chapter 1: first order linear differential...

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Noor Alia Md Zain Semester 1 2014/2015

CHAPTER 1: FIRST ORDER LINEAR DIFFERENTIAL EQUATION

1.1 : Introduction to DE

An ODE is an equation that contains one or several derivatives of an unknown function. Example:

1. 3sin xdx

dy

2. xeyyy 6'2''

3. 022

2

3

3

ydx

dy

dx

yd

dx

yd

Standard Notation

If )(xy is a function of x , then the first order differential equation can be written as

1st order

dx

dy

)(' xy or 'y

2nd order

2

2

dx

yd

)('' xy or ''y

thn order n

n

dx

yd

)(xyn or ny

The order of an ODE is the order of the highest derivative that appears in the DE. Example:

No D.E Order

1 xxy

dx

dy2sintan

2 1

2

22 64 xy

dx

dyx

dx

ydx

3 xexyyy 254'4''

4 0'3''3''' yyyy

Formation of Differential Equations

Differential equations may be formed in practice from a consideration of the physical problems to which they refer. Mathematically, they can occur when arbitrary constants are eliminated from a given function. For expression with n arbitrary constants, it needs n times differentiation to eliminate the constants. Here are several examples.

Example 1: Form ODE from the function2xAxy , A is a constant.

Solution:

2


Example 2: Form ODE from the functionx

Axy 2 , A is a constant.

Solution

Example 3: Form ODE from the function xx BeAey 2 , BA, are constants.

Solution

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Concept of Solution and Initial Value Problem (IVP)

Solution of a DE

Definition: If )(xFy is the solution of an ODE, hence a function )(xF satisfies the given

DE. (when we can show that LHS=RHS)

Example4: Given 062

2

ydx

dy

dx

yd. Show that:

a) xey 2 is the solution.

b) xx eey 32 45 is the solution.

c) xxey 2 is not the solution.

Solution

Example 5: Find the value of m so that mxey is the solution of this DE, 03'5''2 yyy

Solution:

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General and Particular Solution Definition: i) General Solution - solution of the DE (same as solution of DE: show LHS=RHS) ii) Particular Solution – obtained from GS by substituting the given initial condition

Example 6: Show that xx exAey 2)2( is the general solution of this DE xexydx

dy 2)3(

and hence determine the value of A given that 4y when 0x

Solution:

Example 7: Show that x

xBxAy

3sin3cos is the general solution for 092

2

2

xydx

dy

dx

ydx

and hence obtained the particular solution with condition 3)( y when 0)(' y

Solution:

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Example 8: Show that 3

3

By Ax

x is the general solution for 2 '' ' 9 0x y xy y and hence

obtain the particular solution with initial condition 2 1y and ' 1 0y

Solution: 1.2 : Types Of First Order Differential Equations And Method Of Solutions There are four types of first order differential equations, namely:

i. Separable equations ii. Homogeneous equations iii. Linear equations iv. Exact equations

We need different approaches for solving these four different types of differential equations

Separable equations A separable first order differential equation is an equation which may be put into one of the following forms:

Method of Solution - Steps for Solving Separable Equation:

Step 1: Identify the problem as a separable equation.

f xdy

dx g y

Step 2: Separate the equation into the form. Rearrange

f xdy

dx g y to

g f dy f x dx

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Step 3: Integrate both sides of the equation, adding one arbitrary constant, say C , to the x side. This is done because adding an arbitrary constant to both sides after integrating is equivalent to adding just a single arbitrary constant to one side. Evaluate

g f dy f x dx C

Step 4: If there is an initial condition, then substitute it to obtain the value of .C

Example 1: Determine whether or not the following equations are separable:

a) dy

xy xdx

Solution:

b) y xdyxe

dx

Solution:

c) sin cos cos sin 0dy

y x y xdx

Solution:

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d) 2dy

x x ydx

Solution:

Example2: Solve the equation ydx

dyx )2(

Solution:

Example3: Solve the equation 02 xydx

dyex

Solution:

8


Example 4: Solve the following DE 0)1(2 dyxydxx which satisfied the condition 2y and

0x .

