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7/30/2019 Chapter 2-Differential Equation
1/9
10/19/20
CHAPTER 2
DIFFERENTIALEQUATION1
2.1 INTRODUCTION TODIFFERENTIAL EQUATION
OBJECTIVES : At the end of the lesson, students will be
able to
1) define differential equations
2) understand degree, order and solution
3) define differential equations of separable
variable type
4) solve differential equations of separable variable
type2
Introduction
A simple type of differential equation is
We integrate it to produce the required solution.
In many practical situations such as population
growth, radioactive decay, chemical mixture,
temperature cooling, velocity and acceleration
problem and etc. we have much more
interesting differential equations
32
xdx
dy
3
1) Definitions
2) Differential equations with
separable variables
3) Linear first-order differential
equations
4
Definition 1 A differential equation (DE) is one which
relates an independent or dependent variables
with one or more derivatives.
Examples of differential equations:
a) dxxdy 32
b) 523 x
dx
dy
c) xydx
dysin2
2
d)0743
2
2
dx
dy
dx
yd
5
Definition 2
Orderis the highest derivative in a differential
equation.
Degree is the highest power of the highest
derivative which occurs in a differential
equation.
Examples: a) 823
3
2
2
yxdx
dy
dx
yd
This DE has order 2 (the highest derivative appearing is the second
derivative) and degree 1 (the power of the highest derivative is 1.)6
7/30/2019 Chapter 2-Differential Equation
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b)
This DE has order 1 (the highest derivative appearing
is the first derivative) and degree 5 (the power of the
highest derivative is 5.)
yxxdx
dysinsin32
5
35'2" 74 yyy
This DE has order 2 (the highest derivative appearing
is the second derivative) and degree 4 (the power of
the highest derivative is 4.)
c)
7
Definition 3 General solution The general solution
of a differential equation contains an
arbitrary constant c.
Particular solution - The particular
solution of a differential equation
contains a specified initial value and
containing no constant.8
Differential Equations with
Separable Variables
Definition 4
Differential equations with separable variables Differential
equations in which the variables can be algebraically separated.
Some differential equations can be solved by the method of
separation of variables. It is the technique used when integrals were
antiderivatives. A differential equation is separable if it can be
written in the form 0 dyyNdxxMThe method for solving separable differential equations is to
separate the variables so that the yN is with dy and the xMis with the dx, where yN is a function ofy only and xM
is a function of x only. Once we can write it in the above form, all
we do is integrate throughout, to obtain our general solution. 9
Below are examples ofgeneral solution for the
differential equations.
Example 1:
, where and
This is already in the required form, so we
simply integrate:
, c is constant
032 dxxdyy 3xxM
2yyN
cdxxdyy 32
cxy
43
43
10
In this chapter we only deal with first
order, first degree differential equations.
A solution for a differential equation is a
function whose elements and derivatives
may be substituted into the differential
equation. There are two types of solution
for differential equations.11
Example 2: 011
dx
dyyxxy
First we must separate the variables:
011 dyyxdxxy Multiply throughout by dx
01
1
y
dyyxdxx Divide throughout byy
0
11
y
dyy
x
dxxDivide throughout byx
12
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This gives us: 01
11
1
dy
ydx
x
where
xxM
11 and
y
yN1
1
We now integrate,
constantis
11
11
c,cx
ylnyx
cxlnylnyx
cylnyxlnx
cdyy
dxx
13
Example 3 21 y
dx
dye
x
First we must separate the variables:
dxydye x 21 Multiply throughout by dx
dx
e
ydy
x
21
Divide throughout by xe
dx
edy
yx
1
1
12
Divide throughout by 21 y
14
This gives us: dxedyy x 21
We now integrate:
constantis,
1
1
11
1
1
1
2
ccey
cey
dxedyy
x
x
x
15
2.2 Separable Variables and
Integrating Factors
Example 1:
Solve the differential equation
012
dx
dyyexx
y
First we must separate the variables:
cdyyedxxx
dyyedxxxy
y
101
2
2
Consider dxxx 12 : 16
By using substitution,
Let 12 xt x
dtdx
xdx
dt
2
2
cx
ct
ct
dtt
x
dttxdxxx
2
32
2
3
2
3
2
1
2
13
13
1
2
32
1
2
1
21Thus,
17
Consider :
By using integration by parts,
Let
dyye y
yu
dydudy
du
1
and yedy
dv
y
y
y
ev
dyedv
dyedv
1
Thus,
ye
eye
dyeye
vduuvdyye
y
yy
yy
y
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So the general solution is:
23
21
3
1x 01 cyey
19
Example 2
Solve the differential equation ,
First we must separate the variables:
2 yxdx
dy
xdxy
dy
2
We now integrate: xdx
y
dy
2
cx
y 2
2ln2
20
We now use the information which means
at , to find c.
