Exact Differential Equations - (2.4)
In this section, we consider the general solution of the first order differential equation of the form:
M!x, y"dx ! N!x, y"dy " 0
where both M and N are functions in two variables x and y.
1. Differentials of a Function of Two Variables:
Let z " f!x, y" be a function of two variables x and y. Then the differential dz is defined as
dz " !f dx ! !f dy.
!x
!y
If f!x, y" " C where C is a constant, then
df " 0 # !f dx ! !f dy " 0.
!y
!x
Note that this is a differential equation of the form
M!x, y"dx ! N!x, y"dy " 0 M!x, y" " !f and N!x, y" " !f .
!x
!y
So, for a given differential equation: M!x, y"dx ! N!x, y"dy " 0, if we can find a function f!x, y" such that
!f " M!x, y" and !f " N!x, y"
!x
!y
then the general solution of the differential equation is: f!x, y" " C.
Two questions: for a given differential equation M!x, y"dx ! N!x, y"dy " 0
!f " M!x, y" and
!x
?
a. Is there a function f!x, y" such that
!f " N!x, y"
!y
b. In the case where such a f!x, y" exists, how can it be found?
2. Exact Equation:
The answer to the first question: If !M " !N , then f!x, y" exists. The reason is as follows.
!x
!y
2
2
It it known that if f!x, y" has continuous second derivatives, then ! f " ! f . If a function f!x, y"
!x!y
!y!x
!f " M!x, y" and
!x
satisfies equalities
then it must satisfy the equality !M " !N because
!f " N!x, y"
!x
!y
!y
2
!M " ! 2 f
and !N " ! f .
!x!y
!y!x
!x
!y
A first-order differential equation of the form
M!x, y"dx ! N!x, y"dy " 0
is said to be exact if
!M " !N ,
!y
!x
and M!x, y"dx ! N!x, y"dy is called an exact differential.
Example Determine if the differential equation !e 2y " y cos!xy""dx ! !2xe 2y " x cos!xy" ! 2y"dy " 0 is
1
exact.
Here
M!x, y" " e 2y " y cos!xy" and
2y
N!x, y" " 2xe " x cos!xy" ! 2y
!M " !N .
!x
!y
. Check if
!M " 2e 2y " cos!xy" ! yx sin!xy",
!y
Since !M " !N , the differential equation is exact.
!x
!y
!N " 2e 2y " cos!xy" ! xy sin!xy"
!x
Example Find the value of k so that the differential equation
!y 3 ! kxy 4 " 2x"dx ! !3xy 2 ! 20x 2 y 3 "dy " 0
is exact.
Here
M!x, y" " y 3 ! kxy 4 " 2x and
2
2 3
N!x, y" " 3xy ! 20x y
!M " 3y 2 ! 4kxy 3 ,
!y
. Find k so that !M " !N .
!y
!x
!M " !N
!x
!y
!N " 3y 2 ! 40xy 3 ,
!x
$
4kxy 3 " 40xy 3 , k " 10.
Example Find all possible functions N!x, y" so that the equation
1 ! x
xy
x2 ! y
dx ! N!x, y"dy " 0
is exact.
Here M!x, y" " 1 ! 2 x . Know that !M " !N and
xy
!x
!y
x !y
!M " " 1
x
"
2
2
3
!y
!x ! y"
2 xy
N!x, y" "
#
""
!N dx "
!x
#
!M dx "
!y
1 2 x " 1
2
2 y3
#
"
"1
x2 ! y
1
x
"
2
2
3
!x ! y"
2 xy
! g!y" " "
dx
1
x !
! g!y"
2!x 2 ! y"
y3
where g!y" is a function of y.
3. The General Solution of an Exact Differential Equation:
The answer to the second question: f!x, y" can be solved by the following two steps:
a. f!x, y" " # M!x, y"dx ! g!y" (because !f " M!x, y")
!x
#
b. Solve g !y" from the equation
! # M!x, y"dx ! g!y" " ! # M!x, y"dx ! g # !y" " N!x, y"
!y
!y
!f
, and then
because N!x, y" "
!y
g!y" "
# g # !y"dy.
The general solution for the exact differential equation: f!x, y" " C
Note that we can also use N!x, y" to solve f!x, y". Steps are:
2
a. f!x, y" " # N!x, y"dy ! h!x";
b. Solve h # !x" from the equation
! # N!x, y"dy ! h!x"
!x
and then
" !
!x
h!x" "
# N!x, y"dy ! h # !y" " M!x, y"
# h # !x"dx.
The general solution: f!x, y" " C.
Example Solve the differential equation !e 2y " y cos!xy""dx ! !2xe 2y " x cos!xy" ! 2y"dy " 0.
Know that it is exact. Find f!x, y".
a. f!x, y" " #!e 2y " y cos!xy""dx " xe 2y " sin!xy" ! g!y"
b. !f " 2xe 2y " x cos!xy" ! g # !y" " N!x, y" " 2xe 2y " x cos!xy" ! 2y
!y
g # !y" " 2y, g!y" "
# 2ydy " y 2
The general solution: xe 2y " sin!xy" ! y 2 " C
Example Solve the initial value problem:
3y 2 " t 2
y5
dy
! t 4 " 0, y!1" " 1.
dt
2y
The differential equation is not separable and not linear in y. Check if it is exact. Rewrite the equation:
3y 2 " t 2
dy ! t 4 dt " 0.
y5
2y
!M " " 4t " " 2t
M!t, y" " t 4
2y
!y
2y 5
y5
!M " !N .
2
2 , check if
3y " t
!y
!t
!N " " 2t
N!t, y" "
5
!t
y5
y
The differential equation is exact.
a. f!y, t" " # M!t, y"dt ! g!y" " # t 4 dt ! g!y" " " 1 4 t 2 ! g!y"
2y
4y
2
3y 2 " t 2
" 3 13 " t 5
b. !f " ! " 1 4 t 2 ! h!y" " " "45 t 2 ! g # !y" " " 15 t 2 ! g # !y" " N!t, y" "
5
!y
!y
y
y
y
4y
4y
y
3
3
1
3
#
#
g !y" " 3 , g!y" " # g !y"dy " # 3 dy " "
2 y2
y
y
The general solution: " 1 4 t 2 " 3 12 " C
2 y
4y
Solve the initial value problem: when x " 1, y!1" " 1, solve for C :
" 1 " 3 " "7 " C
2
4
4
The solution for the initial value problem: " 1 4 t 2 " 3 12 " " 7
2 y
4
4y
3