# Lesson 29 ```Module 3: Ordinary Differential Equations
Lesson 29
Exact Differential Equations: Integrating Factors
In general, equations of the type M(x, y)dx + N(x, y)dy = 0 are not exact. However, it
is sometimes possible to transform the equation into an exact differential equation multiplying it by a suitable function I(x, y). That is, if I(x, y) is an integrating factor then the
differential equation
I(x, y)M(x, y)dx + I(x, y)N(x, y)dy = 0
becomes exact. A solution to the above equation is obtained by solving the exact differential equation as in the previous lesson. Note that the given equation may have several
integrating factors. This is exactly the procedure we have used for solving linear differential equations in earlier lesson. Here we deal with more general differential equation.
29.1 Rule I: By Inspection
There is not much theory behind finding integrating factor by inspection. This method
works based on recognition of some standard exact differentials that occur frequently in
practice. The following list of exact differentials would be quite useful in solving exact
differential equations:
(i)
(ii)
(iii)
(iv)
(v)
d(xy) = ydx + xdy
y xdy − ydx
ydx − xdy
x
=
=
or
d
d
x
x2
y
y2
y xdy − ydx
ydx − xdy
x
d ln
=
=
or d ln
x
xy
y
xy
xdy − ydx
ydx − xdy
x
y
=
=
or d arctan
d arctan
2
2
x
x +y
y
y 2 + x2
ydx + xdy
d (ln xy) =
xy
29.1.1 Example
Solve the differential equation y(y 2 + 1)dx + x(y 2 − 1)dy .
Exact Differential Equations: Integrating Factors
Solution: The given equation can be rewritten as
y 2(ydx + xdy) + ydx − xdy
This is further rewritten as
(ydx + xdy) +
ydx − xdy
y2
=0
Using standard differential forms given above we get
x
d(xy) + d
=0
y
Integrating the above equation, the desired solution is given as
xy 2 + x = cy
Here c is an arbitrary constant.
29.2 Rule II: Mdx + N dy = 0 is homogeneous and Mx + N y 6= 0
If the equation Mdx+Ndy = 0 is homogeneous and Mx+Ny 6= 0, then I(x, y) =
1
(Mx + Ny)
is an integrating factor. In order to prove the result, we need to show that
Mdx + Ndy
= d (some function x and y)
Mx + Ny
Rewriting Mdx + Ndy as
1
Mdx + Ndy =
2
(Mx + Ny)
dx dy
+
x
y
+ (Mx − Ny)
dx dy
−
x
y
Multiplying by proposed integrating factor we get
Mdx + Ndy
1
=
Mx + Ny
2
dx dy
+
x
y
(Mx − Ny)
+
(Mx + Ny)
dx dy
−
x
y
(29.1)
Given that M(x, y) and N(x, y) are homogeneous functions of some degree n, i.e., M(tx, ty) =
tn M(x, y) and N(x, y) = tn N(x, y). Then
M
x
,1
y
=M
1 1
x, y
y y
1
= n M(x, y) ⇒ M(x, y) = y n M
y
2
x
,1
y
Exact Differential Equations: Integrating Factors
Similarly, we get
n
N(x, y) = y N
x
,1
y
Now consider
y nxM
x
y,1
y nyN
x
y,1
−
(Mx − Ny)
=
=
x
x
(Mx + Ny)
n
n
y xM y , 1 + y yN y , 1
x
yM
x
y,1
−N
x
y,1
x
=f
x
x
x
y
yM y,1 + N y,1
Going back to the Equation (29.1), we have
Mdx + Ndy
1
=
Mx + Ny
2
x
x
d (ln(xy)) + f
d ln
y
y
Rewriting f (x/y) = f (exp(ln(x/y))) and defining g(x) := f (exp(x)), the above equation
becomes
1
Mdx + Ndy
=
Mx + Ny
2
x
d (ln(xy)) + g (ln(x/y)) d ln
y
Hence, we have shown that
Z Mdx + Ndy
x
x
1
1
g ln
d ln
=d
ln(xy) +
Mx + Ny
2
2
y
y
1
is an integrating factor of the homogenous differential equation Mdx +
Mx + Ny
Ndy = 0.
Thus
29.2.1 Example
Solve the differential equation (x2 y − 2xy 2 )dx − (x3 − 3x2 y)dy = 0
Solution: The given equation is a homogeneous differential equation. Comparing it with
Mdx + Ndy = 0, we have M = x2 y − 2xy 2 and N = −(x3 − 3x2 y). Since
Mx + Ny = (x2 y − 2xy 2 )x − y(x3 − 3x2 y) = x2 y 2 6= 0,
the integrating factor is
1
1
= 2 2
(Mx + Ny)
x y
Multiply by the integrating factor, the given differential equation becomes
(1/y − 2/x)dx − (x/y 2 − 3/y)dy = 0
3
Exact Differential Equations: Integrating Factors
This is now exact and can be rewritten as
ydx − xdy
2
3
−
dx
+
dy = 0
y2
x
y
2
x
3
− dx + dy = 0
d
y
x
y
⇒
Integrating the above equation we obtain the desired solution as
x − 2y ln x + 3y ln y = cy
29.3 Rule III: Mdx + N dy = 0 is of the form f1(xy)ydx + f2(xy)xdy = 0
If the equation Mdx + Ndy = 0 is of the form f1 (xy)ydx + f2(xy)xdy = 0, then
1
(Mx − Ny)
is an integrating factor provided Mx − Ny 6= 0. Similar to rule II we now show that
Mdx + Ndy
= d (some function x and y)
Mx − Ny
Again, rewriting Mdx + Ndy as
1
Mdx + Ndy =
2
(Mx + Ny)
dx dy
+
x
y
+ (Mx − Ny)
dx dy
−
x
y
Now dividing by Mx − Ny we get
Mdx + Ndy 1
=
Mx − Ny
2
(Mx + Ny)
Mx − Ny
dx dy
+
x
y
+
dx dy
−
x
y
Using M = f1 (xy)y and N = f2 (xy)x we obtain
Mdx + Ndy
1
=
Mx − Ny
2
f1 (xy) + f2 (xy)
x
d (ln xy) + d ln
f1 (xy) − f2 (xy)
y
f1 (xy) + f2 (xy)
and g(x) := f (exp(x)), the above equation reduces to
f1 (xy) − f2 (xy)
x
1
x
1
Mdx + Ndy
f (xy)d (ln xy) + d ln
=
g(ln xy)d (ln xy) + d ln
=
Mx − Ny
2
y
2
y
Let f (xy) :=
This shows that
Z
1
x
Mdx + Ndy
1
g(ln xy)d (ln xy) +
ln
=d
Mx − Ny
2
2
y
4
Exact Differential Equations: Integrating Factors
29.3.1 Example
Solve y(x2 y 2 + 2)dx + x(2 − 2x2 y 2)dy = 0.
