differential equation ( final product )

7/30/2019 Differential Equation ( FINAL PRODUCT )

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De La Salle University Dasmarias

College of Engineering, Architecture, and Technology

MEET316

S.Y. 2012-2013, 1st Semester

P R O B L E M S E T S I N D I F F E R E N T I A L E Q U A T I O N S W I T H

E N G I N E E R I N G A P P L I C A T I O N S

Pantas, Rowell P.

EEE31

ENGR. JOEVEN V. JAGOCOY, PME

Professor

In partial fulfilment to the requirements in MEET316 Differential Equations with

Engineering Applications


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T A B L E O F C O N T E N T S

TOPIC Page No.

Elimination of Arbitrary Constants ( PLATE NO.1 ) 3

Separation of Variables ( PLATE NO. 2 ) 5

Homogenous Differential Equations ( PLATE NO. 3 ) 8

Exact Differential Equations ( PLATE NO. 4 ) 12

Non-Exact Differential Equations ( PLATE NO. 5 ) 16

Escape Velocity ( PLATE NO. 6 ) 21

Newtons Law of Cooling ( PLATE NO. 7 ) 23

Exponential Growth and Decay ( PLATE NO. 8 ) 26

Mixture Problem ( PLATE NO. 9 ) 28

Undetermined Coefficients ( PLATE NO. 10 ) 31

Variation of Parameters ( PLATE NO. 11 ) 34

Inverse Differential Operator ( PLATE NO. 12 ) 39


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P L A T E N O . 1 E L I M I N A T I O N O F A R B I T R A R Y C O N S T A N T S

1.1 PRINCIPLES

The order of differential equation is equal to the number of arbitrary constants in

the given relation.

The differential equation is consistent with the relation.

The differential equation is free from arbitrary constants.

1.2 PROBLEM SET

a.

b.

1.3SOLUTION

a.) equation (1)

Substitute c to equation (1)


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b.) equation (1)

Divide by dx

Substitute c to equation (1)

Multiply by dx


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P L A T E N O . 2 S E P A R A T I O N O F V A R I A B L E S

2.1 PRINCIPLES

Equation (1)

where and may be functions of both and . If the above equation can betransformed into the form

Equation (2)

where is a function of alone and is a function of alone, equation (1) is calledvariables separable.

To find the general solution of equation (1), simply equate the integral of equation (2) toa constant . Thus, the general solution is:

2.2 PROBLEM SET

a. , when , .

b. , when , .

c. , when , .

2.3 SOLUTION


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a.)

,

b.)


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,

c.)

x = -2, y = 1

Thus,


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P L A T E N O . 3 H O M O G E N O U S D I F F E R E N T I A L E Q U A T I O N

3.1 PRINCIPLES

If the function f(x, y) remains unchanged after replacing x by kx and y by ky,

where k is a constant term, then f(x, y) is called a homogeneous function.

A differential equation

Equation (1)

is homogeneous in x and y if M and N are homogeneous functions of the same degreein x and y.

To solve for Equation (1) let

or

3.2 PROBLEM SET

a.

b.


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3.3 SOLUTION

a.)

Let

Substitute,

Divide by x2,


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b.)

Let

Substitute,


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From


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P L A T E N O . 4 E X A C T D I F F E R E N T I A L E Q U A T I O N S

4.1 PRINCIPLES

The differential equation

is an exact equation if

Steps in Solving an Exact Equation

1. Let .

2. Write the equation in Step 1 into the form

and integrate it partially in terms of x holding y as constant.

3. Differentiate partially in terms of y the result in Step 2 holding x as constant.


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4. Equate the result in Step 3 to N and collect similar terms.5. Integrate the result in Step 4 with respect to y, holding x as constant.6. Substitute the result in Step 5 to the result in Step 2 and equate the result to a

constant c.

4.2 PROBLEM SET

a.

b.

4.3 SOLUTION

a.)

Test for exactness

;

;

; thus, exact!

Step 1: Let

Step 2: Integrate partially with respect to x, holding y as constant


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Equation (1)

Step 3: Differentiate Equation (1) partially with respect to y, holding x as constant

Step 4: Equate the result of Step 3 to N and collect similar terms. Let

Step 5: Integrate partially the result in Step 4 with respect to y, holding x as constant

Step 6: Substitute f(y) to Equation (1)

Equate F to c

b.)

Test for exactness


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Exact!

