First Order Linear Equations and Integrating Factors (1)

7/29/2019 First Order Linear Equations and Integrating Factors (1)

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Copyright Rene Barrientos Page 1

MAP2302 Lecture 4 2011-3

FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS

The equation , is a first order linear differential equation if it can be written in the form where, ,and are arbitrary functions ofx. We call equation (1) its general formand division by in an interval where it is possible gives us itsstandard form For example, the equation /is a first order linear equation because it can be written as 0which has the form of equation (1) with , , and . Its standard form is

1 0; 0Observe that if 0 and either 0 or 0 on some interval I, then (1) is separable and reducesto the familiar equations if 01 if 0Needless to say, these are the rare cases. We will now show that under fairly general assumptions equation (1) canbe solved by introducing anintegrating factor.

Integrating Factors

An integrating factor is one that allows us to integrate (1). Our first step is to write the equation in its standardform which we achieve by dividing its terms by the leading coefficient

on an interval where

0.

Thus, henceforth we will assume that this has been done and (1) has the form where / and /.Now imagine that we are able to find a function such that equation (2) can rewritten as

Then integration leads to a solution: or

1 Any function which allows us to do this is called an integrating factor of equation (2) and our aim now is toobtain it.

Let us multiply equation (2) by the yet undetermined factor:


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In order for this factor to have the desired effect, it must convert the expression into . In other words, the function must be such that Using the product rule on the right-hand side,

Cancelling and, Separating variables ; 0Solving for ln||

This gives us a family of functions that have the desired property, however, we only need one. Setting 1 anddropping the:

When isa complicated expression we use the following alternate notation instead:

where.The procedure used to solve linear equations using integrating factors can be summarized as follows:

In practice it is not a good idea to memorize the formula in step (4) because it is not an obvious one. It is better toremember how to obtain the integrating factor and go through the steps.

To solve

(1) multip ly the differential equation by (2) Then by the way was defined we have(3) Integrate:

(4) Solving fory:


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Let us first compute some integrating factors:

Differential Equation Standard Form p 2 1 2 1 2 /

2

2

2

3 3 3 3 / /cot s e c 2 t a n sec2tan tan 1sec

Remarks:1) It is very important to remember that the differential equation must be standard form.2) by construction, the integrating factor converts the left-hand side of equation (2) into

Initial value problems can be solved via the definite integral instead:

Thus,

or

Example 2 Solve 2 Solution

1) Write the differential equation in standard form, 2 2) 1 so an integrating factor is .3) Multiplying by the integrating factor: 2

Equivalently,

2 4) Integrating (integration by parts is needed),

2 2

Solve fory:


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The figure below corresponds to the slope field of this differential equation.

Here are some members of the family of solutions:

Compare this graph with the slope field above. Theseparatrix , shown as a dotted red line, is a solution ofthe equation corresponding to 0.Example 3 Solve 2 1Solution

1) Write the equation in standard form: 2 1The coefficient ofy is

: 2) Find the integrating factor

/

||

3) Multiply the equation is step 1) by 2 1 multiplying by simplifying rewiting[ verify that

2 ]

-2 -1 0 1 2

-2

-1

0

1

2

-3 -2 -1 1 2 3

-3

-2

-1

1

2

2

Field corresponding to

2 2


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4) integrate:

2 Solve fory:

Done! A quick substitution into the original equation verifies that this function is a solution:

2 1 Example 4 Solve 2 Solution

This equation is already in standard form, and is an integrating factor. Multiplying the equationby this factor:

2 which is equivalent to

Integrating:

To evaluate this integral, we can use a table or use integration by parts. Using the latter,

12 2l e t and

Then Thus,

12

2 12

2

12 Finally we have our solution: 12 Solving fory

Example 5 Solve 1 , 0 0Solution

First write the equation in standard form:

1 1, 0 0An integrating factor of this equations is

It is more convenient to write because of the complexity of the argument .


