first order linear equations and integrating factors

7/29/2019 First Order Linear Equations and Integrating Factors

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Copyright Rene Barrientos Page 1

MAP2302 Lecture 4 2010-3

LINEAR DIFFERENTIAL EQUATIONS

Integrating Factors

A first order equation that can be written in the form

where , ,and are arbitrary functions of x is called a first order linear equation and equation (1) isits general form . For example, the equation / is a first order linear equation because it can be written as

0 which has the form of equation ( 1) with , , and . Let us first consider the simplest possible case: 0, 0on some interval I. In this(1) isseparable and reduces to the familiar :

which we can easily be solved by integration:

where is an arbitrary constant. Needless to say, these are the rare cases for in general 0. We will nowshow that under fairly general assumptions equation ( 1) can be solved by introducing an integrating factor . Ourfirst step is to write the equation in its standard form , which is obtained by dividing its terms by the leadingcoefficient :

where / and /. From this point forward, we will always assume a linearequation has been expressed in its standard form.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 where is an arbitrary constant. Any function which allows us to do this is called an integrating factor of equation ( 2) and our aim now is to obtain it.So let us multiply equation (2) by the yet undetermined factor :

In order for an integration 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,


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Cancelling , Cancelling y and separating variables,; 0

Solving for ln|| This gives us a family of functions that have the desired property, but we only need one. Setting 1anddropping the :When is a 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 obvious and easy to makemistakes with. It is better to remember the how to obtain the integrating factor and go through the steps.

Let us first compute some integrating factors:

Differential Equation Standard Form p (x)

2 1 2 1 2 / 2 2 2 3 3 3 3 /

To solve

(1) multip ly the differential equation by

(2) Then by the way was defined we have

(3) Integrate:

(4) Solving for y :


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

Caution: if other variables are used besides the usual x and y, make sure that you use them. Do not bring into the problemvariables that were not there to begin with.

I 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) 1so an integrating factor is . 3) multiplying by the integrating factor:

2 or2

4) Integrating,

2 Integration by parts tells us that . Thus,

2 Solving for y :

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


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Here are some of the members of the family of solutions:

Compare this graph with the slope field above. The separatrix , shown as a dotted red line, is a solution of the equation; It corresponds to 0.

Example 3 Solve 2 1 Solution

1) Write the equation in standard form:

2 1 The coefficient of y is :

2) Find the integrating factor

/ | | 3) Multiplying the equation is step 1) by 2 1 multiplying

simplifying rewiting

- 2 - 1 0 1 2

- 2

- 1

0

1

2

- 3 - 2 - 1 1 2 3

- 3

- 2

- 1

1

2

3

2 Field corresponding to

2 2


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[ verify that 2]4) We are ready to integrate:

2 Solving for y: Done!

A quick substitution into the original equation tells us that 2 1 Once again, the hardest thing here is to evaluate the last integral. There will be times when this is verydifficult and we might have to use a computer, tables, or even leave the answer in terms of an integral.

Example 4 Solve 2 Solution

This equation is already in standard form, and we found that is an integrating factor.Multiplying the equation by this factor will make the left hand side equal to the derivative of theintegrating factor times the dependent variable:

2 Integrating:

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

12 2

let and

Then

Thus, 12 2 12 2 12 Finally we have our solution: 12 Solving for y

Example 5 Solve 1 ,0 0 SolutionFirst write the equation in standard form:

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


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

Thus,

exp2 exp2 1 Integrating,exp2 exp2 1

Let . Then so the integral is straightforward:

exp2 exp orexp2 exp2 Solving for y :

1 exp2

Applying the initial condition 0 0gives us 1. Therefore,

Example 6 Solve 2 4 2 Solution

First we put this in standard form:

2 14 2 An integrating factor is

Multiplying by this integrating factor:

2 1 4 2 Thus,

4 2 Integrating,

4 2

2 2 1

Evaluating the integral1 2Solving for P ,

1 To evaluate 2 1, let . Then 2 1and the integral is of the form


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Example 7 Solve the IVP 4 8 50exp 10 ; 0 40 Solution

The equation in standard form is:

2252exp 10 ; 0 40

Note: as stated earlier, exp10 .An integrating factor of the differential equation is

Applying the integrating factor,

2252 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 40to obtain 665/16. Thus,2516 66516 Solving for :

Method II: (Use the definite integral ) 252 Integrating,

0 2516 1 Since 0 40, 2516 66516 Hence Example 8 Solve the IVP 4 2 0; 1 5

SolutionSince this equation is given in differential form, we must first decide which variable represents theindependent variable and which one is the dependent. The presence of t 2tells us immediately that t cannot be the dependent variable if we are interested in viewing this as a linear equation. Thus, wechoose t to be the independent variable. Then by default z is the dependent variable and the equation canbe written as


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4 2 0; 1 5 Thus, the general form of the equation is

4 2; 1 5 You 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 5,1in the t - z plane.With this in mind we proceed to put the equation in standard form :

14 2; 5 1So that We are ready to find the integrating factor:

| |/ Note 1 : 0is a singularity of this equation since 0 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.

/ /14 /2; 5 1 This simplifies to/ 14 /12 /; 5 1 Hence,

/ 12 /; 5 1 Integrating,/ 12 / ; 5 1/ 1249 / ; 5 129 /; 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 1

integrate from the point 5,1on the t-z plane to some arbitrary point ,:


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12 / 551249 /

529 / 5/

29 /295 / 29 / 5419

Example 9 Find a solution of the equation 1/ Solution

If we take to be the independent variable, then this equation is not linear. However, there is nothingspecial about the variable y and 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 is x. Writing it in standard form,

An integrating factor is . Thus,

or

Integrating,

Integration by parts gives us

Solving for ,

Example 10 (an application) A large tank contains a 15-liter mixture in which 10 gramsof salt are dissolved.Salt water of concentration 35 gm/Lenters the tank at a rate of 1L/min and is quickly mixed. The mixture leavesthe tank at rate of 2 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 3 min. 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.


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We can summarize the given information in the following figure:

The rate at which salt enters the mixture of leave it is given by the general formula

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

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

135

Observe that we obtain the rate at which salt goes in by multiplying the rate at which solution goes in timesthe concentration of that solution . Similarly,

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

The concentration of the solution coming out of the tank is changing in time and is given by

Thus, if we know the volume of solution in the tank present at any time t ,we can obtain anexpression for the concentration of the mixture coming out. We can write a formula for the volume as afunction of time t : 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/minand 2 L/min. Therefore the mixtures volume ischanging at a rate of 1 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

15 Hence, the differential equation governing is

1 35 2 15

Salt water of concentration 35 gm/Lgoes in at a rateof 1

L/min

Mixture comes out at a rate of 2 L/min

Initially : 15 L, 10 gm. of salt


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Since the initial amount of salt in the mixture is 10 gm, we have the following initial value problem:35 215; 0 10 We are ready to calculate:

(a) from the volume equation, the tank will be empty when

0,that is, when

(b) To answer the second question, we solve the initial value problem.Writing it in standard form :

215 35 ; 0 10 An integrating factor for this equation is 15 Multiplying ( a ) by the I.F.

15 215 15 35 15 Thus, 15 35 15 Integrating,

15 35 1515 150 35 15

or

15 10225 3515 115 2453511511515 351510345 Solving ,

At 3, 3 90.4 gmand 3 12 L. the salinity is approximately 7.53 gm/L.

first order linear equations and integrating factors

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