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CCCChapterhapterhapterhapter 2 2 2 21st ORDER DIFFERENTIAL
EQUATIONS
2
2.1 Types of 12.1 Types of 12.1 Types of 12.1 Types of 1stststst ODE ODE ODE ODE
� Separable Equations
�Homogeneous Equations
� Linear Equations
� Exact Equations
Separable equationsSeparable equationsSeparable equationsSeparable equations
( ) ( ) ( )( )
( )( )
( ) ( )
1. Reform the given differential equation into the general
form of separable equation (above)
2. Rearrange the equation to
3
General form:
Method of solut
. Inte
ion:
grat
, ,
1
g x h ydy dy dyg x h y
dx dx h y dx g x
dy g x dxh y
= = =
=
( ) ( )e both side of equation in step 2
4. Obtain the general solution H y G x CÞ = +
Example 2.1Example 2.1Example 2.1Example 2.1
Solve the differential equation
3
4
3
dy xdx y
=
Separate the equation into the form,
Integrate the both sides,
Rearranging, this equation becom
e s:
3
3
4 2
42 4
:
: 3 4
:
3 4
3 44 2
:
3 82
4 3
1
2
3
Solution
y dy xdx
y dy
Step
Step
S
xdx
y xC
yx c y
tep
=
=
æ ö æ ö÷ ÷ç ç÷ ÷= +ç ç÷ ÷ç ç÷ ÷ç çè ø è ø
= + Þ =
ò ò
General solution is,
where
2
142
43
:
8
4
4,
3 3
x C
S
y x A A
t
C
ep
+ +
æ ö÷ç= + =÷ç ÷çè ø
Example 2.2Example 2.2Example 2.2Example 2.2
( )
Solve the differential equation
and given that
24
0 2
y x y
y
¢ =
=
General solution is,
wher
e
3 3
3
2
2
3
4 43 3
43
:
: 4
: 4
4ln
3
:
:
,
1
2
3
4
x xC C
xC
Solution
dyx dx
y
dydy x dx
y
xy C
y e y e e
y De
Step
Step
Step
Step
D e
+
=
=
= +
= Þ = ×
= =
ò ò
( )( )
Given that
Particular soluti
on:
3
3
4 0
3
43
: 0 2
2 2
5
2x
y
De D
e
Ste
y
p =
= Þ\ =
=
HOMOGENEOUS equationsHOMOGENEOUS equationsHOMOGENEOUS equationsHOMOGENEOUS equations
Homogeneous functions are functions where the sum of
the powers of every term are the same.
( ) ���� ���� ����
( ) ���� ����
( ) ���� ���� ����
3 2 3
3
deg3 deg3 deg3
deg3 deg1
deg3 deg0deg3
3 3
1) , 4 3
2) ,
3) , 4 4
f x y x x y y
f x y x y
f x y x y
= + +
= +
= + +
( )
( ) ( )1. Rearrange the equation in the general form
2. Show that (Test the homogeneity)
3. Substitute and into the general
form of homoge
General form:
Method of solution:
neou
,
, ,
dyf x y
dx
f x y f x y
dy dvy xv x v
dx dx
l l
=
=
= = +
s equation.
4. Separate variables and to form separable equation.
5. Integrate both side of equation in step 4
6. Substitute into solution in step 5 and simplify the solution.yv
x v
x=
Example 2.3Example 2.3Example 2.3Example 2.3
Solve the differential equation
2 2dy x y
dx xy+
=
( )
( ) ( ) ( )( )( )
( )( )
( )
Test homogeneity,
hence, the equation is homogeneous.
2 2
2 2
2 2 2
2
2 2
:
:
,
,
,
1
Solution
x yf x y
xy
x yf x y
x y
x y
xy
x yf x y
xy
Step
l ll l
l l
l
l
+=
+Þ =
+=
+= =
( )( )
Substitute and into
homegeneous equation,
Rearranging, this equation becomes: (separable
form)
22
2 2 2 2
2
2
:
1
:
1
1
1
2
3
dy dvy xv x v
dx dx
x xvdvx vdx x xv
x x v vvx v
dv vx vdx vdvxdx v
vdv dx
Step
S ep
x
t
= = +
++ =
+ += =
+= -
=
=
( )
( )
( )
( )
Integrate the both sides,
General solution is,
by substituting
4
2
2
2
2
2
2 2
:
1
ln2
2 ln
:
2 ln ( )
2 ln
2 ln
5
vdv dxx
vx C
v x C
y yx C v
x
Step
Step
x
yx C
xy x x C
=
= +
= +
æ ö÷ç = + =÷ç ÷çè ø
= +
= +
ò ò
Example 2.4Example 2.4Example 2.4Example 2.4
( )Solve the differential equation
2 24y dx xy x dy= -
( )
( )
( ) ( )( )( ) ( )
( )( )
( )
Test homogeneity,
hence, the equation is homogeneous.
