Exact and Inexact Differential Equation

ENGINEERING SCIENCE 21
MATHEMATICAL METHODS
IN ENGINEERING
Exact and Inexact Differential Equation
Prepared by: JRTLustro

EXACT DIFFERENTIAL
EQUATION
Recall: Complete Derivatives
If F(x,y) = C is a function in two variables
x and y that is differentiable over a given
interval, the total or complete derivative
of F(x,y) applying chain rule is
𝜕𝐹 𝜕𝐹
𝑑𝐹 = 𝑑𝑥 + 𝑑𝑦
𝜕𝑥 𝜕𝑦
And by the assumption of continuity the
second partial derivatives
2 2
𝜕 𝐹 𝜕 𝐹
and
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
Should be equal, so that
2
𝜕 𝐹 2
𝜕 𝐹
=
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
This suggests the test for exactness
Exact Differential Equations
A given differential equation M(x,y)dx +
N(x,y)dy = 0 is an exact differential
equation if the following condition is
satisfied:
𝜕𝑀 𝜕𝑁 𝜕𝑀 𝜕𝑁
− =0 ↔ =
𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥
Exact differential equation is solved by
reversing the method in getting the
complete derivative of the function
F(x,y). Thus we can let
𝜕𝐹 𝜕𝐹
𝑀= and 𝑁 =
𝜕𝑥 𝜕𝑦
Methods of Solutions
Method 1 – Working on M(x,y)
Step 1. Solve for F(x,y) by integrating Mdx
with respect to x
𝜕𝐹
𝐹 𝑥, 𝑦 = 𝑑𝑥 = 𝑀 𝑥, 𝑦 𝑑𝑥
𝜕𝑥
= 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑔(𝑦)
Note: Integrating with respect to x yields a function g(y)

instead of a constant C since the partial derivative of g(y) with
respect to x is equal to zero
Method 1 – Working on M(x,y)
Step 2. Differentiate F(x,y) with respect to y
and equate to N
𝜕𝐹
= 𝑁(𝑥, 𝑦)
𝜕𝑦
Step 3. Determine g(y) by integrating g’(y)
with respect to y
The complete solution now is the equation
F(x,y) = C.
Method 2 – Working on N(x,y)
Step 1. Solve for F(x,y) by integrating Ndy
with respect to y
𝜕𝐹
𝐹 𝑥, 𝑦 = 𝑑𝑦 = 𝑁 𝑥, 𝑦 𝑑𝑦
𝜕𝑦
= 𝑁 𝑥, 𝑦 𝑑𝑥 + 𝑕(𝑥)
Note: Integrating with respect to y yields a function h(x)

instead of a constant C since the partial derivative of h(x) with
respect to y is equal to zero
Method 2 – Working on N(x,y)
Step 2. Differentiate F(x,y) with respect to x
and equate to M
𝜕𝐹
= 𝑀(𝑥, 𝑦)
𝜕𝑥
Step 3. Determine h(x) by integrating h’(x)
with respect to x
F(x,y) = C.
Method 3 – Working Simultaneously
on M(x,y)and N(x,y)
Step 1. Solve for F(x,y) by integrating Mdx with
respect to x and by integrating Ndy with
respect to y
𝜕𝐹
𝐹 𝑥, 𝑦 = 𝑑𝑥 = 𝑀 𝑥, 𝑦 𝑑𝑥 = 𝑀 𝑥, 𝑦 𝑑𝑥
𝜕𝑥
𝜕𝐹
𝐹 𝑥, 𝑦 = 𝑑𝑦 = 𝑁 𝑥, 𝑦 𝑑𝑦 = 𝑁 𝑥, 𝑦 𝑑𝑦
𝜕𝑦
Method 3 – Working Simultaneously
on M(x,y)and N(x,y)
Step 2. Copying unique terms will give the
desired general solution

F(x,y) = C
INEXACT DIFFERENTIAL
EQUATION
Consider the differential equation in
the form
𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0
If the given differential equation is not
exact, in theory, there exists a function Ф
such that if this function is multiplied to
both sides of the equation, it will yield a
new differential equation that is now exact.
This function Ф is what we call an
integrating factor. Hence, we have
the exact differential equation
Ф𝑀𝑑𝑥 + Ф𝑁𝑑𝑦 = 0
Method of Solutions
Step 1. Determination of the integrating
factor Ф using some known methods based
on the assumptions on the nature of Ф
Step 2. Multiplying the integrating factor Ф
to both sides of the equation and solving
the resulting exact differential equation
ФMdx + ФNdy = 0
Determination of the Integrating Factor
Ф𝑀𝑑𝑥 + Ф𝑁𝑑𝑦 = 0
Applying the condition for exactness
𝜕(Ф𝑀) 𝜕(Ф𝑁)
=
𝜕𝑦 𝜕𝑥
Since Ф, in general, is not a constant, then
we can differentiate the test for exactness
as follows
𝜕𝑀 𝜕Ф 𝜕𝑁 𝜕Ф
Ф +𝑀 =Ф +𝑁
𝜕𝑦 𝜕𝑦 𝜕𝑥 𝜕𝑥
This equation can now be written as
𝜕𝑀 𝜕𝑁 𝜕Ф 𝜕Ф
Ф − =𝑁 −𝑀
𝜕𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦
or
𝜕𝑀 𝜕𝑁 𝜕Ф 𝜕Ф
Ф − =− 𝑀 −𝑁
𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥
Case 1: Ф = Ф(x)
𝑓 𝑥 𝑑𝑥
Ф 𝑥 =𝑒
1 𝜕𝑀 𝜕𝑁
If − = 𝑓(𝑥)  a function of x only
𝑁 𝜕𝑦 𝜕𝑥
Case 2: Ф = Ф(y)
𝑔 𝑦 𝑑𝑦
Ф 𝑦 =𝑒
1 𝜕𝑀 𝜕𝑁
If − − = 𝑔(𝑦)  a function of y only
𝑀 𝜕𝑦 𝜕𝑥

Exact and Inexact Differential Equation

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Exact and Inexact Differential Equation

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ENGINEERING SCIENCE 21

Prepared by: JRTLustro

Note: Integrating with respect to x yields a function g(y)

Note: Integrating with respect to y yields a function h(x)

The complete solution now is the equation

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