CHAPTER 2: DE

Lecture Note: 1 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Topic: 2.0 FIRST ORDER DIFFERENTIAL EQUATIONS

Sub-Topic: 2.1 Separable Variables
2.2 First Order Linear Differential Equations

Learning Outcomes: At the end of this lesson, students should be able to
(a) distinguish between general and particular solutions.

*state the degree and order.
(b) solve separable differential equations.

*use the boundary conditions to find a particular solution.
(c) solve first order linear differential equations by means of an integrating factor.

*must show all the necessary steps.

Definition

 A simple type of differential equation is dy  x2  9. We integrate it to produce the
dx

required solution. In many practical situations such as population growth, radioactive decay,
chemical mixture, temperature cooling, velocity, acceleration problem, etc. We have much
more interesting differential equations.
 A differential equation, DE is one which relates an independent or dependent variable with
one or more derivatives.
 Examples:

dy  sin xdx,

ds  2t3  5,
dt

 dy 2  y2  sin x,
dx

3 d2y  4 dy 7  0,
dx2 dx

y  2y  5y  4,

 Order is the highest derivative in a differential equation.
 Degree is the highest power of the highest derivative which occurs in a differential equation.

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Lecture Note: 1 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 1
Determine the order and degree of differential equations below.

Differential Equations Order Degree

y5 dy  x2  5 11
dx 21
34
7 d2y  dy 3 13
dx2 dx

 dy 6   d3y 4 1
dx dx3

 dy 3  3x  ln x
dx

y4  2y7  5y  3 2 4

 dy 5  2x  3 sin x  sin y 1 5
dx

Note
*In this chapter we only deal with first order and first degree differential equations.

How do we solve first order differential equations?

There are two methods which can be used to solve first order differential equations.
They are Separable Variables and Integrating Factor.

Differential Equations with Separable Variables
The separation of variables method is used when a differential equation can be separated
algebraically.

Steps to solve the differential equations with separable variables

 Separate the variables x and y (x by the side of dx and y by the side of dy).
 Integrate both sides independently.

Types of Solutions
General Solution of a differential equation contains an arbitrary constant C.
Particular Solution of a differential equation contains a specified Initial Value Problem, IVP
and containing no constant.

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Lecture Note: 1 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 2 (b) sin x dy  cos x
Solve each of the following. 1  y dx

(a) y2dy  x3dx  0  1 dy   cos x dx
1y sin x
 y2dy   x3dx
ln 1  y  ln sin x  C
y3  x4  C
34 e  eln 1 ln sin x C
y3   3x4  C 1y

4

1  A sin x, A  eC
1y

1y  1
A sin x

y 1 1
A sin x

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Lecture Note: 1 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 3

Find the particular solution of the differential equation 1  sin2 x dy  e2y sin 2x, y 0  1.

dx

 e2ydy  1 sin 2x x dx
 sin2

e2y  2 sin x cos x dx u  1  sin2 x
1 sin2 x
2 du  2 sin x cos x
dx
e2y  1 du du  2 sin x cos xdx

2u

e2y  ln u  C
2

e2y  ln 1  sin2 x  C
2

When x  0, y  1,

e2  ln 1  sin2 0  C
2
C  e2

2

e2y  ln 1  sin2 x  e2
22
e2y  2 ln 1  sin2 x  e2

2y  ln 2 ln 1  sin2 x  e2 

y  1 ln 2 ln 1 sin2 x  e2 
2

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Lecture Note: 1 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Linear First-Order Differential Equations

 A linear first-order differential equation is one of the form dy  P xy  Qx.

dx

 The solution must be on an interval where both P x and Qx are continuous.

 The usual solution to the differential equation is to change it to an exact equation by means

of an integrating factor, V x.

Steps to solve the differential equations by means of Integrating Factor

 S: Write the equation in the form of dy  P xy  Qx.

dx

 I: Find the integrating factor, V x  e .Pxdx

 M: Multiply both sides of equation with V x,

V x dy  P xV xy  V xQx

dx

 P: Change the equation into product rule d V x y  V xQ x.
dx

 I: Integrate both sides  d V x y dx   V xQ x dx.
dx

 Y: State y as subject if necessary.