2
2
2
2
2
2
2 2
2 2
2
2
:
: ,4
:
,
1
2
4
,4
4
,4
Solution
dy yf x y
dx xy x
yf x y
xy x
yf x y
x y x
y
xy x
yf x y
xy
Step
Step
x
ll l
l l l
l
l
= =-
=-
Þ =-
=-
= =-
( )( )
( )( )
( )
2
2
2 2 2
2
2
2
:4
44
:4
4
4 44
44
2
14
3
xvdvx vdx x xv x
x v vvx v
dv vx vdx v
v vdv vxdx v vdv vxdx vv
dv dxv x
Step
Step
+ =-
= =--
= --
-= -
- -
=-
-=
( )
( )
Integrate the both sides,
4
:
4 141 1 14
1ln ln
41
ln4
4 ln 4
vdv dx
v x
dv dxv x
v v x C
v xv C
v xv C
Step
-=
æ ö÷ç - =÷ç ÷çè ø
- = +
= +
= +
ò ò
ò ò
General solution is,
by substituting
wher
e
4ln 4
4 4 4 4
:
4 ln 4 ( )
4 ln 4
5
,
yy Cx
y yC Cx x
y y yx C v
x x x
yy C
x
e e e
e y e y A
Step
e A e-
æ ö÷ç= × + =÷ç ÷çè ø
= +
= ×
= × Þ\ = =
Example 2.5Example 2.5Example 2.5Example 2.5
Verify that the differential equation
is homogeneous. Then find the general solution.
2
2
3 4
2 2
dy y xydx xy x
+=
+
Example 2.6Example 2.6Example 2.6Example 2.6
( )
Find the solution for the following differential
equation
2 22 4x y dy xydx- =
Example 2.7Example 2.7Example 2.7Example 2.7By using the substitution and ,
show that the given differential equation
can be reduced to a homogeneous equation.
Then, solve it.
1 2
22
x X y Y
dy x ydx y
= + = -
+=
+
linear linear linear linear equationsequationsequationsequations
( ) ( ) ( )
( ) ( )
( ) ( )( ) ( ) ( )
( )
( )
1. Write the equation in the
General form:
Method of solution:
form:
2. Find the integrating factor :
;
p x dx
dya x b x y c x
dx
dyp x y q x
dxb x c x
p x q xa x a x
er r
+ =
+ =
= =
ò=
linear equationslinear equationslinear equationslinear equations
( ) ( )
{ }
( )
3. Multiply both sides by
The left side of the equ. can be simplified as,
4. Integrate both sides of the equ. to obtain
If given an initial condition, substitute into the
:
5.
dyp x y q x
dxd
ydx
y q x dx
r r r
r
r r
r + =
= òsolution.
Example 2.8Example 2.8Example 2.8Example 2.8
Solve the following differential equation:
32 5dyx y xdx
+ =
( ) ( )
2
3
32
2
2 ln
ln
2
:
2 5: ,
2 5;
1
:2
5
dxx
x
x
Solution
dy xy
dx x x
xp x q x x
x x
e
e
e
x
Step
Step r
r
æ ö÷ç+ =÷ç ÷çè ø
= = =
ò=
=
=
=
( )
{ }
Multiply both sides by
2
2 2 2 2
2 4
2 4
2 4
52
32
: :
25
2 5
5
:
5
55
3
4
x
dyx x y x xdx x
dyx xy xdx
dx y
Step
Ste
xdx
x y x dx
xx y C
Cy x
x
p
r =
æ ö÷ç+ =÷ç ÷çè ø
+ =
=
=
= +
= +
ò
Example 2.9Example 2.9Example 2.9Example 2.9
( ) ( )
Solve the following differential equation:
1 , 0 1x xdye e y x y
dx+ + = =
( ) ( )
Let
1
:
: ,1 1
;1 1
1
2 : , 1
x
x
x
x
x x
x
x x
edx
xe
e
Solution
dy e xy
dx e e
e
Step
St
xp x q x
e
ep
e
u
e
e er +
æ ö÷ç ÷+ =ç ÷ç ÷ç + +è ø
= =+ +
ò= = +
= x
duu e
ln
ln 1
1
x
xx
u
e
x
du due dx
dx e
e
e
er
+
ò= Þ =
=
=
= +
( ) ( ) ( )
( )
( )( ){ }
( )
( )
( ) ( )
Multiply both sides by
2
2
: 1 :
1 1 1 11 1
1
1
:
1
12
2
3
1 1
4
x
xx x x
x x
x x
x
x
x
x x
e
dy e xe e y e
dx e e
dye e y x
dxd
e y xdx
e y xdx
xe y C
x Cy
Step
S
e e
tep
r = +
æ ö æ ö÷ç ÷ç÷+ + + = + ÷ç ç÷ ÷ç ç÷ç è ø+ +è ø
+ + =
+ =
+ =
+ = +
= ++ +
ò
( )
( )( ) ( )
( ) ( )
( )
When
Particular solution is
2
0 0
2
2
: 0 1 :
01
2 1 1
1 22
2
2 1 1
1 12
21
5
x x
x
y
C
e e
CC
xy
e
St p
x
e
e
ye
=
= ++ +
= Þ\ =
= ++ +
æ ö÷ç= + ÷ç ÷çè ø+
Exercise 2.1Exercise 2.1Exercise 2.1Exercise 2.1
( )
Solve the following differential equation:
2 cos
cos: sin
x
x
dyx x y e xdx
e x CAnswer y x
x x x
+ - =
æ ö÷ç= + + ÷ç ÷çè ø
exact equationsexact equationsexact equationsexact equations
( ) ( )
( ) ( )
( )
1.