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SolutionThe equation in standard form is: 2 252 exp 10; 0 40Note: as stated earlier,

exp10 .

An integrating factor of the differential equation is Applying the integrating factor, 2 252

Thus, 252 We are ready to integrate, but we have a choice; this is an initial value problem, therefore, we eitherintegrate and then find the arbitrary constant, or we use a definite integral to dispense with that extra step.

Method I: (Use the indefinite integral and then apply the initial conditions)

252 2516 Apply 0 40 to obtain 665/16. Thus,

2516 66516 Solving for : Method II: (Use the definite integral)

252

Integrating, 0 2516 1Since0 40, 2516 66516 Hence

Example 8 Solve the IVP 4 2 0 ; 1 5SolutionSince this equation is given in differential form, we must first decide which variable represents theindependent variable and which one is the dependent. The presence oft2 tells us immediately that tcannot be the dependent variable if we are interested in viewing this as a linear equation. Thus, wechoosetto be the independent variable. Then by defaultzis the dependent variable and the equation canbe written as 4 2 0 ; 1 5


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Thus, the general form of the equation is

4 2 ; 1 5You might object to the choice of initial condition; does it not look backwards? But all this conditionsays is that the integral curve that we are interested in is the one that has the property that

when

, that is, the solution curve contains the point 5,1 in thet-

zplane.With this in mind we proceed to put the equation instandard form: 14 2 ; 5 1So that and an integrating factor is ||/Note 1: 0 is asingularity of this equation since0 0. Thus, we need avoid intervals containingthe origin.

Note 2: since we are working with initial conditions specified at 5, we can assume we are working onan interval in which

0, say (0,), and dispense with the absolute value. Thus, we can take

/ is our integrating factor. Then,/ / 14 / 2 ; 5 1Simplifying / 14 / 12 /; 5 1Hence, / 12 /; 5 1Integrating, / 1

2/ ; 5 1

/ 12 49 / ; 5 1 29 /; 1 5Applying the initial condition, 1 5 5/ 5/ Substituting for c: 29 5/ 419 /

Using the definite integral reduces the amount of work by a few steps. Starting with / 12 /; 5 1integrate from the point5,1 on the t-z plane to some arbitrary point,:

12 /


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55 12 49 / 5 29 / 5/

29 /

29 5/

29 / 5 419

Example 9 Find a solution of the equation 1 /

SolutionIf we take to be the independent variable, then this equation is not linear. However, there is nothingspecial about the variableyand in fact we learn in calculus that

/

provided that/0. Thus, the differential equation under consideration may be written as which is a linear differential equation whose dependent variable isx. Writing it in standard form, An integrating factor is . Thus,

or

Integrating, Integration by parts gives us Solving for ,

Applications

Newtons Second Law

Newtons Second Law allows us to model motion if all the forces acting on a particle or system of particles areknow. It states: where the momentum of the center of mass of the system and the vector sum of all external forces actingon the system. Momentum is defined by


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where is the mass of the particle or total mass of the system and the velocity of its center of mass. In mostapplications is constant so Newtons Second Law becomes

Furthermore, if we are dealing with a single particle we usually use

instead of

.

Example 10 A 2-Kg object is projected upwards from ground level with a initial velocity of 10 (m/sec) ina medium which produces a resistive force (drag) given by1.5. Find (a) the velocity of the object at time, and (b) the time it takes to get to the top of its flight and its maximum altitude.Solution

We refer to the following figure:

At any time, Newtons Second Law tells us that

Let . Substituting, 2 2 1.5Using 10 / as an approximation and equating components,

2 201.5Hence, the differential equation that governs the velocity function of this object is

1 0 34 ; 0 10This is a first order linear equation and it is also separable, but we will solve it using an integratingfactor. Writing it in standard form, 34 1 0 ; 0 10