, gives
0x 5y
c2
025ln
2
3lnc
So the particular solution is:
3ln22ln2
xy
21
Separable Variables and Integrating Factors
Examples of DE with separable variables. These
equations can easily be solved after we separate the
variables
xydxdyxx
ydx
dyx
xydx
dy
11c)
1b)
1xa)
2
22
Examples of DE with non-separable variables.
These equations cannot be solved by separating
the variables, because the variables are unseparable.
These are called linear first-order DE.
xydx
dy
xydx
dyx
xx
y
dx
dy
3c)
02b)
1a)
23
Linear First-Order Differential Equations
A linear first-order differential equation is
one of the form .
The solution must be on an interval where
both and are continuous. The
usual solution to the differential equation
is to change it to an exact equation by
means of an integrating factor. This
integrating factor is
xQyxPdx
dy
xP xQ
.xV24
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By multiplying both sides of the equation by the integrating
factorV(x), you change the left-hand side of the equation into
the derivative of the product V(x)y:
xQxVyxVxPdx
dyxV
Once we have chosen V(x), (we shall do so in a moment) and
carried out the multiplication ,
xVxQxVyxVdx
d
xQxVyxVxPdx
dyxV
ofChoice
we can solve the above equation by integrating both sideswith respect to x.
25
To find , we find the functions ofx that satisfies
the equations xV
ln
dintergrateEquation
separatedVariables
cancelledThe
cancelledtermsThe
sderivativeforRuleProduct
dxxPCexV
dxxPxV
dxxPxV
xdV
dxxPxV
xdV
yxVxPdx
xdV
dx
dyxVyxVxP
dx
xdVy
yxVxP
dx
dyxV
dx
xdVy
dx
dyxV
yxVxPdx
dyxVyxV
dx
d
26
Since we do not need the most general function V(x), we
may take Cto be 1 and we get dxxPexV
.
By multiplying both sides of the equation by the integrating
factor
dxxPexV , you obtain the exact differential
equation:
xQeyxPe
dx
dye
dxxPdxxPdxxP
xQe
dx
yeddxxP
dxxP
dxxQeyeddxxPdxxP
To solve this differential equation, integrate both sides andsolve for y. 27
Example 1:
Solve
This is already in the required form
with and
The integrating factor is
0 ydx
dy
xQyxPdx
dy 1xP 0xQ
dxxP
exV dx
e
Thus the integrating factor is xe . Multiplying bothsides of the equation by xe
28
0dxyed
x
0yed x
dxyed x 0
cyex
xe
cy
gives the solution: 0 yedx
dye
xx
29
Example 2:
Solve1
12
x
yxdx
dy
This is already in the required form xQyxPdx
dy
with x
xP2
and 1
1
x
xQ
.
The integrating factor is dx
xexV2
xe
ln22ln xe2x
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Multiply by to both sides of the equation gives2x
1
22
2
x
xx
x
y
dx
dy
1
22
2
x
xyx
dx
dyx
1
22
x
x
dx
yxd
dxx
xyxd
1
22
dxxx
yxd1
2
2
31
Consider divide the denominator into
the numerator to obtain
dxxx
1
2
dxx
x
1
11
Let
dx
xxdx
xx
111
1
2
Thus
dxx
xyxd1
112
cxxx
yx 1ln2
22
32
Solve differential equations using integrating
factors (particular solutions)
Example 1:
Solve and that when .23 xydx
dyx
5
2y 1x
First we change the equation to the required form:
xQyxPdx
dy by multiplying it throughout by dx.
xx
y
dx
dy
3with
xxP
3 and xxQ
.