2. Test for exa
Write down your and
Let be the solution,
i) integrate with respect to while holding constant,
to ob
General form
ta
c
:
Method of solution:
tness:
, , 0
, ,
3. ,
M x y dx N x y dy
M x y N x y
M N
F x y
M x y
y x
+ =
¶ ¶=
¶ ¶
( )
( )where is an arbi
in
trary function of
,
( )
.
F i or
y y
Mdx y
f
f= +ò ⋯⋯⋯⋯⋯⋯⋯⋯
( )
( )
ii
where is an arbitrary function
integrate with respect to while holding constant,
to obtain,
Differentiate from with respect to , and equate
the result to and
of
i ( )
)
( )
.
4. )
F Ndy ii
x
N y x
N
x
x
F i y
f
f= +ò ⋯⋯⋯⋯⋯⋯⋯⋯
( )
( )
find
Differentiate from with respect to , and equate
the result to and find
ii
( )
)
y
or
F ii x
M
FN
y
FM
x
x
f
f
¢
¢
¶=
¶
¶=
¶
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
( )
Integrate with respect to to find
Integrate with respect to to find
i)
ii)
6. Put back the value of in
or
7. The solution of the equati
equation
on is:
5.
,
F
y y dy
x x dx
F x y
y y y
or
x x x
y x
f f
f f
f
f
f
f
f f
¢=
¢
¢
=
¢
=
ò
ò
C
Example 2.10Example 2.10Example 2.10Example 2.10
( ) ( )
Solve the following differential equation:
2 3 2 2 0dy
x ydx
+ + - =
( ) ( )
( )
( ) ( )
( )
( )
Test exactnes
s:
2
2
:
: 2 3 2 2 0
2 3; 2 2
:
2
2
:
2 3
23
2
2
1
3
3
Solution
x dx y dy
M x N y
MM Ny Exacty xN
xF Mdx y
F x dx y
xF x y
Step
Step
Step
F x x y
f
f
f
f
+ + - =
= + = -
ü¶ ïï= ï¶ ¶ï¶ ï = Þýï ¶ ¶¶ ï= ïïï¶ þ= +
= + +
= + +
= + +
òò
( )( )
( ) ( )( ) ( ) ( )
( )
( )( )
( )
Find
Integrate to get
Put back the value in equation:
where
2
2
2 2
2 2
:
0 0 2 2
2 2
: :
2 2
22
22
:
,
3 2
3 2 ,
4
5
6
FN
y
y y
y y
y y
y y dy y dy
yy y A
y y y A
y F
F x y C
x x y y A C
x x y y B B C A
Step
Step
Step
f
f
f f
f f
f
f
f
¶=
¶¢+ + = -
¢\ = -
¢
¢= = -
= - +
= - +
=
+ + - + =
+ + - = = -
ò ò
Example 2.11Example 2.11Example 2.11Example 2.11
( ) ( )( )
Solve the following differential equation:
Given that,
2 3 23 1 6 0
0 3.
