1. The document discusses using exact differentials to solve integration problems. 2. It provides examples of using exact differentials and integrating terms to find solutions. 3. The solutions found are particular solutions for the given values of x and y in each problem.

2. This section will use the following four
exact differentials that occurs
frequently :
( )d xy xdy ydx
2
x ydx xdy
d
y y
2
y xdy ydx
d
x x
2 2
arctan
y xdy ydx
d
x x y

3. 2 2
arctan
y xdy ydx
d
x x y From 2
(arctan )
1
du
d u
u
where :
y
u
x 2
xdy ydx
du
x
2
2
1
xdy ydx
x
y
x
2
2
2
1
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
y
d
x
Answer

4. 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.

5. Examples :
1. (2 1) 0y xy dx xdy
Group terms of like degree
Divide by y2
2 0
2
ydx xdy
xdx
y
2
2 0xy dx ydx xdy
2
2 ( ) 0xy dx ydx xdy
2
2
2 ( ) 0xy dx ydx xdy
y

6. 2 0
x
xdx d
y
Integrate each term
From
2 0
x
xdx d
y
2
x ydx xdy
d
y y
By power formula
2
2
2
x x
c
y
....2 0
2
ydx xdy
xdx
y

7. 2
x y x cy
( 1)x xy cy Answer
2 x
x c y
y
Multiply each terms by y to
eliminate fractions
2
......2
2
x x
c
y

8. 3 3
2. ( ) ( ) 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

9. 3 ( )
0
x d xy
x 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 xdy
d
y y
( )d xy xdy ydx
3
...... ( ) ( ) 0x ydx xdy y ydx xdy

10. Assume an integrating factor is xkyn.
3
0k n k nd xyx
x d x y x y
y y
3
0 k nd xyx
x d x y
y y
3
1
( ) 0
k
k n
n
x x
d x y d xy
y y
Distribute to each term
3 ( )
0
x d xy
x d
y y
Since we can’t integrate directly

11. Solve for k and n.
@
x
d
y
3x k
y n
Equate : 3 k n
@d xy x k
1y 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
3
1
( ) 0
k
k n
n
x x
d x y d xy
y y

12. Thus; Substitute (n = -1 & k = -2) to
3 ( 2)
2 1 1
( 1)
( ) 0
x x
d x y d xy
y y
2
( )
0
( )
x x d xy
d
y y xy
2 2
( )
0
x x d xy
d
y y x y
Integrate each term2
( )
0
x x d xy
d
y y xy
3
1
( ) 0
k
k n
n
x x
d x y d xy
y y
By power formula

13. 2
( )
...... 0
( )
x x d xy
d
y y xy
2
1
0
2 1
x
xyy
c Multiply by 2
2
1 1
0 2
2 2
x c
y xy
where :
2
c
c
2
2
2
0
x
c
y xy
Answer

14. 2 2 2 2
3. ( 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 ydx
ydx xdy
x y

15. Integrate each term
2 2
......( ) 0
xdy ydx
ydx xdy
x y
( ) arctan
y
d xy d
x
arctan
y
xy c
x
Answer
( ) arctan
y
d xy d
x
2 2
( )
arctan
from
d xy ydx xdy
y xdy ydx
d
x x y

16. 2 2 2
4. ( 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

17. 2
...... ( ) 2 0xy ydx xdy xydx dy
2
( ) 2
0
xy ydx xdy xydx dy
y y y
Divide by y to integrate each term
( ) ( ) 2 0
dy
xyd xy xd x
y
( )from d xy ydx xdy
Integrate each term
( ) ( ) 2 0
dy
xyd xy xd x
y
By power formula

18. ...... ( ) ( ) 2 0
dy
xyd xy xd x
y
2 2
2 ln
2 2
xy x
y c Multiply by 2 to
eliminate fractions
2 2
( ) 4(ln ) 2xy x y c
2 2
2 ln 2
2 2
xy 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

19. 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

20. 2 2 2 2
5. ( ) ( ) 0x x y x dx y x y dy x = 2 , y = 0
3 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

21. 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 formula
2 4 3 4
( )
2 4 3 4
xy x x y
c

22. 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 y
c
Multiply by their LCD = 12 to eliminate the fractions
2 4 3 4
6( ) 3 4 3 12xy x x y c where 12c = c
2 4 3 4
6( ) 3 4 3xy x x y c general solution

23. 2 4 3 4
6( ) 3 4 3xy x x y c When x = 2 , y = 0
Solve for c :
2 4 3 4
6(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 3
6 3 3 4 16x y x y x
2 2 4 4 3
3(2 ) 4( 4)x y x y x particular solution