Solution: Example 5: Find the general solution of the following equation:

a) 2 0x dye xy

dx

Solution:

b) 2secdy

ydx

Solution:

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Example 6: Find the particular solution for the differential equation 3 2dy dy

x ydx dx

which satisfies the initial condition 0y when 3.x

Solution:

Example 7: Using the substitution ,z x y convert 1

5

dy x y

dx x y

to a separable equation.

Hence, solve the original equation which satisfies the condition 1 1.y

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Homogeneous equations

The first order differential equation ,dy

f x ydx

is denoted as the homogeneous equation if

, ,f x y f x y

Method of Solution - Steps for Solving Homogeneous Equation:

Step 1: Write to the general form , .dy

f x ydx

Make sure this equation is homogeneous. Need to

show , , .f x y f x y

Step 2: Use substitution y xv and ,dy dv

x vdx dx

into homogeneous equation in Step 1.

Step 3: Separate the variables x and v in the resulting equation.

Step 4: Integrate both sides of the equation and then put only a constant, say C on the right integration. When this equation is integrated, obtain a relationship between x and .v

Step 5: Then by substituting y

vx

we have the required solution.

Step 6: If there is an initial condition, and then use it to obtain the value for .C

Example 8: Determine whether or not the following equation are homogenous

1. dy y x

dx y x

Solution:

2. 2 2

dy xy

dx x y

Solution:

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3. dy

x ydx

Solution:

Example 9: Show that the equation dy x y

dx x y

is homogeneous and find the general equation.

Solution:

12


Example 10: Show that the equation 2

2 2

2dy y

dx x y

is homogeneous and find the general equation.

Solution:

13


Example 11: Find the particular solution of the differential equation 22 yx

xy

dx

dy

with condition

2)0( y

Solution:

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Example 12: Find the solution of the differential equation 2 2dy x y

dx xy

which satisfies the initial

condition 2y when 2.x

Solution:

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Example 13: Using the substitution x X and 2,y Y show that the equation 2

2

dy x y

dx x y

can be reduced to a homogeneous equation. Hence, find the solution of the original equation.

Solution:

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Linear equations

General form of a linear differential equation is ( ) ( )dy

P x y Q xdx

where )(),( xQxP is any

functions that contains variable x and coefficient of dx

dyis 1.

Method of Solution - Steps for Solving Linear Equation:

Step 1: write in general form ( ) ( )dy

P x y Q xdx

and determine ( )P x and ( )Q x .

Step 2: Determine ( )P x and evaluate ( ) .P x dx

Step 3: Obtain integrating factor, dxxP

e)(

Step 4: Multiply the general form throughout by the integrating factor, and write down in the

form of d

y Q xdx

Step 5: Integrate both side and solve for .y

dxxQy )(

Step 6: If there is an initial condition then use it to obtain the value for .C

Example 14: Find the particular solution of the differential equation tan cos .dy

y x xdx

Given

0 1y

Solution:

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Example 15: Find the general solution of the differential equation 23 .xdyy e

dx

Solution:

Example 16: Find the general solution of the differential equation 2.dy

y xdx

Solution:

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Example 17: Find the general solution of the differential equation 2 cos .dy

x y xdx

Solution:

Exact equations

General form of exact differential equation is 0),(),( dyyxNdxyxM . Condition for the exact

differential equation is x

N

y

M

Method of Solution - Steps for Solving Exact Equation:

Step 1: Write in general form and test for the exactness : x

N

y

M

Step 2: Integrate M w.r.t x while holding y as a constant : U M dx y

Step 3: Take partial derivative U w.r.t y and equate to N in order to find the value of y'

UN

y

Step 4: Integrate ' y w.r.t y to find y :

