Engineering Mathematics2

2008_O.D.E(1)

Naval Architecture & Ocean Enginee ring

Engineering Mathematics 2

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering

[2008][01-2]

September, 2008

2008_O.D.E(1)

Naval Architecture & Ocean Enginee ring

Ordinary Differential Equations (1)

Simple Integration Separable Variables

Spring/Mass Systems : Driven Motion

2008_O.D.E(1)

Differential Equation

*Zill D.G., Cullen M.R., Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p5

3 /364

2008_O.D.E(1)

Differential Equation

Differential Equation

An equation containing the derivatives of one or more dependent variables, with respect to one or more

independent variables, is said to be a Differential Equation(DE)

Definition 1.1*

dx xy x

dy ( ) 0 . 2



*Zill D.G., Cullen M.R., Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p5

4 /364

2008_O.D.E(1)

Differential Equation

Differential Equation

An equation containing the derivatives of one or more dependent variables, with respect to one or more

independent variables, is said to be a Differential Equation(DE)

Definition 1.1*

dx xy x

dy ( ) 0 . 2



How to solve a

Differential Equation?

*Zill D.G., Cullen M.R., Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p5

5 /364

2008_O.D.E(1)

Simple Integration



6 /364

2008_O.D.E(1)

Simple Integration



How to solve a

Differential Equation?

7 /364

2008_O.D.E(1)

Simple Integration



How to solve a

Differential Equation?

Integration!

8 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

How to solve a

Differential Equation?

Integration!

9 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

How to solve a

Differential Equation?

Integration!

10 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

How to solve a

Differential Equation?

Integration!

X=0 X=L

w₀

y

x

11 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

Deflection of beam How to solve a

Differential Equation?

Integration!

X=0 X=L

w₀

y

x

12 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

Deflection of beam

We need 4 initial condition

because we‟ll integrate 4 times (4 integral coefficient)

How to solve a

Differential Equation?

Integration!

X=0 X=L

w₀

y

x

13 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

There is no vertical

deflection at end points so Deflection of beam

We need 4 initial condition

because we‟ll integrate 4 times (4 integral coefficient)

How to solve a

Differential Equation?

Integration!

X=0 X=L

w₀

y

x

14 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

There is no vertical

deflection at end points so Deflection of beam

We need 4 initial condition

because we‟ll integrate 4 times (4 integral coefficient)

How to solve a

Differential Equation?

Integration!

X=0 X=L

w₀

y

x

15 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the

Beam if a constant load w

is uniformly distributed along its length

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

There is no vertical

deflection at end points so Deflection of beam

We need 4 initial condition

because we‟ll integrate 4 times (4 integral coefficient)

(4 initial conditions) How to solve a

Differential Equation?

Integration!

X=0 X=L

w₀

y

x

16 /364

2008_O.D.E(1)

Simple Integration



17 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

X=0 X=L

w₀

y

x

18 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

X=0 X=L

w₀

y

x

19 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

X=0 X=L

w₀

y

x

20 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

0 3 2

4 3

2

2 3 6

)

( x

EI x w

c x

c c

x

y     

X=0 X=L

w₀

y

x

21 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

0 ) 0 ( 

y

y  ( 0 )  0

0 3 2

4 3

2

2 3 6

)

( x

EI x w

c x

c c

x

y     

X=0 X=L

w₀

y

x

22 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

0 ) 0 ( 

y

y  ( 0 )  0

0 ,

0

1

 c 

c

0 3 2

4 3

2

2 3 6

)

( x

EI x w

c x

c c

x

y     

X=0 X=L

w₀

y

x

23 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

0 ) 0 ( 

y

y  ( 0 )  0

0 ,

0

1

 c 

c 0 ) ( L 

y

^{y  ( L )  0}

0 3 2

4 3

2

2 3 6

)

( x

EI x w

c x

c c

x

y     

X=0 X=L

w₀

y

x

24 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

0 ) 0 ( 

y

y  ( 0 )  0

0 ,

0

1

 c 

c 0 ) ( L 

y

^{y  ( L )  0}

6 0 3

3

24 0

0 3 2

4 3

0 4 3

4 2

3













EI L L w

c L

c

EI L L w

c L

c

0 3 2

4 3

2

2 3 6

)

( x

EI x w

c x

c c

x

y     

X=0 X=L

w₀

y

x

25 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

) ) (

(

4 0 4

x dx w

x y

EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2 3 2

1

24

)

( x

EI x w

c x

c x c c x

y     

0 ) 0 ( 

y

y  ( 0 )  0

0 ,

0

1

 c 

c 0 ) ( L 

y

^{y  ( L )  0}

6 0 3

3

24 0

0 3 2

4 3

0 4 3

4 2

3













EI L L w

c L

c

EI L L w

c L

c

0 3 2

4 3

2

2 3 6

)

( x

EI x w

c x

c c

x

y     

EI L c w

EI L c w

, 12 24

0 4

2 0

3

  