x y dx x y y dy
y
- + + - =
=
( ) ( )
( )
( ) ( )
( )
( )
Test exactness:
2 3 2
2 3 2
2
2
2
3
3
:
: 3 1 6 0
3 1; 6
:
3
3
:
3
2
33
3
1
1
Solution
x y dx x y y dy
M x y N x y y
Mx
M Ny Exacty xN
xx
F Mdx
Step
St
y
F x y dx y
x yF x y
F x y x
ep
Step
y
f
f
f
f
- + + - =
= - = + -
ü¶ ïï= ï¶ ¶ï¶ ï = Þýï ¶ ¶¶ ï= ïïï¶ þ= +
= - +
= - +
= - +
òò
( )( )
( )
( ) ( )( ) ( )
( ) ( )
( )
Find
Integrate to get
3 3 2
3 2 3
2
2
3 2 32
4 :
6
6
6
: :
6
63
3
5
2 3
Step
S
FN
y
x y x y y
y x y y x
y y y
y y
y y dy
y y y dy
y y yy A y A
tep
f
f
f
f f
f f
f
f
¶=
¶
¢+ = + -
¢ = + - -
¢\ = - +
¢
¢=
= - +
= - + + = - + +
òò
( )( )
( )
( ) ( ) ( )
Put back the value in equation:
where
When
33 2
33 2
33 2
33 2
:
,
33
3 ,3
: 0 3 :
30 3 0 3 3
39 27
18
3 183
6
7
y F
F x y C
yx y x y A C
yx y x y B B C A
y
C
C
C
yx
Step
St
y
e
x
p
y
f
=
- - + + =
- - + = = -
=
- - + =
- + =
\ =
\ - - + =
Exercise 2.2Exercise 2.2Exercise 2.2Exercise 2.2
( ) ( )Solve the following differential equation:
cos cos sin 0
: cos sin ,
y y x dx x x y dy
Answer xy x y x B B A C
+ - + - =
+ - = = -
2.2 applications of 12.2 applications of 12.2 applications of 12.2 applications of 1stststst ODE ODE ODE ODE
�Newton’s Law of Cooling
�Modeling of Population Growth
2.2.1 newton2.2.1 newton2.2.1 newton2.2.1 newton’’’’s law of coolings law of coolings law of coolings law of cooling
Definition (by word):Definition (by word):Definition (by word):Definition (by word):
“Newton's Law of Cooling states that the rate the rate the rate the rate
of change of the temperature of an object of change of the temperature of an object of change of the temperature of an object of change of the temperature of an object is
proportional to the difference between its the difference between its the difference between its the difference between its
own temperature and the ambient own temperature and the ambient own temperature and the ambient own temperature and the ambient
temperaturetemperaturetemperaturetemperature (the surroundings temperature)”
Mathematically:
sdT
T Tdt
µ -The rate of change
Its own temp.
Ambient temp.
( ) ( ) 0, 0sdT
k T T T Tdt
= - - =
( ) ktsT t T Ae= +
Separable Equ.
Solution: Represent the temperature of an object at time t.
is decreasing
is increasing
s
s
dTT T Cooling
dtdT
T T Heatingdt
> Þ Þ
< Þ Þ
( )
Integrate the both sides,
ln
ln
11 :
:
1
ln
2
,
s
s
s
s
s
s
T T kt C
T T kt C
kts
kt Cs
dTk T T
dt
dT k dtT T
dT k dtT
Step
S
T
T T kt C
e e
e e e
T T Ae
te
T Ae A e
p
T
- - +
- -
-
-
= - -
= --
= --
- = - +
=
= ×
- =
\ = + =
ò ò
Example 2.14Example 2.14Example 2.14Example 2.14
constant temperature of At 12 noon
A body of an apparent homicide victim was found in a room that
was kept at a . , the
and at 1temp pm erature it was of the body was
Assume that tem th
e
7
7
.8 50
0
F
F
F
°
° °
Based on the given information, determine the ti
Surrou
perature of the body at the time of death was
98
Initi
me of
al tem
de
Temp. after 1 hour
p. (12
nding temp.: =
noon): =
a.6 th.
0 8
7
0
.
0
:
s
Soluti
F
n
T
o
F
F
T °
°
°
Temp. of normal
(1 pm): =
body: = 1 75
98.6
T F
T F
°
°
( ) ( )
( )
( )
When
Whe
n
0
0
1
:
: , 0
: 0, 80, 70 :
80 70
10
70 1
1
0
: 1, 75
2
:
75 70 10
3
s
kts
s
k
kt
k
Step
Step
S
Solution
dTk T T T T
dt
tep
T T Ae
t T T
Ae
A
T e
t T
e
-
-
-
-
= - - =
Þ = +
= = =
= +
\ =
Þ = +
= =
= +
( )
Determine the time of death, when :
hours
The negative sign indicates t
0.6931
0.6931
0.6931
75 70101
ln2
0.6931
70 10
: 98.6
98.6 70 10
98.6 7010
ln 2.86 0.6931
1.51
4
61
k
t
t
t
e
k
k
T e
T
e
e
t
t
Step
-
-
-
-
-=
æ ö÷ç = -÷ç ÷çè ø
\ =
Þ = +
=
= +
-=
= -
\ = -
hat the time is 1.5161 hrs before 12 noon.