The integrating factor is dx
xexV
3xe ln3
3lnxe
3x33
Multiply by to both sides of the equation
gives
3x
333 xxxx
y
dx
dy
423 3 xyxdx
dyx
43x
dx
yxd
dxxyxd 43
dxxyxd43
cx
yx 5
53
34
We now use the information which means
at , to find c.1x5
2y
c
5
1
5
21
53 , gives
5
1c
So the particular solution is:
5
1
5
53
xyx
35
Example 2
Solve 20,costan 2 yxyxdx
dy
xxQxxPyxdxdy
22 cosandtanx withcostan
The integrating factor is dxxPexV xdxe tan
x
e
e
e
e
x
x
dxx
x
dxx
x
sec
cos
1ln
cosln
cos
sin
cos
sin
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Multiplying both sides of the equation by gives :xsec
x
cxy
cxyx
xdxyxd
xdxyxd
xdxxyxd
xxdx
yxd
xxyxxdx
dyx
sec
sin
sinsec
cossec
cossec
cossecsec
cossecsec
cossectansecsec
2
2
2
37
We now use the information which means at ,
to find c.
0x
2y
0sec
0sin2
c , gives 2c
So the particular solution is:
x
xy
sec
2sin
xx cos2sin
38
Applications of Differential Equations Now we will apply the methods in this section to the solution
of some practical situations.
A) population growth model
The simplest growth model has a constant relative growth
rate. If we denote the population we are considering by ,
then the rate of change of the population is . To say that
the rate of change is proportional to the population is just
saying that there is a constant of proportionality ksuch that
ty
dt
dy
39
kydt
dy
Since kis constant, this can be immediately separated and
integrated to yieldkdt
y
dy
kdtydy
ckty lncktey ckt
eey ktAey
40
Example 1:
Assume that the population of the earth changes
at a rate proportional to the current population.
a) Write the differential equation satisfied bythe population
b) In 1650 the population is estimated to havebeen about 600 million and in 1950 about 2800million. By solving the differential equation andfitting this data, estimate the population at timet(years AD).
c) Using this fitted solution, and assuming thatthe greatest population the earth can support ispeople, in what year will this limit be reached?
41
Solution
a)
b)
42
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43
c)
Radioactive decay models, on the other hand, are very
accurate over long periods of time. They are theprimary method for determining age of prehistoric
fossils and ancient artifacts. If we denote the decay we
are considering by , then the decreasing rate of
the decay is . To say that the decreasing rate is
proportional to the decay is just saying that there is a
constant of proportionality ksuch that
tC
B) Radioactive decay models
dt
dC
44
Since kis constant, this can beimmediately separated and integrated toyield
kCdt
dC
kdtC
dC
kdtC
dC
cktC lnckteC
ckt
eeC ktAeC
45
The rate of decay for a radioactive substance is
proportional to the remaining concentration. Write
down the differential equation satisfied by the decay.
If one third of the substance still remains within 12
years, how much left is still available after 3 years.
Solution:
concentration of radioactive substance at t
time
tC
46
Example :
47
C) Newton's Law of Cooling When an object has a temperature greater than the
ambient temperature, it cools according to Newton's
Law of cooling which states that the rate of cooling is
proportional to the difference in the temperatures, that
is: , where is the temperature of
the object at any time tand a is the ambient temperature.
The solution to this separable differential equation is:
akdt
d
t
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dtakd
kdta
d
kdtad
ckta ln
ae ckt
ae ckt
aAe kt 49
Example :
A turkey is taken from the oven at 300 o F and
placed at room temperature of 70o F. In t= 2
minutes, the turkey's temperature is 200o F. Find
how long it takes the turkey to cool to 150 o F.
Solution
temperature of object at t time t
50
51