Solution: Comparing with Mdx+Ndy = 0, we have M = y(x2 y 2 +2) and N = x(2−2x2 y 2 ).
The given equation is of the form
f1 (xy)ydx + f2 (xy)xdy = 0
and we have
Mx − Ny = xy(x2 y 2 + 2) − xy(2 − 2x2 y 2 ) = 3x3 y 3 6= 0
Therefore, multiplying the equation by 1/3x3 y 3 , we obtain
(1/3x + 2/(3x3 y 2))dx + (2/(3x2 y 3) − 2/3y)dy = 0
This is an exact differential equation which can be solved with the technique discussed in
previous lesson.
29.4 Rule IV: Most general approach
Now we discuss the most general approach of finding integrating function. The idea is to
multiply the given differential equation
M(x, y)dx + N(x, y)dy = 0
(29.2)
by a function I(x, y) and then try to choose I(x, y) so that the resulting equation
I(x, y)M(x, y)dx + I(x, y)N(x, y)dy = 0
(29.3)
becomes exact. The above equation is exact if and only if
∂(IM)
∂(IN)
=
∂y
∂x
(29.4)
If a function I(x, y) satisfying the partial differential Equation (29.4) can be found, then
(29.3) will be exact. Unfortunately, solving Equation (29.4), is as difficult to solve as the
original Equation (29.2) by some other methods. Therefore, while in principle integrating
factors are powerful tools for solving differential equations, in practice they can be found
5
Exact Differential Equations: Integrating Factors
only in special cases. The cases we will consider are: (i) an integrating factor I that is
either as function of x only, or (ii) a function of y only.
Let us determine necessary conditions on M and N so that (29.2) has an integrating factor
I that depends on x only. Assuming that I is a function of x only, then Equation (29.4)
reduces to
IMy = INx + N
dI
dx
⇒
IMy − INx
dI
=
dx
N
(29.5)
If (My − Nx )/N is a function of x only, say f(x), then there is an integrating factor I
R
that also depends only on x which can be found by solving (29.5) as I(x) = e f (x)dx . A
similar procedure can be used to determine a condition under which Equation (29.2) has
an integrating factor depending only on y . To conclude, we have:
R
∂N
1 ∂M
is function of x alone say f (x), then I(x) = e f (x)dx is an I.F.
−
If
N ∂y
∂x R
∂M
1 ∂N
is function of y alone say f (y), then I(y) = e f (y)dy is an I.F.
−
If
M ∂x
∂y
29.5 Example Problems
29.5.1 Problem 1
Find an integrating factor of (x2 + y 2 + x)dx + xydy = 0 Solution: Comparing with
Mdx + Ndy = 0, we have
M = (x2 + y 2 + x) and N = xy
Further, note that
1
N
∂M
∂N
−
∂y
∂x
=
1
x
R
is a function of x alone. Hence, the integrating factor of the given problem is e
1/xdx
29.5.2 Problem 2
Find an integrating factor of (2xy 4 ey + 2xy 3 + y)dx + (x2 y 4 ey − x2 y 2 − 3x)dy = 0
Solution: Compare with Mdx + Ndy = 0, we get
M = (2xy 4 ey + 2xy 3 + y) and N = (x2 y 4 ey − x2 y 2 − 3x)
6
= x.
Exact Differential Equations: Integrating Factors
Also, note that
1
M
∂N
∂M
−
∂y
∂x
=−
4
y
R
is a function of y alone. Hence the integrating factor of the given problem is e
−4/ydy
= 1/y 4.
Boyce, W.E. and DiPrima, R.C. (2001). Elementary Differential Equations and Boundary
Value Problems. Seventh Edition, John Willey &amp; Sons, Inc., New York.
Dubey, R. (2010). Mathematics for Engineers (Volume II). Narosa Publishing House.
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McQuarrie, D.A. (2009). Mathematical Methods for Scientist and Engineers. First Indian
Edition. Viva Books Pvt. Ltd. New Delhi.
Raisinghania, M.D. (2005). Ordinary &amp; Partial Differential Equation. Eighth Edition. S.
Chand &amp; Company Ltd., New Delhi.
Kreyszig, E. (1993). Advanced Engineering Mathematics. Seventh Edition, John Willey
&amp; Sons, Inc., New York.
Arfken, G.B. (2001). Mathematical Methods for Physicists. Fifth Edition, Harcourt Academic Press, San Diego.
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Publishilers, New Delhi.
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Publisher, Moscow.
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