Bdyyy ' , B is constant

Step 5: Substitute back the value of y into the second step. The solution is ,U x y c

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Example 18: Determine whether or not the following differential equation are exact.

a) 0)26()32( 2 dyyxydxyx

Solution:

b) 3

2( 3 ) (1 ) 0y x

x dx dyx y

Solution:

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Example 19: Solve the differential equation 0)356()3106( 2222 dyyxxydxyxyx

Solution:

Example 20: Solve the differential equation,

cos 2 ( cos 1) 0y ye x y x dx xe x y dy

Solution:

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Example 21: Solve the DE (sin ) ( cos sin ) 0x dy y x x x dx

Solution:

1.3 : Application of First Order Differential Equations Modelling with differential equations is the process of writing a differential equation to describe a physical situation. We will look at three different situations which involve applications of first order differential equations, namely Newton’s Law of Cooling and population growth.

1. Newton’s Law of Cooling Used to model the temperature change of an object of some temperature placed in an environment of a different temperature. The law states that:

s

dTk T T

dt ….(i)

where T is the temperature of the object at time t, sT is the temperature of the surrounding

environment (constant) and k is a constant of proportionality. This law says that the rate of change of temperature is proportional to the difference between the temperature of the object and that of the surrounding environment in which it is placed. In order to get the equation (i) to something that we can use, we must solve the differential equation. The steps are as follow:

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I. Separate the variables: s

dTkdt

T T

II. Integrate both sides: s

dTkdt

T T

ln sT T kt C

III. Solve for T: kt c

sT T e e

;kt c

sT Ae T A e

We now have a useful equation. When you are working with Newton's Law of Cooling, remember that t is the variable. The other letters, M, k, A, are all constants. In order to find the temperature of the object at a given time, all of the constants must first have numerical values.

Example 1: According to Newton’s Law of cooling, the rate at which body cools is given by the equation

s

dTk T T

dt ,where sT is the temperature of surrounding medium, k is a constant and t is the

time in minutes. If the body cools from 80°C to 50°C in 20 minutes which the surrounding temperature of 10 °C, how long does it need for the body to cool from 80°C to 30°C and also time taken for the body to cool from 50°C to 30°C.

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Example 2: A boiling (100°C) solution is set on a table where room temperature is assumed to be constant 20°C. Find the solution cooled to 60°C after five minutes. a. Find a formula for the temperature (T) of the solution, t minutes after it is placed on the table. b. Determine how long it will take for the solution to cool to 22°C

Answer: a. 0.1386320 80 tT e ; b. t=26.6 minutes

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Example 3: Time of Death

Suppose that a corpse was discovered in a motel room at midnight and its temperature was F80 .

The temperature of the room is kept constant at F60 . Two hours later the temperature of the

corpse dropped to F75 . Find the time of death. Normal body temperature is F6.98 (assuming the dead person was not sick!). Answer: The death happened around 7:26 P.M.

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Example 4: Crime Scene

A detective is called to the scene of a crime where a dead body has just been found. She arrives on the scene at 10:23 pm and begins her investigation. Immediately, the temperature of the body is taken and is found to be 80o F. The detective checks the programmable thermostat and finds that the room has been kept at a constant 68o F for the past 3 days. After evidence from the crime scene is collected, the temperature of the body is taken once more and found to be 78.5o F. This last temperature reading was taken exactly one hour after the first one. The next day the detective is asked by another investigator, “What time did our victim die?” Assuming that the victim’s body temperature was normal (98.6o F) prior to death, what is her answer to this question? Newton's Law of Cooling can be used to determine a victim's time of death. Answer: 3.22 pm

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2. Population Growth

Differential equations whose solutions involve exponential growth or decay are discussed.