EXAMPLE 4

Find the general solution of the differential equation dy  2y  1  0.
dx x

dy  2y  1  0 2 dx
dx x x

S : dy  2 y  1 I : V (x)  e
dx x
 e2 ln x
M : x2  dy  x2  2 y  x2 1
 x2
dx x

P : d x2y  x2

dx

I: d x2ydx   x2dx

dx

x2y  x3  C
3

Y :y  x  C
3 x2

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Lecture Note: 1 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 5

The gradient of the curve is given by x  y . Find the equation of the curve which passes
x 1

through the point 0,2.

dy  x  y
dx x  1

S : dy  1 y  x
dx x  1

I : V (x)  e  1 dx
x1

 eln(x1)

1
x1

M :  x 1 1 dy  x 1 1 x 1 y   x 1 1 x
 dx  1 

P : d  x y 1  x
dx  x1

I :  d  x y 1dx   x dx
dx  x1

 x y 1   1  x 1 1 dx
 

= x  ln x  1  C

Y : y  x  1x  ln x  1  C

The curve passes through point 0,2,

2  (1)(0  ln1  C)
C 2

 y  x  1x  ln x  1  2

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Topic: 2.0 FIRST ORDER DIFFERENTIAL EQUATIONS

Sub-Topic: 2.3 Applications

Learning Outcomes: At the end of this lesson, students should be able to
(a) solve problems involving first order differential equations.

*Any applications including Newton’s Law of Cooling, electric circuits, population growth, radioactive
decay.
**Exclude: Formulation of differential equations.

Applications of Differential Equations

Population Growth Model
The simplest growth model has a constant relative growth rate. If we denote the population

we are considering by y t, then the rate of change of the population is dy . To say that the

dt
rate of change is proportional to the population is just saying that there is a constant of
proportionality k such that

dy  ky.
dt

 1 dy   kdt
y

ln y  kt  C

y  ektC

y  ekteC

y  Aekt, A  eC

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 6
In a particular bacteria culture, the rate of increase of bacteria is proportional to the number of

bacteria, N, present at time, t hours after the experiment and follows the equation dN  kN.
dt

Given that the number of bacteria at the beginning is 106, and after 1 hour is 109. Find
(a) the number of bacteria after 5 hours.

dN  kN
dt

 1 dN   kdt
N

ln N  kt  C

N  ektC

N  Aekt, A  eC

When t  0, N  106,
106  Ae0

A  106

When t  1, N  109,
109  106 ek1

k  ln 1000

 N  106 et ln1000

When t  5,
N  106 e5 ln1000

 1021

(b) the time taken for the number of bacteria to be 3 times the original.
When N  3 106,

 3 106  106 et ln1000

et ln1000  3
t ln1000  ln 3

t  ln 3
ln 1000

 0.159 hour

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Radioactive Decay Models

Radioactive decay models, on the other hand, are very accurate over long periods of time.
They are the primary method for determining age of prehistoric fossils and ancient artefacts.

If we denote the decay that we are considering by C t, then the decreasing rate of the decay

is dC . To say that the decreasing rate is proportional to the decay is just saying that there
dt

is a constant of proportionality k such that
dC  kC.
dt

 1 dC   kdt
C

ln C  kt  D

C  ektD

C  ekteD

C  Aekt, A  eD

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 7
Radium decomposes at a rate which is proportional to the amount present at any time and

follows the equation dR  kR. If 10% decomposes in 200 years, what percentage of the original
dt

amount of radium will remain after 1000 years?

dR  kR
dt

 1 dR   kdt
R

ln R  kt  C

R  ektC

R  Aekt

When t  0, R  100%,

100  A 1

A  100

When t  200, R  90%,

90  100ek200
e200k  9

10
200k  ln 9

10
k   1 ln 9

200 10

t ln 9

 R  100e200 10

When t  1000,

1000 ln 9

R  100e 200 10
 59.049%

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

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 d  k   a, where  t is the temperature of the

dt
object at any time t and a is the ambient temperature. The solution to this separable
differential equation is

d  k   a.