Let

Integrate partially in x, holding y as constant

Equation (1)

Differentiate partially in y, holding x as constant

Let

Integrate partially in y, holding x as constant

Substitute f(y) to Equation (1)

Equate F to c


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P L A T E N O . 5 N O N E X A C T D I F F E R E N T I A L E Q U A T I O N S

2.1 PRINCIPLES

All the techniques we have reviewed so far were not of a general nature since in

each case the equations themselves were of a special form. So, we may ask, what to

do for the general equation

Let us first rewrite the equation into

This equation will be called exact if

,

and nonexact otherwise. The condition of exactness insures the existence of afunction F(x,y) such that


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When the equation (E) is exact, we solve it using the following steps:

1. Check that the equation is indeed exact;

2. Write down the system

3. Integrate either the first equation with respect of the variable x or the second withrespect of the variable y. The choice of the equation to be integrated will depend on howeasy the calculations are. Let us assume that the first equation was chosen, then we get

The function should be there, since in our integration, we assumed that thevariable y is constant.

4. Use the second equation of the system to find the derivative of . Indeed, wehave

,

which implies

Note that is a function of y only. Therefore, in the expression giving thevariable, x, should disappear. Otherwise something went wrong!

5. Integrate to find ;

6. Write down the function F(x,y);

7. All the solutions are given by the implicit equation


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8. If you are given an IVP, plug in the initial condition to find the constant C.

2.2 PROBLEM SET

a.

b.

2.3SOLUTION

a.) is not exact, since

However, note that

is a function of x alone. Therefore, by Case 1,

will be an integrating factor of the differential equation. Multiplying both sides of thegiven equation by = x yields


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which is exact because

Solving this equivalent exact equation by the method described in the previous

section, M is integrated with respect to x,

and N integrated with respect to y:

(with each constant of integration ignored, as usual). These calculations clearly give

b.) is not exact, since

However, note that


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is a function of y alone (Case 2). Denote this function by ( y); since

the given differential equation will have:

as an integrating factor. Multiplying the differential equation through by = (siny)

1 yields

which is exact because:

To solve this exact equation, integrate M with respect to x and integrate N with respectto y, ignoring the constant of integration in each case:

These integrations imply that


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P L A T E N O . 6 E S C A P E V E L O C I T Y

6.1 PRINCIPLES

Escape velocity is defined as the smallest speed that we need to give an object

in order to allow it to completely escape from the gravitational pull of the planet on which

it is sitting. To calculate it we need only realize that as an object moves away from the

center of a planet, its kinetic energy gets converted into gravitational potential energy.

Thus we need only figure out how much gravitational potential energy an object gains

as it moves from the surface of the planet off to infinity. According to the above

discussion for a planet with mass M and radius R, this gain in gravitational potential

energy is GmM/R. For an object to just barely escape to infinity (without any residual

speed), all its initial kinetic energy must go into this increase in gravitational potential

energy. Thus, the initial kinetic energy must be equal to GmM/R. Since kinetic energy

is mv2/2, equating these two expressions tells us that the square of the initial velocity

must be equal to twice the gravitational potential energy divided the inertial mass of the

object. However, since gravitational potential energy is proportional to inertial mass, we

find finally that the square of the escape velocity depends only on the mass and radius

of the planet (and of course Newton's gravitational constant):


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6.2 PROBLEM SET

A. Mars: mass 6.46 x kg; radius 3.39 x m

B. Mercury: mass 3.35 x kg; radius 2.44 x m

6.3 SOLUTION

A.

B.


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P L A T E N O . 7 N E W T O N S L A W O F C O O L I N G

7.1 PRINCIPLES

Newton's Law of Cooling states that the rate of change of the temperature of anobject is proportional to the difference between its own temperature and the ambienttemperature (i.e. the temperature of its surroundings).

Newton's Law makes a statement about an instantaneous rate of change of thetemperature. We will see that when we translate this verbal statement into a differentialequation, we arrive at a differential equation. The solution to this equation will then bea function that tracks the complete record of the temperature over time. Newton's Lawwould enable us to solve the following problem.

7.2 PROBLEM SET

a. We start with a tank containing 50 gallons of salt water with the saltconcentration being 2 lb/gal. Salt water with a salt concentration of 3 lb/gal isthen poured into the top of the tank at the rate of 3 gal/min and salt water is atthe same time drained from the bottom of the tank at the rate of 3 gal/min. Wewill consider the water and salt mixture in the tank to be well-stirred and at alltimes to have a uniform concentration of salt. Find the function S(t) that gives theamount of salt in the tank as a function of time (t) since we began pouring in salt


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water at the top and simultaneously draining salt water from the bottom of thetank. How long before there will be 120 pounds of salt in the tank?

b. A pot of liquid is put on the stove to boil. The temperature of the liquidreaches 170oF and then the pot is taken off the burner and placed on a counter inthe kitchen. The temperature of the air in the kitchen is 76oF. After two minutesthe temperature of the liquid in the pot is 123oF. How long before thetemperature of the liquid in the pot will be 84oF?