X=0 X=L

w₀

y

x

26 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

4 0 4

dx w y EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

X=0 X=L

w₀

y

x

27 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

4 0 4

dx w y EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2

3 2

1

24

)

( x

EI x w

c x

c x c c

x

y     

, 0 ,

0

1

 c 

c EI

L c w

EI L c w

, 12 24

0 4

2 0

3

  

X=0 X=L

w₀

y

x

28 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

4 0 4

dx w y EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2

3 2

1

24

)

( x

EI x w

c x

c x c c

x

y     

, 0 ,

0

1

 c 

c EI

L c w

EI L c w

, 12 24

0 4

2 0

3

  

0 4 0 3

2 2 0

24 12

) 24

( x

EI x w

EI L x w

EI L x w

y   



X=0 X=L

w₀

y

x

29 /364

2008_O.D.E(1)

Simple Integration

 Example 1

An Embedded Beam

4 0 4

dx w y EI d 

Find the deflection of the Beam

0 )

0 (  y

0 )

0 (  y 

0 )

( L  y

0 )

( 

 L y

After integrate four times,

0 4 3

4 2

3 2

1

24

)

( x

EI x w

c x

c x c c

x

y     

, 0 ,

0

1

 c 

c EI

L c w

EI L c w

, 12 24

0 4

2 0

3

  

0 4 0 3

2 2 0

24 12

) 24

( x

EI x w

EI L x w

EI L x w

y   



2

0 2

( )

) 24

( x x L

EI x w

y  

or

X=0 X=L

w₀

y

x

30 /364

2008_O.D.E(1)

Separable Variables

31 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

How to solve a

Differential Equation?

32 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

How to solve a

Differential Equation?

Integration!

33 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x

dP  Integration!!

How to solve a

Differential Equation?

Integration!

34 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x

dP  Integration!!

C x

P dx dx

x

dP  

 ^{( ) ( )}

L.H.S:

dx x P k dx

x

kP 

 ^{( )  ( )}

R.H.S:







 P ( x ) C k P ( x ) dx

How to solve a

Differential Equation?

Integration!

35 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x

dP  Integration!!

C x

P dx dx

x

dP  

 ^{( ) ( )}

L.H.S:

dx x P k dx

x

kP 

 ^{( )  ( )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

How to solve a

Differential Equation?

Integration!

36 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x

dP  Integration!!

C x

P dx dx

x

dP  

 ^{( ) ( )}

L.H.S:

dx x P k dx

x

kP 

 ^{( )  ( )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

How to solve a

Differential Equation?

Integration!

Then, how?

37 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

Integration!!

C x

dx P x

dP  

 ^{( ) ( )}

L.H.S:



 ^{kP ( x )  k P ( x )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

38 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

Integration!!

C x

dx P x

dP  

 ^{( ) ( )}

L.H.S:



 ^{kP ( x )  k P ( x )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

transform

39 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

Integration!!

C x

dx P x

dP  

 ^{( ) ( )}

L.H.S:



 ^{kP ( x )  k P ( x )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

P kdx x dP ( ) 

transform

40 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

Integration!!

C x

dx P x

dP  

 ^{( ) ( )}

L.H.S:



 ^{kP ( x )  k P ( x )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

Separable Variables

P kdx x dP ( ) 

transform

41 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

Integration!!

C x

dx P x

dP  

 ^{( ) ( )}

L.H.S:



 ^{kP ( x )  k P ( x )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

Separable Variables

P kdx x dP ( ) 

kx kdx 



R.H.S:

kx c

x

P ( )   ln

c x

p P x

dP  

 ^{( ) ln ( )}

L.H.S:

transform

42 /364

2008_O.D.E(1)

Separable Variables

Ex) Population dynamics

) ) (

( kP x

dx x dP 

Integration!!

C x

dx P x

dP  

 ^{( ) ( )}

L.H.S:



 ^{kP ( x )  k P ( x )}

R.H.S:







 P ( x ) C k P ( x ) dx

solved?

Separable Variables

P kdx x dP ( ) 

kx kdx 



R.H.S:

kx c

x

P ( )   ln

c x

p P x

dP  

 ^{( ) ln ( )}

L.H.S:

transform

43 /364

2008_O.D.E(1)

Separable Variables

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

44 /364

2008_O.D.E(1)

Separable Variables

) , ( x y dx f

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

45 /364

2008_O.D.E(1)

Separable Variables

) , ( x y dx f

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

46 /364

2008_O.D.E(1)

Separable Variables

 ^





 g x y g x dx c dx

dy ( ) ( )

) , ( x y dx f

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

47 /364

2008_O.D.E(1)

Separable Variables

 ^





 g x y g x dx c dx

dy ( ) ( )

) , ( x y dx f

dy 

) , ( x y dx f

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

48 /364

2008_O.D.E(1)

Separable Variables

 ^





 g x y g x dx c dx

dy ( ) ( )