dt

  1 d   kdt
a

ln   a  kt  C

  a  ektC

  Aekt  a, A  eC

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 8
A pie is baked at 180 C is removed from an oven and put into the kitchen with a constant
temperature of 70 C. The pie is cooled off to 160 C after 5 minutes. Use Newton’s law of

cooling given by d  k    to find the time taken for the pie to cool off to 135 C.

dt

d  k    Given that   135 C at time t,

dt 135  70  110e0.0401t
65  110e0.0401t
 d  k dt e 0.0401t 65

  110
t  ln16150
  d   k dt
0.0401
   13.12

  d   k dt Therefore, the pie takes 13.12 minutes to cool off to
135 C.
 70

ln  70  kt  C

  70  ektC

  70  ektC

  70  Aekt

When t  0,   180 C,

180  70  A
A  110

   70  110ekt

When t  5,   160 C,

160  70  110e5k

90  110e5k

e5k  90
110

k  1 ln 191
5

 0.0401

   70  110e0.0401t

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Electric Circuits

RL circuit is one with a constant resistant, R, and a constant inductance, L. The electromotive
force, EMF, a resistor and an inductor connected in series. The EMF source which is usually
a battery or generator supplies voltage that causes a current flow in the circuit.
According to Kirchhoff’s Second Law, if the circuit is closed when t  0, then the applied
electromotive force is equal to the sum of the voltage drops in the rest of the circuit. It can be

shown that this implies that the current I t, that flows in the circuit at time t must satisfy

the first-order linear differential equation

L dI  RI  E
dt
dI  R I  E
dt L L

where P t  R and Qt  E are constants.

LL

The simplest model of the amount of current I in a simple electrical RL circuit is given by a
linear first-order differential equation,

dI  P tI  Qt

dt
where I is amount of current and t is time.

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

EXAMPLE 9
The basic equation governing the amount of current I in a simple RL circuit is given by
dI  50I  5. When t  0, I  0, find the current at any time t.
dt

dI  50I  5
dt

1 dI  dt
5  50I

 1 dI   dt
5  50I

ln 5  50I  t  C
50

ln 5  50I  50t  50C

5  50I  e50t50C

5  50I  Ae50t

 I  1 5  Ae50t

50

When t  0, I  0,

0  1 5  Ae500 
50

A5

  I  1 1  e50t

10

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Lecture Note: 2 of 2 Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Velocity-Acceleration

EXAMPLE 10
An object moves from rest in a straight line so that its acceleration after t seconds is given by
8  2v ms2, where v is its velocity.
(a) Express v in terms of t.

dv = 8  2v
dt

1 dv  dt
8  2v

 1 dv   dt
8  2v

ln 8  2v  t  C
2

ln 8  2v  2t  2C

8  2v  e2t2C

v  4  A e2t
2

When t  0, v  0,

0  4  A e20
2

A8

 v  4  4e2t

(b) Estimate the value of v after a long time.
As t  , e2t  0, then

v  4  40

 4 ms1

(c) Show that the distance travelled after 2 seconds is 6.0366 meter.
ds  4  4e2t
dt

   ds  2 4  4e2t dt
0
s  4t  2e2t 02
 6.0366 meter

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

1. (a) Solve y x  1  x2  2x dy . (b) Solve cos2 x dy  xy2, y 0  2.

dx dx

 x  1 dx   1 dy  1 dy   x x dx
x2  2x y y2 cos2

1  2x  1 dx   1 dy  y2dy  x sec2 xdx
2 y
x2  2x

1 ln x2  2x  ln y  C u  x,  dv   sec2 xdx
2
du  dx v  tan x

ln x2  2x  ln eC  ln y

ln A x2  2x  ln y  y2dy  x tan x   tan xdx

y  A x2  2x 1  x tan x   sin x dx
y cos x

u  cos x

du   sin x
dx
du   sin xdx

1  x tan x   1 du
y u

 1  x tan x  ln u  C
y

 1  x tan x  ln cos x  C
y

When x  0, y  2,

 1  0tan 0  ln cos 0  C

2
C 1

2

 1  x tan x  ln cos x  1
y2
y 2
1  2x tan x  2 ln cos x

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

2. (a) Solve the differential equation x x2  1  yey dy  0. u  x2  1
dx u2  x2  1
2u du  2x
x x2  1  yey dy dx
dx udu  xdx