Consider the differential equation

Pkdt

dP

Using the method of separation of variables we find

dtkP

dP

Integrating both sides to obtain

CktP ln , hence ktAeP

Example 4 Around thousands citizens are living in a small country and this amount has been changed to 450 thousands within 2 years. The population grows at rate that is proportional to the number of the citizens present at the time. Assume that the 50 thousands is the initially, determine

a) The equation of related the size of population to the time b) How many citizen of the small country in this population after 5 years

Answer: tetPa 0986.150)() ; 12,149.25m)5() Pb

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Example 5 Suppose that the population of a colony of bacteria increases exponentially. At the start of an experiment, there are 6,000 bacteria, and one hour later, the population has increased to 6,400. How long will it take for the population to reach 10,000? Round your answer to the nearest hour.

Answer: hourst 8

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EXERCISE 1. Form the ODE for the following

a) BxAey

b) 214

6

1xDxCxy

c) 2Dxxy

2. Given that 025'' yy . Show that,

a) xy 5cos and xy 5sin is a solution for the DE

3. If x

BAxy 2 , prove that yyx 2''2

4. If that xx BeAey 3 is the general solution for 03'2'' yyy , find the particular solution

if the initial condition 0)0( y and 4)0(' y

5. If that 33 BxAxy is the general solution for 09'''2 yxyyx , find the particular solution

if the boundary condition 1)2( y and 0)1(' y

6. Find the solution of each of the following DE:

a) y

x

dx

dy 12

b) y

x

dx

dy 2

c) 3)3(,)3(

yy

xx

dx

dy

d) 0)2(')1( xyyyx

e) yydx

dyx 2

f) 2)0(,01 2 yxy

dx

dy

ex

g) 222 xy

dx

dyxy

h) 2)1(,222 yxxyydx

dyx

i) )(' xInyInyxy

j) 22 ')( yxyyxyx

k) yx

yx

dx

dy

3

3

l) xeydx

dy 2

m) xx

y

dx

dy62

n) xydx

dyx cos2

o) xxydx

dyx sin2

p) 0)sin2()cos( 2 dyyxydxxy

q) 0)cos()2( 2 dyyexdxxe yy

r) 0)22()3224( 223 dyxyxdxyxyx

s) 1)1(,022 22 ydyyxdxxy

t) 0)()24( 222 dyyxdxyxye x

u) 2)2(,22

yxy

yx

dx

dy

v) 1)1(),(2 yxInxydx

dyx

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7. It was noon on a cold December day in Cameron Highland C16 . Detective Ismail arrived at the crime scene to find Sergeant Normah leaning over a body. Sergeant Normah said that there were several suspects. If only they knew the exact time of death, then they could narrow down the list. Detective Ismail took out a thermometer and measured the temperature of the body

C5.34 . He then left for lunch. Upon returning at 1:00 pm, he found the body temperature to be C7.33 . When did the murder occur? Hint: Normal body temperature is C37 . Answer: 9.08 AM

8. In a murder investigation, a corpse was found by a detective at exactly 8pm. Being alert, the

detective also measured the body temperature and found it to be F70 . Two hours later, the

detective measured the body temperature again and found it to be F60 . If the room

temperature is F50 and assuming that the body temperature of the person before death was

F6.98 , at what time did the murder occur? Answer: 5.26 PM

9. Suppose that a cup of coffee is left on a table. A thirsty mathematician finds this cup of (cold)

coffee and measures its temperature to be C43 . The room the cup was in is kept at a

constant C30 . Considering the coffee too cold to drink, the mathematician walks away. Three

hours later, he realizes the cup is still sitting on the table, having dropped in temperature to

C36 . If coffee comes out of the coffee maker at C70 , how many hours ago was the cup left

on the table. Answer: 7.3567ht

10. A body of an apparent homicide victim was found in a room that was kept at a constant

temperature of F70 . At 12 noon, the temperature of the body F80 and at 1 pm it was

F75 . Assume that the temperature of the body at the time of death was F6.98 . Determine

the time of death. Answer: 10.29 AM

chapter 1: first order linear differential...

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