This equation has the integrating factor /. Hence, /1 0 /Integrating from

0to some later time

,

/ 1 0 / Hence, / 0 1 0 43 / 1Solving for , 103 7/ 4

1.5 2

View when the object ismoving up


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(a) The velocity vector is given by 103 7/ 4 (b) The object attains its highest point on its trajectory when , that is, when10

3 7/

4 0

Solving for , 43 ln4/7 0.75 secTo find how high it goes, we must find its position function:

0 103 7/ 4 10

3 7 / 4

103 283 283 4

The maximum altitude is reached when 0.75 sec,0.75 103 283 283 . 3 3.38 m

The mass reaches a maximum altitude of approximately3.38 meters above ground.Mixture Problems

Mixture problems are also called compartmental analysis problems and involve, as the name implies, mixtures.

Example 11 (an application) A large tank contains a15-liter mixture in which 10 grams of salt are dissolved.Salt water of concentration35 gm/L enters the tank at a rate of1 L/min and is quickly mixed. The mixture leavesthe tank at rate of2 L/min. (a) How long does it take for the tank to be completely empty? (b) Find the salinity(measured by the amount of salt per liter of mixture) of the tank at 3min.Solution

First we need to derive a mathematical model of how the amount of salt (or whatever chemical is beingconsidered) changes in the mixture. Let be the amount of salt in the mixture at time and let bethe volume of this mixture.

We can summarize the given information in the following figure:

Salt water of concentration

35 gm/Lgoes in at a rate

of

1

L/min

Mixture comes out atarate of2 L/min

Initially: 15 L,10 gm. of salt


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The rate at which salt enters or leaves the mixture is given by

rate of salt entering or leaving rate of mixture entering or leaving the tank the concentration ofthemixture.

If we assume that the mixture is well stirred at all times so that there are no significant variations inconcentration, it is reasonable to assume that the rate at which the amount of salt changes is given by the

conservation equation2

Similarly,

rate of salt out = rate of mixture coming out times the concentration of the mixture

The problem here is that we do not know the concentration of the mixture leaving the tank because in

general it is changing all the time. What we do know is that it is given by the quotient of the amount ofsalt per volume of mixture:

Thus, if we know the volumeof solution in the tank present at any timet, we can obtain an expressionfor the concentration of the mixture coming out. We can write a formula for this volume as a function oftimet:

where is the initial volume of the mixture in the tank, and are the rates at which mixtureenters and exist the tank, respectively.

In this example, 15 L, 1 L/min and 2 L/min. Therefore the mixtures volume ischanging at a rate of1 L/min, that is, it is decreasing by one liter every minute: The concentration in the tank, and also that of the out-flowing mixture, is given by

1 5 Hence, the differential equation governing is

1 35

2

1 5

Since the initial amount of salt in the mixture is10 gm, we have the following initial value problem: ;

We are ready to calculate:

(a) from the volume equation, the tank will be empty when 0, that is, when .2 It is called a conservation equation because is presumes that what goes i n must come out.

concentrationrate


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(b) To answer the second question, we solve the initial value problem.

Writing it instandard form: 21 5 35 ; 0 10 An integrating factor for this equation is

1 5 Multiplying (a) by the I.F.

15 215 3515Thus, 1 5 3515Integrating,

1 5 3 5 1 5 15 150 3 5 1 5 or

15 10225 3 5 1 5 115 245 3 5 11 5 115

Solving,

At 3, 3 90.4gm and3 12L. the salinity is approximately 7.53gm/L.Example 12 (three fluxes) A large tank containing210 liters of pure water receives a chemical mixture containinga solvent of concentration20% by volume at a rate of five L/min. The mixture is well stirred and delivered to twotanks, tank A and tank B, at rates of3 L/min and4 L/min, respectively . Determine the amount of solvent in thelarge tank when its volume is 100 L.

