Integrating Factors Found by Inspection.pptx

8/10/2019 Integrating Factors Found by Inspection.pptx

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This section will use the following four

exact differentials that occurs

frequently :

( )d xy xdy ydx

2

x ydx xdydy y

2

y xdy ydxd

x x

2 2arctan

y xdy ydxd

x x y


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2 2arctan

y xdy ydxd

x x y

From 2(arctan ) 1

dud u

u

where :

yu

x

2

xdy ydxdu

x

2

2

1

xdy ydx

x

y

x

2

2

21

xdy ydx

x

y

x

2

2 2

2

xdy ydx

x

x y

x

2

2 2 2

xdy ydx x

x x y

2 2

xdy ydx

x y

arctan

yd

x

Answer


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Process :

1. Regroup terms of like degree to form

equations from those exact differentials

given.

4. Simplify further.

3. Integrate.

2. Substitute the terms with their

corresponding equivalent of exact

differentials.


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Examples :

1. (2 1) 0y xy dx xdy

Group terms of like degree

Divide by y2

2 0

2

ydx xdyxdx

y

2

2 0xy dx ydx xdy

22 ( ) 0xy dx ydx xdy 2

22 ( ) 0xy dx ydx xdy

y


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2 0x

xdx dy

Integrate each term

From

2 0x

xdx dy

2

x ydx xdyd

y y

By power formula

2

22

x xc

y

....2 02

ydx xdyxdx

y


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2x y x cy

( 1)x xy cy Answer

2 xx c yy

Multiply each terms by y to

eliminate fractions

2

......22

x xc

y


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3 32. ( ) ( ) 0y x y dx x x y dy 3 2 4 0x ydx y dx x dy xydy

3 4 2( ) ( ) 0x ydx x dy y dx xydy

3 ( ) ( ) 0x ydx xdy y ydx xdy


Form one of the 4 exact differentials given

by factoring common factors on each term


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3 ( ) 0x d xy

x d

y y

3

2

( ) ( ) 0x ydx xdy y ydx xdy

y

Divide each term by y2to form an exact differential

From :

2

x ydx xdyd

y y

( )d xy xdy ydx

3...... ( ) ( ) 0x ydx xdy y ydx xdy


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Assume an integrating factor is xkyn.

3 0k n k nd xyx

x d x y x yy y

3 0 k nd xyx

x d x yy y

31 ( ) 0

kk n

n

x xd x y d xy

y y

Distribute to each term

3 ( ) 0x d xy

x d

y y

Since we cant integrate directly


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Solve for k and n.

@ x

d

y

3x k

y n

Equate : 3 k n

@ d xy x k1y n

Equate : 1k n

1

2

Substitute 2 in 1 .

3 ( 1)n n

1n

Substitute n in 1 .

3 ( 1)k

2k 2 2n 1 3k

31 ( ) 0

kk n

n

x xd x y d xy

y y


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Thus; Substitute (n = -1 & k = -2) to

3 ( 2 )2 1 1

( 1) ( ) 0

x xd x y d xy

y y

2

( )0

( )

x x d xyd

y y xy

2 2

( )0

x x d xyd

y y x y

Integrate each term 2

( )

0

x x d xy

dy y xy

31 ( ) 0

kk n

n

x xd x y d xy

y y

By power formula


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2

( )...... 0

( )

x x d xyd

y y xy

2

1

02 1

x

xyy

c

Multiply by 2

2

1 10 2

2 2

x c

y xy

where :2

cc

2

2

20

xc

y xy Answer


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2 2 2 23. ( 1) ( 1) 0y x y dx x x y dy 2 2 2 2( ) ( ) 0y x y dx x x y dy xdy ydx

Divide by (x2+y2)

2 2 2 2

2 2 2 2 2 2

( ) ( )0

( ) ( ) ( )

y x y dx x x y dy xdy ydx

x y x y x y

2 2( ) 0

xdy ydxydx xdy

x y


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Integrate each term

2 2......( ) 0

xdy ydx

ydx xdy x y

( ) arctan y

d xy d x

arctan y

xy cx

Answer

( ) arctan y

d xy d x

2 2

( )

arctan

from

d xy ydx xdy

y xdy ydxd x x y


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2 2 24. ( 1) ( 2) 0xy y dx x y dy 3 2 2 2 0xy dx xydx x y dy dy

3 2 2( ) 2 0xy dx x y dy xydx dy


2 ( ) 2 0xy ydx xdy xydx dy

Form one of the 4 exact differentials given by

factoring common factor on the grouped term

x = 1 , y = 1


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2...... ( ) 2 0xy ydx xdy xydx dy

2 ( ) 20

xy ydx xdy xydx dy

y y y

Divide by y to integrate each term

( ) ( ) 2 0dy

xyd xy xd xy

( )from d xy ydx xdy

Integrate each term

( ) ( ) 2 0dy

xyd xy xd xy

By power formula


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...... ( ) ( ) 2 0dy

xyd xy xd xy

2 2

2 ln2 2

xy xy c Multiply by 2 to

eliminate fractions

2 2( ) 4(ln ) 2xy x y c

2

22 ln 2

2 2xy x y c

where 2c = c

2 2 2 4(ln )x y x y c

general solution2 2( 1) 4(ln )x y y c


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When x=1 , y=1

2 2 2(1 1 ) 1 4(ln1) c

Solve for c :

2 0 c

2c

Substitute c in the general solution

2 2( 1) 4(ln )x y y c

2 2( 1) 4(ln ) 2x y y 2 2( 1) 2 4(ln )x y y particular solution


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2 2 2 25. ( ) ( ) 0x x y x dx y x y dy x = 2 , y = 03 2 2 2 3 0x dx xy dx x dx x ydy y dy

2 2 3 3 2 0xy dx x ydy x dx y dy x dx


Multiply by -1

2 2 3 3 2

0xy dx x ydy x dx y dy x dx 2 2 3 2 3 0xy dx x ydy x dx x dx y dy


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Form one of the 4 exact differentials given by

factoring common factor on the grouped term(s)

2 2 3 2 3

...... 0xy dx x ydy x dx x dx y dy

3 2 3 0xy ydx xdy x dx x dx y dy 3 2 3( ) 0xyd xy x dx x dx y dy ( )from d xy ydx xdy

Integrate each term

3 2 3( ) 0xyd xy x dx x dx y dy

By power formula2 4 3 4( )2 4 3 4

xy x x yc


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2 4 3 4( )

...... 2 4 3 4

xy x x y

c

2 4 3 4( )12

2 4 3 4

xy x x yc

Multiply by their LCD = 12 to eliminate the fractions

2 4 3 46( ) 3 4 3 12xy x x y c where 12c = c2 4 3 46( ) 3 4 3xy x x y c general solution


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2 4 3 46( ) 3 4 3xy x x y c When x = 2 , y = 0

Solve for c :2 4 3 46(2 0) 3(2) 4(2) 3(0) c

48 32 c 16c

Substitute c in the general solution

2 4 3 4

6( ) 3 4 3 16xy x x y 2 2 4 4 36 3 3 4 16x y x y x

2 2 4 4 33(2 ) 4( 4)x y x y x particular solution


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Integrating Factors Found by Inspection.pptx

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