x x2  1dx  yeydy

 yeydy   x x2  1dx u  y  dv   eydy
yey   eydy   uudu
du  dy v  ey

yey  ey  u2du

yey  ey  u3  C
3

yey  ey   3 C
x2  1

3

(b) Solve xydx  1  x2 dy  0, y 1  2.

1  x2 dy  xydx

 1 dy   x dx
y 1  x2

ln y   1 ln 1  x2  C
2

When x  1, y  2,

ln 2   1 ln 1  1  C
2

3

C  ln 22

 ln y 1 3
 ln 1  x2 2
 ln 22

 1 
ln y  ln  
82

1
1  x2 2 

y2 2
1  x2

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

3. Given that 2 dy  y2  4. Find the particular solution if given y ln 2  0.

dx

 2 dy   dx
4  y2

  1 y  1 y dy   dx 2  2
4  y2
22  22  2  y2  y

ln 2  y  ln 2y xC  AB
2 2 2y 2y
1

1 ln 2  y  1 ln 2  y  x  C 2  A2  y  B 2  y
22
When y  2, When y  2,
1 ln 2  y  x  C
2 2y 2  A 4 2  B4

A1 B1
2 2

When x  ln 2, y  0,  2  22 1 y  1
4  y2 
1 ln 2  0  ln 2  C 22  y
2 20

C   ln 2

1 ln 2  y  x  ln 2
2 2y

ln 42  y  2x

2y

42  y  e2x

2y

8  4y  e2x 2  y

8  4y  2e2x  ye2x

 y 4  e2x  2e2x  8

y  2e2x 8
4 e2x

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

4. Solve the following differential equations.

(a) tan yex dy  1.

dx

 tan ydy   exdx u  cos y
 sin y dy  ex  C du   sin y
dy
cos y du  sin ydy

 1 du  ex  C
u
ln u  ex  C
ln cos y  ex  C

ln cos y  ex  C

(b) x2 dy  3xy  1, y 2  0.

dx

dy  3 y  1
dx x x2

x3 dy  x3 3 y  x3 1 e 3 dx
dx x x2 x
V 
d x3y  x
 e3 ln x
dx
 x3
 d x3ydx   xdx

dx

x3y  x2  C
2

When x  2, y  0,

23 0  22  C

2
C  2

x3y  x2  2
2

y  1  2
2x x3

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

5. Solve dy  2y  cos t , y   0.
dt t t2

dy  2 y  cos t V  e 2 dt
dt t t2 t

t2 dy  t2 2y  t2 cos t  e2 ln t
dt t t2
 t2
d t2y  cos t

dt

 d t2ydt   cos tdt

dt

t2y  sin t  C

When t  , y  0,

2 0  sin   C

C 0

t2y  sin t  0

y  sin t
t2

6. Solve y  1 dx  x  2y  1, given that x  1, y  1.

dy

dx  1 x  2y  1 V  e 1 dy
dy y  1 y  1 y 1

y  1 dx  y  1 1 x  y  1 2y  1  eln(y1)