SolutionLet be the amount of solvent (in liters) in the large tank at time and let be the its volume.We can summarize the given information in the following figure:

20%solvent mixture is

delivered at a rate of

5

L/min

Initially: 210 L,pure water so

0 0

mixture is delivered intotank A at a rate of3L/min mixture is delivered intotank B at a rate of4L/min

Tank A Tank B


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The volume in the original tank is given by 2 1 0 7

5 0.20 3 2107 4 2107 Hence, 5 0.20 72107 ; 0 0Simplifying and rewriting the equation in standard form, 13 0 1 ; 0 0

Using 3 0 as an integrating factor,

3 0 3 0

Integrating,

3 0 3 0 00 13 0 Since0 0, solving for:

3 0 13 0 When the volume of the large tank is 100L, 2 1 0 7 1 0 0L. Hence, 15.71 min. At thistime,

15.71 3015.71 13 0 . Evaluating the definite integral and performing the indicated operations results in. . L.

Exercise 1 The graph of is shown below. Use it to guide you in establishinglim .

Exercise 2 It is clear from the graph that attains a maximum value sometime in the neighborhood of 3 0.Find the exact time and exact value of this maximum.

Example 13 (Environmental Science) A lake with an average capacity of one trillion gallons receives 100 milliongallons of water from tributaries and rains every year and supplies an equal amount to local communities also on ayearly basis. A manufacturing company plans to operate a plant nearby and wants to utilize water from the lake

5 10 15 20 2 30

2

4

6

8

10


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for its operation. As part of the agreement with the local water authority, it proposes to use one million gallonsevery year and return the same amount of water to the lake after it has been treated for contaminants.

A quality control test of a20-gallon sample of the returned water during the first month of operation shows that itcontains 0.05% of certain contaminant by volume. The manufacturing company assures the local authority that itwill take the necessary steps to ensure that the pollutant levels in the returned water never exceed this value andpromises to monitor it every three months. Assuming that the company abided by its promise, determine the

levels of contaminant (in gallons) in the lake ten years after the plant begins its operation.

SolutionThere are some assumptions that are implicit in this problem. First, such a large lake cannot be quicklystirred so a more accurate picture would be presented by adelay differential equation, something thatwill be studied in more advanced applied math course. We will assume that the lake attains ahomogeneous state within a year period. Secondly, when the water returns to the plant, it does so withincreasing amount of the pollutant. We assume that the filters do what is needed in order to ensure thatthe promised value is never exceeded.

We summarize this information as follows:

Let represent the amount of pollutant in the lake at time in gallons. Then, the tanks profile lookslike this:

In Out

Rain and tributaries 100 million gal/yr 0Plant 1 million gal/yr 1 million gal/yrCommunity 0 100 million gal/yrAmount of pol lut ant 500 gal/yr 1 1 0 10 10010 10

The conservation equation tells us that

5 0 0 110

6 10010

6 ; 0 0or 101106 500 ; 0 0

This equation has an integrating factor given by /106 from which it follows that /1065 0 0 /106Integrating over the interval 0,10,

pure water enter the lake ata rate of

100million gal. per

year

100 million gallons/year of lake water usedby local communities

Initially: 1 trillion galand0 gal. ofcontaminant

1 million gallons/year of lake water used bythe plant

1 million gallons of 0.05%contaminated water returned tothe lake per year


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/106 500 /106

This gives us

/10510 0 50010101 /105 1Solving for

10,

10 50010101 1 /105. In a one-trillion gallon lake, this constitutes a very small quantity indeed (about 5 parts per billion).However, it is interesting to observe, however, that the pollution function is given by

/andits graph looks like this:

Observe that the limiting value of this function is ,,. gal. The good news is that it will nothappen for at least25,000 years!Obviously the lake will far outlive our fictitious firm. However, there are geological processes for which25,000years are like a blink of an eye and extreme changes can take place within that time span as a consequence ofchemical processes that appear insignificant in the short term.

5000 10000 15000 20000 25000

1106

2106

3106

4106

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First Order Linear Equations and Integrating Factors (1)