7.3 SOLUTIONS

a)


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b.)


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P L A T E N O . 8 E X P O N E N T I A L G R O W T H A N D D E C A Y

8.1 PRINCIPLES

In this set of supplemental notes, I will provide more worked examples of a typeof differential equations that their solutions are exponential functions. These kinds ofproblems can represent the exponential growth or decay of a substance. I will first statethe law of exponential change.

Law of Exponential Change

y = y0ek t

y0 is the initial amount of the substance present at t = 0. k is the rate constant. If k

> 0, then it is a growth constant. If k < 0, the it is a decay constant.

8.2 PROBLEM SET

a. The charcoal from a tree killed in the volcanic eruption that formed that

formed Crater Lake in Oregon contained 44.5% of the carbon-14 found in

living matter. About how old is Crater Lake?

b. A painting attributed to Vermeer (1632 - 1675), which should contain no

more than 96.2% of its original carbon-14, contains 99.5% instead. About

how old is the forgery?

8.3SOLUTION

a. The half-life of carbon-14 is 5700 years. The decay constant, k, is defined tobe


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Let A 0 be the amount of carbon-14 that is found in living matter (t = 0years). We want to determine when there is 44.5% of A 0 is left.

t = 6658.299725 years

b. Using the value for k from example 4, and A = A 0 when t = 0, we will solvethe following equation for t.

t = 41.22 years

The forgery is only 41.22 years old.


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P L A T E N O . 9 M I X T U R E P R O B L E M

9.1 PRINCIPLES

In these problems we will start with a substance that is dissolved in a liquid.

Liquid will be entering and leaving a holding tank. The liquid entering the tank may or

may not contain more of the substance dissolved in it. Liquid leaving the tank will of

course contain the substance dissolved in it. If Q(t) gives the amount of the substance

dissolved in the liquid in the tank at any time t we want to develop a differential equation

that, when solved, will give us an expression for Q(t). Note as well that in many

situations we can think of air as a liquid for the purposes of these kinds of discussions

and so we dont actually need to have an actual liquid, but could instead use air as the

liquid.

The main assumption that well be using here is that the concentration of the

substance in the liquid is uniform throughout the tank. Clearly this will not be the case,but if we allow the concentration to vary depending on the location in the tank theproblem becomes very difficult and will involve partial differential equations, which is notthe focus of this course.The main equation that well be using to model this situation is :

Rate ofchange of

Q(t)=

Rate atwhich Q(t)enters the

tank

Rate atwhich Q(t)exits the

tank

Where,

Rate of change of Q(t) =

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x(concentration of substance in liquid

entering)Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)


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9.2 PROBLEM SET

a. How many liters of 20% alcohol solution should be added to 40 liters of a50% alcohol solution to make a 30% solution?

b. John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a2% alcohol solution with a 7% alcohol solution. What are the quantities of each ofthe two solutions (2% and 7%) he has to use?

9.3SOLUTION

A. Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a

50% alcohol. Let y be the quantity of the final 30% solution. Hence

x + 40 = y

We shall now express mathematically that the quantity of alcohol in x litersplus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in yliters. But remember the alcohol is measured in percentage term.20% x + 50% * 40 = 30% y

Substitute y by x + 40 in the last equation to obtain.

20% x + 50% * 40 = 30% (x + 40)

Change percentages into fractions.

20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100

Mutliply all terms by 100 to simplify.

20 x + 50 * 40 = 30 x + 30 * 40

Solve for x.

x = 80 liters


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80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solutionto make a 30% solution.

B.

Let x and y be the quatities of the 2% and 7% aclohol solutions to be used tomake 100 ml. Hence

x + y = 100

We now write mathematically that the quantity of alcohol in x ml plus the quantityof alcohol in y ml is equal to the quantity of alcohol in 100 ml.

2% x + 7% y = 5% 100

The first equation gives y = 100 - x. Substitute in the last equation to obtain

2% x + 7% (100 - x) = 5% 100

Multiply by 100 and simplify

2 x + 700 - 7 x = 5 * 100

Solve for x

x = 40 ml

Substitute x by 40 in the first equation to find y

y = 100 - x = 60 ml


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P L A T E N O . 1 0 U N D E T E R M I N E D C O E F F I C I E N T S

10.1 PRINCIPLES

The method of undetermined coefficients is a technique for determining the

particular solution to linear constant-coefficient differential equations

for certain types of nonhomogeneous terms f(t).

10.2 PROBLEM SET

a.

b.

10.3 SOLUTION


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a. The most general linear combination of the functions in the family

is y = Ax2 + Bx + C (whereA, B, and C are the undetermined coefficients).