dy y  1 y 1  y1

d y  1 x  2y  1
dy

 d y  1xdy   2y  1dy
dy

y  1x  y2  y  C

When x  1, y  1,

1  11  12  1  C

C 2

y  1x  y2  y  2

x  y2  y  2
y 1

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

7. (a) Solve x dy  y  x3.
dx

dy  1 y  x2 V  e 1 dx
dx x x

x dy  x 1 y  xx2  eln x

dx x x

d xy  x3

dx

 d xydx   x3dx

dx

xy  x4  C
4

y  x3  C
4x

(b) Solve x dy  y  x2 cos x, given that y  0 when x  .
dx

dy  1 y  x cos x V  e 1 dx
dx x x

 1  dy   x1  1 y   1  x cos x  eln x
x dx x x
1
d  1 y  cos x x
dx x

 d xy  dx   cos xdx
dx

y  sin x  C
x

When x  , y  0,

0  sin   C

C 0

y  sin x  0
x
y  x sin x

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

8. Solve dy  y cot x  3 .
dx cos x

dy  cot xy  3 V  e cotxdx

dx cos x

sin x dy  sin xcot xy  sin x 3  e cos x dx
sin x
dx cos x

d y sin x  3 sin x  elnsin x

dx cos x  sin x

 d y sin xdx   3 sin x dx u  cos x
cos x
dx du   sin x
dx
y sin x    3 du du  sin xdx
u

y sin x  3 ln u  C

y sin x  3 ln cos x  C

y  3 ln cos x  C
sin x sin x

9. Solve the differential equation x dy  y  x sin x, y   1.

dx

dy  1 y  sin x V  e 1 dx
dx x x

x dy  x 1 y  x sin x  eln x
dx x
x
d xy  x sin x

dx

 d xydx   x sin xdx u  x,  dv   sin xdx

dx du  1 v   cos x
dx
xy  x cos x   cos xdx du  dx

xy  x cos x  sin x  C

When x  , y  1,

   cos   sin   C
C 0

xy  x cos x  sin x
y   cos x  sin x
x

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

10. The number of a certain bacterial culture has increases from 5000 to 15000 within 10 hours.

The rate of increase of the population is given by dx  kx.
dt

(a) Find the formula for the number of bacteria at time t.

dx  kx
dt

 1 dx   kdt
x

ln x  kt  C

x  ektC

x  Aekt

When t  0, x  5000,

5000  Aek0
A  5000

When t  10, x  15000,

15000  5000ek10
3  e10k
k  ln 3
10

t ln 3

 x t  5000e 10

(b) Find the number of bacteria when t  20 hours. When will the number of bacteria

become 50 000? Let  xt the number of bacteria at time t

When t  20 hours,

20 ln 3

x  5000e 10
 45000

When x  50000,

t ln 3

50000  5000e 10
t  20.96 hours

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

11. Consider a simple electric circuit with the resistance of 3 Ω and inductance of 2 H. If a

battery gives a constant voltage of 24 V and the switch is closed when t  0, the current

I t, after t seconds is given by dI  3 I  12, I 0  0.

dt 2

(a) Obtain I t.

dI  3 I  12
dt 2

3t dI  3 e 3 t 3t
2
e2 I  12e2
dt 2

d e 3 t I   12e 3 t
dt 2 2

V  e 3 dt
2

 d e 3 t I dt  3t 3t
dt 2
12e2 dt  e2

3t  3t C

e2 I 12e 2
3

2

3 t

I  8  Ce 2

When t  0, I  0,

0  8  Ce320
C  8

 I t  8  3 t

8e 2

(b) Determine the difference in the amount of current flowing through the circuit from the
fourth to the eight seconds. Give your answer correct to 3 decimal places.

I 8  I 4  8  38  8  34 

8e 2 8e 2

 7.9999508  7.98017

 0.020

(c) If current is allowed to flow through the circuit for a very long period of time, estimate I t.

As t  , e3 t  0,
2

I  8  80

8

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Tutorial 2: 4 Hours Mathematics 2 SM025
Ch. 2 Differential Equations Session 2021/2022

Answers 2
1. (a) y  A x2  2x (b) y 

1  2x tan x  2 ln cos x

2. (a) yey  ey   3 C 2
x2  1 (b) y  2

3 1  x2

2e2x  8
3. y 

4  e2x

4. (a) ln cos y  ex  C 12
(b) y  
sin t
5. y  2x x3

t2
y2  y  2
6. x 

y 1

x3 C (b) y  x sin x
7. (a) y  

4x

8. y  3 ln cos x  C
sin x sin x

sin x
9. y   cos x 

x

t ln 3

10. (a) x t  5000e 10 (b) x  45000, t  20.96 hours

3
t

11. (a) I t  8  8e 2 (b) 0.020 (c) 8

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