) , ( x y dx f

dy 

) , ( x y dx f

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

49 /364

2008_O.D.E(1)

Separable Variables

 ^





 g x y g x dx c dx

dy ( ) ( )

) , ( x y dx f

dy 



 ^{ }



 g x dx c

y h y dy

h x dx g

dy ( )

) ) (

( ) ) (

, ( x y dx f

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

50 /364

2008_O.D.E(1)

Separable Variables

 ^





 g x y g x dx c dx

dy ( ) ( )

) , ( x y dx f

dy 



 ^{ }



 g x dx c

y h y dy

h x dx g

dy ( )

) ) (

( ) ) (

, ( x y dx f

dy 

Separable Equation A first order differential equation of the form

is said to be separable or to have separable variables Definition 2.1*

) ( )

( x h y dx g

dy 

*Dennis G. Zill, Michael R. Cullen, Advanced Engineering Mathematics 3rd Edition, Johns and Bartlett, 2006, p45

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Separable Variables





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Separable Variables

 Population Dynamics

 Modeled by English economist Thomas Malthus in 1798

① Population varies in time P=P(t)

② Population of country grows at a certain time is proportional to the total population

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Separable Variables

 Population Dynamics

 Modeled by English economist Thomas Malthus in 1798

① Population varies in time P=P(t)

② Population of country grows at a certain time is proportional to the total population

dP dP

P or kP

dt  dt 

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Separable Variables





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Separable Variables

 Newton‟s Law of Cooling/Warming

 The rate at which the temperature of a body changes is proportional to the difference between temperature of the body and the temperature of the surrounding medium

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Separable Variables

 Newton‟s Law of Cooling/Warming

 The rate at which the temperature of a body changes is proportional to the difference between temperature of the body and the temperature of the surrounding medium

( )

m m

dT dA

T T or k T T

dt   dt  

m

T :body temperature

T :Surronding medium temperature

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Separable Variables

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2008_O.D.E(1)

Separable Variables

t

T

o

T

T

T

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

t

T

o

T

T

T

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

t

T

o

T

T

T

Great Idea !!

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

Ex. )Newton’s law of cooling

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Separable Variables

t

T

o

T

T

T

Relation between and ^{T  T}

Great Idea !!

dt t dT ) (

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

t A T

o

T

T

T

Relation between and ^{T  T}

Great Idea !!

dt t dT ) (

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0 ),

(  

 k T T k dt

dT

t A T

o

T

T

T

Relation between and ^{T  T}

Great Idea !!

dt t dT ) (

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

) ( T T _A dt k

dT  

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0 ),

(  

 k T T k dt

dT

t A T

o

T

T

T

Relation between and ^{T  T}

Great Idea !!

dt t dT ) (

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

) ( T T _A dt k

dT  

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0 ),

(  

 k T T k dt

dT

t A T

o

T

T

T

Relation between and ^{T  T}

Great Idea !!

dt t dT ) (

T T

T

T

 ,

 ratures Body tempe

Initial :

t) re(constan temperatu

outside :

rature body tempe

:

2 1

,T T T T

A

) ( T T _A dt k

dT  

dt T k

T

dT

A



 

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

T A

T Y  

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

 1 dT dY

T A

T Y  

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

 1 dT dY

T A

T Y  

dT dY 

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

 1 dT dY

T A

T Y  

dT dY 

dt T k

T dT

A



 

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

dt Y k

dY  

 1 dT dY

T A

T Y  

dT dY 

dt T k

T dT

A



 

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

dt Y k

dY  

 1 dT dY

T A

T Y  

dT dY 

dt T k

T dT

A



 



 ^dY _Y ^{ k  dt}

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

dt Y k

dY  

 1 dT

dY ^{L R}

c kt

c

Y   

A ln

T T

Y  

dT dY 

dt T k

T dT

A



 



 ^dY _Y ^{ k  dt}

Ex. )Newton’s law of cooling

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Separable Variables

0 ),

(  

 k T T k dt

dT

A

dt T k

T

dT

A



 

dt Y k

dY  

 1 dT

dY ^{L R}

c kt

c

Y   

A ln

T T

Y  

dT dY 

dt T k

T dT

A



 



 ^dY _Y ^{ k  dt}

c kt

c c

kt

Y _{R L}









 ln

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0

), (





 k

T T

dt k dT

A

c kt

Y   ln

① ② ③

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0

), (





 k

T T

dt k dT

A

c kt

Y   ln

c Y kt

e e ^ln  ^

① ② ③

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0

), (





 k

T T

dt k dT

A

c kt

Y   ln

c Y kt

e e ^ln  ^

c

e kt

Y  ^

① ② ③

Ex. )Newton’s law of cooling

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2008_O.D.E(1)

Separable Variables

0

), (





 k

T T

dt k dT

A

c kt

Y   ln

c Y kt

e e ^ln  ^

c

e kt

Y  ^

e kt

c Y  ~

① ② ③

Ex. )Newton’s law of cooling

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