solution curve for f (x, y) = −x/y. This wasthe easy problem identified earlier.

Analytic Solution:

dy/dx = −x/y =⇒ y dy = −x dx =⇒∫

y dy = −∫

=⇒ y2/2 = −x2/2 + c

=⇒ x2 + y2 = 2c

=⇒ x2 + y2 = r2 (let 2c = r2)

So the solutions are circles of arbitrary radius.Slide 15/24 — Dr. John Ehrke — Lecture 1 — Fall 2012

Isocline Diagram

First Order Linear EquationsIn this lecture, we will consider first order linear ODEs of the form

a(x)y′ + b(x)y = c(x) (11)

The standard form for a first order linear ODE is obtained by dividingthrough by a(x) to create new coefficients p(x) and q(x). Equation (11)becomes

y′ + p(x)y = q(x) (12)

Remarks

• The equation (11) can always be solved!

• If q(x) = 0, (11) is called homogeneous.

• If q(x) 6= 0, (11) is called inhomogeneous.

• The equation (11) arises in a variety of physical situations, i.e. theprocess of conduction and diffusion.

a(x)y′ + b(x)y = c(x) (11)

y′ + p(x)y = q(x) (12)

Remarks

a(x)y′ + b(x)y = c(x) (11)

y′ + p(x)y = q(x) (12)

Remarks

a(x)y′ + b(x)y = c(x) (11)

y′ + p(x)y = q(x) (12)

Remarks

a(x)y′ + b(x)y = c(x) (11)

y′ + p(x)y = q(x) (12)

Remarks

Conduction Diffusion ModelThe conduction diffusion model is based on a physical law known asNewtons law of cooling which states that the rate of change of thetemperature of an object placed into a medium (air, liquid, gel, etc...) isdirectly proportional to the difference in temperature between that objectand its environment.

• In this model, we let T(t) be thetemperature of the object at time tplaced into a medium havingtemperature given by Te(t).

• If T > Te, then dTdt < 0 and so T(t) is a

decreasing function and describeshow the body cools.

• Conversely, if T < Te, then dTdt > 0

and so T(t) is an increasing functionand describes how the body heats.

Newton’s Law of CoolingNewton’s law of cooling states that the rate of change in the temperatureT(t) of a body is proportional to the difference between the temperature ofthe object’s environment, Te and the temperature of the body. That is,

= k (Te(t)− T(t)) (13)

where k is the conductivity constant specific to the medium in which theobject is placed.

1 The units for k are inverse time. Why?

2 The value of k is positive. Why?3 The conduction-diffusion model, (13), is a classic example of a first

order linear model. Written in standard form, it is given by

T′(t) + k · T(t) = k · Te(t) (14)

The big question at this point, is how do we solve such equations? Theanswer involves a technique called the integrating factor method.

= k (Te(t)− T(t)) (13)

1 The units for k are inverse time. Why?2 The value of k is positive. Why?

3 The conduction-diffusion model, (13), is a classic example of a firstorder linear model. Written in standard form, it is given by

T′(t) + k · T(t) = k · Te(t) (14)

= k (Te(t)− T(t)) (13)

1 The units for k are inverse time. Why?2 The value of k is positive. Why?3 The conduction-diffusion model, (13), is a classic example of a first

T′(t) + k · T(t) = k · Te(t) (14)

= k (Te(t)− T(t)) (13)

1 The units for k are inverse time. Why?2 The value of k is positive. Why?3 The conduction-diffusion model, (13), is a classic example of a first

T′(t) + k · T(t) = k · Te(t) (14)

The Integrating Factor MethodOur goal is to solve the general linear first order equation

y′ + p(x)y = q(x). (15)

We seek a function u(x) to multiply through (15) simplifying the left hand side into asingle expression. What conditions, should such a u(x) satisfy?

Solution:Multiply both sides of (15) by u(x) and we obtain,

u(x)y′ + u(x)p(x)y = u(x)q(x).

Using the product rule in reverse, we would like to combine the left hand side as

(u(x) · y)′ = u(x)q(x)

but in order to do so, u(x) must satisfy the separable differential equation

u′(x) = u(x) · p(x).

Solving this equation yields the general form of the integrating factor

u(x) = exp[∫

p(x) dx]

The Integrating Factor MethodOur goal is to solve the general linear first order equation

y′ + p(x)y = q(x). (15)

We seek a function u(x) to multiply through (15) simplifying the left hand side into asingle expression. What conditions, should such a u(x) satisfy?

Solution:Multiply both sides of (15) by u(x) and we obtain,

u(x)y′ + u(x)p(x)y = u(x)q(x).

Using the product rule in reverse, we would like to combine the left hand side as

(u(x) · y)′ = u(x)q(x)

but in order to do so, u(x) must satisfy the separable differential equation

u′(x) = u(x) · p(x).

Solving this equation yields the general form of the integrating factor

u(x) = exp[∫

p(x) dx]

Solution of the Conduction-Diffusion Model

ExampleObtain a particular solution for the conduction-diffusion model,

+ k · T(t) = k · Te(t) (17)

where T(t) and Te(t) are both functions of time t, and k > 0 is the conductivityconstant. Assume T(0) = T0. Consider what happens as t→∞.

Solution: Our integrating factor is given by

u(x) = exp[∫

k dt]= exp(kt) = ekt.

Multiplying through by u(x) we obtain,

ektT(t) + kektT(t) = kTe(t)ekt.

Simplifying the left hand side of the above equation via the product rule yields,(ektT(t)

)′= kTe(t)ekt.

Solution of the Conduction-Diffusion Model

ExampleObtain a particular solution for the conduction-diffusion model,

+ k · T(t) = k · Te(t) (17)

where T(t) and Te(t) are both functions of time t, and k > 0 is the conductivityconstant. Assume T(0) = T0. Consider what happens as t→∞.

Solution: Our integrating factor is given by

u(x) = exp[∫

k dt]= exp(kt) = ekt.

Multiplying through by u(x) we obtain,

ektT(t) + kektT(t) = kTe(t)ekt.

Simplifying the left hand side of the above equation via the product rule yields,(ektT(t)

)′= kTe(t)ekt.

Transient and Steady-State SolutionsContinuing our solution from the previous slide, we integrate both sides of theequation to obtain,

ektT(t) =∫ t

0keksTe(s) ds + c.

Note that we include the constant of integration so we might satisfy the initial valueT(0) = T0 explicitly. Exponentiating both sides gives

T(t) = e−kt∫ t

0keks · Te(s) ds︸︷︷︸

steady-state solution

+ ce−kt︸︷︷︸transient solution

1 The value of c is clearly T0 when t = 0, so our full solution is given by

T(t) = e−kt∫ t

0keks · Te(s) ds + T0e−kt.

2 The steady-state solution is the part of the solution which is a response to thesystem being forced externally. The transient solution (which eventually goesaway) is the baseline behavior of the system as it is the solution of the equationin the case when Te(t) = 0.

Polling Question #1Solve the first order linear equation

x′(t) +3x

90 + tx = 8.

A. c(90 + t)3 + 2t

B. 2(90 + t)4 + c

C. 180 + 2t + c(90 + t)−3

D. 720t + 2t4 + c

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