Sec 2.4: Exact Differential Equations
z  f ( x, y )
differenti al :
Example :
f
f
dz 
dx  dy
x
y
Find the differential
z  f ( x, y) where f ( x, y)  x 2 y  y
special case :
Example :
f ( x, y )  c

f
f
dx  dy  0
x
y
2 xydx  ( x 2  1)dy  0
Sec 2.4
Sec 2.4: Exact Differential Equations
Definition 2.3 (part 1)
An expression
M ( x, y )dx  N ( x, y )dy
is an exact if it corresponds to the
differential of some function f(x,y)
Definition 2.3 (part 2)
A 1st order DE M ( x, y)dx  N ( x, y)dy  0
is an exact if LHS is an exact differential
Example :
1
2 xydx  ( x 2  1)dy  0
f  x2 y  y
Sec 2.4
How?? : check exact or not
Theorem 2.1
Given a DE:
M N

y
x
If
Example :
1
2
3
M ( x, y )dx  N ( x, y )dy  0

----- (1)
(1) exact
2 xydx  ( x 2  1)dy  0
x 2 ydx  y 2 xdy  0
(e 2 y  y cos( xy))dx  (2 xe2 y  x cos( xy)  2 y)dy  0
Sec 2.4
How to Solve ?
M ( x, y )dx  N ( x, y )dy  0
Given a DE:
----- (1)
Method of Solution:
Step 1
Step 2
Step 3
Example :
Check if (1) is an exact
Find
f ( x, y )
The family of solutions is:
1
M N

y
x
(such that :
f
M ,
x
f
 N)
y
f ( x, y )  c
2 xydx  ( x 2  1)dy  0
f  x2 y  y
Sec 2.4
How to find f(x,y) ?
Given an exact DE:
Step 1
Step 2
M ( x, y )dx  N ( x, y )dy  0
Integrate wrt x:
M 

f
x
Differentiate (2) wrt y and equate to N
----- (1)
f   Mdx  g ( y ) ----- (2)

find
g ' ( y)
Check point: g ' ( y ) function of y only
Step 3
Step 3
Example :
Find:
g ( y)
----- (3)
Use (2) and (3) to write:
1
2
f ( x, y )
2 xydx  ( x 2  1)dy  0
f  x2 y  y
(e 2 y  y cos( xy))dx  (2 xe2 y  x cos( xy)  2 y)dy  0
Sec 2.4
How to find f(x,y) ?
Given an exact DE:
Step 1
Step 2
M ( x, y )dx  N ( x, y )dy  0
Integrate wrt x:
M 

f
x
Differentiate (2) wrt y and equate to N
----- (1)
f   Mdx  g ( y ) ----- (2)

find
g ' ( y)
Check point: g ' ( y ) function of y only
Step 3
Step 3
Example :
Find:
g ( y)
----- (3)
Use (2) and (3) to write:
1
f ( x, y )
2 xydx  ( x 2  1)dy  0
f  x2 y  y
Sec 2.4
Made Exact
Given an non-exact DE:
Find u so that :
Mdx  Ndy  0
uMdx  uNdy  0
----- (1)
Exact
This u is called integrating factor
case 1 ( M y  N x ) / N  function of x
u ( x)  e 
case 2  ( M y  N x ) / M  function of y
u ( x)  e 
Example :
xydx  (2 x 2  3 y 2  20)dy  0
( M y  N x ) / Ndx
 ( M y  N x ) / Mdy