Therefore, the time of death is approximately at 10.29 am.
example 2.15example 2.15example 2.15example 2.15
( ) ( )
Temp. after 1.5 hour (10 pm):
Temp. of
Ini
nor
tial t
mal bo
emp. (8.30 p
dy: =
Wh
Surrounding temp.
=
en
m): =
: =
0
1.5
0
98
:
: , 0
: 0, 78.3
74
1
78.
7
2
.6
3
3
s
t
s
ks
Solution
dTk T T T T
dtT
T F
Step
St
T
T F
T Ae
F
p t Te
T F
-
= - - =
Þ = +
°
= =
°
°
°
( )
0
, 73 :
78.3 73
5.3
70 5.3
s
k
kt
T
Ae
A
T e
-
-
=
= +
\ =
Þ = +
( )
( )
When
Determine the time of death, wh en :
1.5
1.5
1.1118
1.1118
1.1118
: 1.5, 74 :
74 73 5.3
74 735.310
ln 1.553
1.1118
73 5.3
: 98.6
98.6 73 5.3
98.6 735.3
ln 4.8
3
3
4
k
k
t
t
t
t T
e
e
k
Step
S
k
T e
Te
e
p
e
t
-
-
-
-
-
= =
= +
-=
æ ö÷ç = -÷ç ÷çè ø
\ =
Þ = +
=
= +
-=
= -
hours
So, death occurred about 7.05pm.
1.1118
1.42
t
t\ = -
Exercise 2.3Exercise 2.3Exercise 2.3Exercise 2.3
A pie is removed from an oven with temperature and
cooled in a room with temperature . In minutes, the pie
has a temperature of Determine the time required to cool
the pie to tempe
350
75 15
150 .
F
F
F
°
°
°
rature of
minutes
80 .
: 46
F
Answer t
°
=
2.2.1 the population growth and 2.2.1 the population growth and 2.2.1 the population growth and 2.2.1 the population growth and decaydecaydecaydecay
Definition (by word):Definition (by word):Definition (by word):Definition (by word):
“The rate of change of the population is
proportional to the existing population”
Mathematically:
( )dPP t
dtµThe rate
of change
( ) ( )
( )
dPb m P t
dtdP
kP tdt
= -
=
ktP Ae=
Separable Equ.
Solution: Represent the temperature of an object at time t.
0
0
k population growth
k decay of radioactive
> Þ
< Þ
56
Example 2.17:population growthExample 2.17:population growthExample 2.17:population growthExample 2.17:population growth
The population of a country is growing at a rate that is
proportional to the population of the country. The
population in 1990 was 20 million and in 2000, the
population was 22 million. Estimate the population in 2020.
( )
( )
Initial (1990): =
Population after 10years (2000): =
Population after 3
0years (2020):
When
0
10
30
0
0
:
20
22
?
: , 0
: 0, 20 :
20
1
20
2
2
0
kt
k
kt
Solution
P m
P m
P
dPP P P
dtStep
S
P Ae
t P
Ae
te
P e
p
A
-
= =
Þ =
= =
=
\ =
Þ =
( )
( )
When
when :
So, in 2020 the population is about 26.62 million.
10
10
0.009531
0.009531 30
3
4
: 10, 22 :
22 20
2220
22ln 10
20
0.009531
20
: 30
20
26.62
k
k
t
t P
e
e
k
k
P e
t
P e
P million
Step
Step
= =
=
=
æ ö÷ç =÷ç ÷çè ø
\ =
Þ =
=
=
=
radioactive decayradioactive decayradioactive decayradioactive decay
� Radioactive substances randomly emit protons, electors, radiation, and they are transformed in another substance.
� It can be seen that the time rate of change of the amount N of a radioactive substances is proportional to the amount of radioactive substance.
� The half-life is the time, k needed to get
� Using the half-life, we get
1 12 2
0 0 12 1
2
1 1 1 1 1ln ln
2 2 2 2
kt ktP e P e kt k
t
æ ö æ ö÷ ÷ç ç= Þ = Þ = Þ\ =÷ ÷ç ç÷ ÷ç çè ø è ø
( ) 01
02
P P=
12
1 1ln
2
0( )
tt
P t P e
æ ö÷ç ÷ç ÷÷çè ø=
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