Substituting this into the given differential equation gives

Now, combinbing like terms yields

In order for this last equation to be an identity, the coefficients of like powers of xon both

sides of the equation must be equated. That is, A, B, and C must be chosen so that

The first equation immediately gives . Substituting this into the second equation

gives , and finally, substituting both of these values into the last equation

yields . Therefore, a particular solution of the given differential equation is


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b. Since the family of d = sin x is {sin x, cos x}, the most general linear combination

of the functions in the family is y = A sin x + B cos x (where A and B are the

undetermined coefficients). Substituting this into the given differential equation

gives

Now, combining like terms and simplifying yields

In order for this last equation to be an identity, the coefficients A and B must be chosen

so that

These equations immediately imply A = 0 and B = . A particular solution of the given

differential equation is therefore

According to Theorem B, combining this y with the result of Example 12 yields the

complete solution of the given nonhomogeneous differential equation: y = c1 ex +c2 xe

x +

cos x.


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P L A T E N O . 1 1 V A R I A T I O N O F P A R A M A T E R S

11.1 PRINCIPLES

For the differential equation

the method of undetermined coefficients works only when the coefficients a, b, andc are


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constants and the right-hand term d( x) is of a special form. If these restrictions do not

apply to a given nonhomogeneous linear differential equation, then a more powerful

method of determining a particular solution is needed: the method known as variation of

parameters.

11.2 PROBLEM SET

a. y + y = tan x

b.

11.3 SOLUTION

a.) Since the nonhomogeneous right-hand term, d = tan x, is not of the special form the

method of undetermined coefficients can handle, variation of parameters is required.The first step is to obtain the general solution of the corresponding homogeneous

equation, y + y = 0. The auxiliary polynomial equation is

whose roots are the distinct conjugate complex numbers m = i = 0 1 i. The generalsolution of the homogeneous equation is therefore

Now, vary the parameters c1 and c2 to obtain


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Differentiation yields

Nest, remember the first condition to be imposed on v1 and v2:

that is,

This reduces the expression for y to

so, then,

Substitution into the given nonhomogeneous equation y + y = tan x yields


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Therefore, the two conditions on v1 and v2 are

To solve these two equations for v1 and v2, first multiply the first equation by sinx; then

multiply the second equation by cos x:

Adding these equations yields

Substituting v1 = sin x back into equation (1) [or equation (2)] then gives

Now, integrate to find v1 and v2 (and ignore the constant of integration in each case):


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and

Therefore, a particular solution of the given nonhomogeneous differential equation is

Combining this with the general solution of the corresponding homogeneous equation

gives the general solution of the nonhomogeneous equation:


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In general, when the method of variation of parameters is applied to the second-order

nonhomogeneous linear differential equation

with y = v1( x) y1 + v2( x) y2 (where y h = c1 y1 + c2 y2 is the general solution of thecorresponding homogeneous equation), the two conditions on v1 and v2 will always be

So after obtaining the general solution of the corresponding homogeneous equation

( y h = c1 y1 + c2 y2) and varying the parameters by writing y = v1 y1 + v2 y2, go directly to

equations (1) and (2) above and solve for v1 and v2.

b.) Because of the In x term, the right-hand side is not one of the special forms that the

method of undetermined coefficients can handle; variation of parameters is required.

The first step requires obtaining the general solution of the corresponding

homogeneous equation, y 2 y + y = 0:

Varying the parameters gives the particular solution

and the system of equations (1) and (2) becomes


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Cancel out the common factor of e x in both equations; then subtract the resulting

equations to obtain

Substituting this back into either equation (1) or (2) determines

Now, integrate (by parts, in both these cases) to obtain v1 and v2 from v2 and v2:

Therefore, a particular solution is


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Consequently, the general solution of the given nonhomogeneous equation is

PLATE NO. 12 INVERSE DIFFERENTIAL OPERATORS

12.1 PRINCIPLES

Consider a linearnon-homogeneousordinary differential equation with constant

coefficients
http://www.efunda.com/math/ode/generalterms.cfm#Linearhttp://www.efunda.com/math/ode/generalterms.cfm#Homogeneoushttp://www.efunda.com/math/ode/generalterms.cfm#Homogeneoushttp://www.efunda.com/math/ode/generalterms.cfmhttp://www.efunda.com/math/ode/generalterms.cfm#Homogeneoushttp://www.efunda.com/math/ode/generalterms.cfmhttp://www.efunda.com/math/ode/generalterms.cfm#Linear


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where are all constants and . Let

the ODE can be rewritten as

Thus, the particular solution is

The particular solution can be easily obtained, if the effects of the inverse

operator have been studied in advance.

12.2 PROBLEM SET

12.3 SOLUTION

differential equation ( final product )

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