integrating factors found by inspection
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Integrating factors found by inspectionTRANSCRIPT
Integrating Factors Found by Inspection
This section will use the following four exact differentials that occurs frequently :
( )d xy xdy ydx
2
x ydx xdydy y
2
y xdy ydxdx x
2 2arctan
y xdy ydxd
x x y
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 ydxxyx
2
2
21
xdy ydxxyx
2
2 2
2
xdy ydxx
x yx
2
2 2 2
xdy ydx x
x x y
2 2
xdy ydx
x y
arctan
yd
x
Answer
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.
Examples :
1. (2 1) 0y xy dx xdy
Group terms of like degree
Divide by y2
2 02
ydx xdyxdx
y
22 0xy dx ydx xdy
22 ( ) 0xy dx ydx xdy 2
2
2 ( ) 0xy dx ydx xdy
y
2 0x
xdx dy
Integrate each term
From
2 0x
xdx dy
2
x ydx xdydy y
By power formula
2
22
x xc
y
....2 02
ydx xdyxdx
y
2x y x cy
( 1)x xy cy Answer
2 xx c y
y
Multiply each terms by y to eliminate fractions
2
......22
x xc
y
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
Group terms of like degree
Form one of the 4 exact differentials given by factoring common factors on each term
3 ( )0
x d xyx d
y y
3
2
( ) ( ) 0x ydx xdy y ydx xdy
y
Divide each term by y2 to form an exact differential
From :
2
x ydx xdydy y
( )d xy xdy ydx
3...... ( ) ( ) 0x ydx xdy y ydx xdy
Assume an integrating factor is xkyn.
3 0k n k nd xyxx d x y x y
y y
3 0 k nd xyxx d x y
y y
31 ( ) 0
kk n
n
x xd x y d xy
y y
Distribute to each term
3 ( )0
x d xyx d
y y
Since we can’t integrate directly
Solve for k and n.
@x
dy
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
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 xydy y xy
2 2
( )0
x x d xydy y x y
Integrate each term 2( )
0x x d xydy y xy
31 ( ) 0
kk n
n
x xd x y d xy
y y
By power formula
2
( )...... 0
( )
x x d xydy y xy
2
1
02 1
xxyy
c
Multiply by 2
21 1
0 22 2
x c
y xy
where :
2
cc
2
2
20
xc
y xy Answer
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
Integrate each term
2 2......( ) 0
xdy ydxydx xdy
x y
( ) arctany
d xy dx
arctany
xy cx
Answer
( ) arctany
d xy dx
2 2
( )
arctan
from
d xy ydx xdy
y xdy ydxd
x x y
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
Group terms of like degree
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
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
...... ( ) ( ) 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 2
2 ln 22 2
xy xy c
where 2c = c
2 2 2 4(ln )x y x y c
general solution2 2( 1) 4(ln )x y y c
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
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
Group terms of like degree
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
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 term3 2 3( ) 0xyd xy x dx x dx y dy By power formula
2 4 3 4( )
2 4 3 4
xy x x yc
2 4 3 4( )......
2 4 3 4
xy x x yc
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 = c
2 4 3 46( ) 3 4 3xy x x y c general solution
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 solution2 4 3 46( ) 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
