integratingfactorsfoundbyinspection-131008062306-phpapp01

8/20/2019 integratingfactorsfoundbyinspection-131008062306-phpapp01

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Integrating FactorsFound byInspection


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This section will use the followingfour exact dierentials that occurs

frequently :

( )d xy xdy ydx• = +

2

x ydx xdyd y y

−• = ÷

2

y xdy ydxd

x x

−

• = ÷

2 2arctan

y xdy ydxd

x x y

− • = ÷ +


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From

where :

Answer

2 2arctan

y xdy ydxd

x x y

− = ÷ + 2(arctan ) 1du

d uu

=+

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

÷


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Process :

1. Regroup terms of like degree to form

equations from those exact differentials

given.

4. Simplify further.

3. Integrate.

. Su!stitute the terms "ith their

corresponding equivalent of exact

differentials.


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Examples :

Group terms of like degree

Divide by y

1. (2 1) 0 y xy dx xdy+ − =

2 02

ydx xdy xdx

y

−+ =

2

2 0 xy dx ydx xdy+ − =22 ( ) 0 xy dx ydx xdy+ − =

2

22 ( ) 0 xy dx ydx xdy

y+ − =


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Integrate each term

From

#y po"er formula

2 0 x

xdx d y

+ = ÷

2 0 x

xdx d y

+ = ÷

∫ ∫

2

x ydx xdyd

y y

−= ÷

2

22

x xc

y

+ = ÷

....2 02

ydx xdy

xdx y

−

+ =


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Answer

Multiply each terms by y to

eliminate fractions

2 x y x cy+ =

( 1) x xy cy+ =

2 x x c y y

+ =

2

......22

x xc

y

+ = ÷


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orm one of the ! exact differentials given

by factoring common factors on each term

3 32. ( ) ( ) 0 y x y dx x x y dy− − + =3 2 4

0 x ydx y dx x dy xydy− − − =

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

3( ) ( ) 0 x ydx xdy y ydx xdy− − + =


9/28

Divide each term by y" to form an exact differential

From

$3 ( ) 0

x d xy x d

y y

− = ÷

3

2

( ) ( ) 0 x ydx xdy y ydx xdy

y

− − + =

2

x ydx xdyd

y y

−= ÷

( )d xy xdy ydx= +

3...... ( ) ( ) 0 x ydx xdy y ydx xdy− − + =


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Assume an integrating factor is xkyn#

Distribute to each term

$ince we can%t

integrate directly

( )3 0k n k nd xy x

x d x y x y y y

− = ÷

( )3 0 k nd xy x

x d x y y y

− = ÷

31 ( ) 0

k k n

n

x xd x y d xy

y y

+−

−

− = ÷

3 ( ) 0 x d xy

x d

y y

− = ÷


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$olve for k and n#

E&uate :

E&uate :

1

$ubstitute " in ' # $ubstitute n in ' #

@ x

d

y

÷

3 x k = + y n= −

3 k n+ = −

( )@ d xy x k =1 y n= − 1k n= −

3 ( 1)n n+ − = −

1n = −

3 ( 1)k + = − −

2k = −2 2n = − 1 3k = −

31 ( ) 0

k k n

n

x xd x y d xy

y y

+−

−

− = ÷


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(hus) $ubstitute *n + ,' - k + ,". to

Integrate each term

#y po"er formula

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

− = ÷

( ) 2

( )

0

x x d xy

d y y xy

− = ÷

31 ( ) 0

k k n

n

x xd x y d xy

y y

+−

−

− = ÷


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Multiply by "

"here $

Answer

2

( )...... 0

( )

x x d xyd

y y xy

− =

÷ ∫ ∫

( )

2

1

02 1

x

xy y

c

− ÷

− + =−

2

1 10 2

2 2

x c

y xy

+ + = ÷

2

cc =

2

2

20

xc

y xy+ + =


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Divide by *x"/y".

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

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 ydx ydx xdy

x y

−+ + =+


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Integrate each term

Answer

2 2......( ) 0

xdy ydx

ydx xdy x y

−

+ + =+

( ) arctan y

d xy d x

+ ÷

arctan y

xy c x

+ = ÷

( ) arctan y

d xy d x

+ ÷ ∫ ∫

2 2

( )

arctan

from

d xy ydx xdy

y xdy ydxd x x y

= +−

= ÷ +


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Divide by y to integrate each term

Integrate each term

#y po"er formula

2...... ( ) 2 0 xy ydx xdy xydx dy+ + − =

2( ) 2

0 xy ydx xdy xydx dy

y y y

++ − =

( ) ( ) 2 0dy

xyd xy xd x y

+ − = ( ) from d xy ydx xdy→ = +

( ) ( ) 2 0dy

xyd xy xd x y

+ − =∫ ∫ ∫


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'ultiply !y to

eliminate fractions

"here c % c

general solution

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

xyd xy xd x y+ − =∫ ∫ ∫

( )( )

2 2

2 ln2 2

xy x y c+ − =

2 2( ) 4(ln ) 2 xy x y c+ − =

( ) ( )2

22 ln 2

2 2 xy x y c + − =

2 2 2 4(ln ) x y x y c+ − =2 2( 1) 4(ln ) x y y c+ − =


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(hen x%1 & y%1

Solve for c $

Su!stitute c in the general solution

particular solution

2 2 2(1 1 ) 1 4(ln1) c• + − =

2 0 c− =

2c =

2 2( 1) 4(ln ) x y y c+ − =

2 2( 1) 4(ln ) 2 x y y+ − =2 2( 1) 2 4(ln ) x y y+ = +


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x % & y % )


'ultiply !y *1

2 2 2 25. ( ) ( ) 0 x x y x dx y x y dy− − − − =3 2 2 2 3 0 x dx xy dx x dx x ydy y dy− − − + =

2 2 3 3 2 0 xy dx x ydy x dx y dy x dx− − + + − =

2 2 3 3 2

0 xy dx x ydy x dx y dy x dx+ − − + =( )2 2 3 2 3 0 xy dx x ydy x dx x dx y dy+ − + − =


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orm one of the ! exact differentials given by

factoring common factor on the grouped term*s.

Integrate each term

#y po"er formula

( )2 2 3 2 3

...... 0 xy dx x ydy x dx x dx y dy+ − + − =

( ) 3 2 3 0 xy ydx xdy x dx x dx y dy+ − + − =3 2 3( ) 0 xyd xy x dx x dx y dy− + − = ( ) from d xy ydx xdy→ = +

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

− + − =∫ ∫ ∫ ∫ 2 4 3 4( )2 4 3 4

xy x x yc− + − =


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'ultiply !y their +,- % 1 to eliminate the fractions

"here 1c % c

general solution

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

− + − =

2 4 3 46( ) 3 4 3 12 xy x x y c− + − =2 4 3 46( ) 3 4 3 xy x x y c− + − =


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(hen x % & y % )

Solve for c $

Su!stitute c in the general solution

particular solution

2 4 3 46( ) 3 4 3 xy x x y c− + − =

2 4 3 46(2 0) 3(2) 4(2) 3(0) c• − + − =

48 32 c− + =16c = −

2 4 3 4

6( ) 3 4 3 16 xy x x y− + − = −2 2 4 4 36 3 3 4 16 x y x y x− − = − −

2 2 4 4 33(2 ) 4( 4) x y x y x− − = − +


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