06. Integrating Factor Technique
Dated: 08-11-2024
If the following equation is not exact
\[
M(x,y)dx+N(x,y)dy=0
\]
then
\[
\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}
\]
Therefore, we will look for a function \(u(x, y)\) such that
\[
u(x,y)M(x,y)dx+u(x,y)N(x,y)dy=0
\]
This function \(u(x, y)\) is called the integrating factor.
\[
\frac{\partial M}{\partial y}u+\frac{\partial u}{\partial y}M=\frac{\partial N}{\partial x}u+\frac{\partial u}{\partial x}N
\]
Example
Show that the following
\[\frac 1 {(x^2 + y^2)}\]
is an integrating factor for the following
\[
(x^{2}+y^{2}-x)dx-ydy=0
\]
\[
M = x^{2}+y^{2}-x, \quad N = -y
\]
\[
\frac{\partial M}{\partial y} = 2y, \quad \frac{\partial N}{\partial x} = 0
\]
\[
\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}
\]
If we multiply the following
\[\frac 1 {(x^2 + y^2)}\]
then the following equation is exact.
\[
\left(1-\frac{x}{x^2+y^2}\right)dx-\frac{y}{x^2+y^2}dy=0
\]
Checking again
\[
M=1-\frac{x}{x^{2}+y^{2}} \text{ and } N=-\frac{y}{x^{2}+y^{2}}
\]
\[
\frac{\partial M}{\partial y}=\frac{2xy}{(x^{2}+y^{2})^{2}}=\frac{\partial N}{\partial x}
\]
We can write the original equation as
\[
dx-\frac{xdx+ydy}{x^{2}+y^{2}}=0
\]
\[
dx-\frac{1}{2}d\left[\ln(x^{2}+y^{2})\right]=0
\]
\[
d\left[x-\frac{\ln(x^{2}+y^{2})}{2}\right]=0
\]
By integration, we have
\[
x-\ln\sqrt{x^{2}+y^{2}}=k
\]
Case 1
If the given expression
\[
\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}
\]
is a function of \(x\) only then the integrating factor \(u(x, y)\) is given by
\[
u = \exp\left(\int \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}dx\right)
\]
Case 2
If the given expression
\[
\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}
\]
is a function of \(y\) only then the integrating factor \(u(x, y)\) is given by
\[
u = \exp\left(\int \frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}dy\right)
\]
Case 3
If the given equation is homogeneous and
\[xM + yN \ne 0\]
then
\[
u=\frac{1}{xM+yN}
\]
Case 4
If the equation is of the form
\[
yf(xy)dx+xg(xy)dy=0
\]
and
\[
xM-yN\ne0
\]
then
\[
u=\frac{1}{xM-yN}
\]
Once \(u(x, y)\) is found, multiply the old equation by \(u(x, y)\) to get a new one which is exact.
Solve it and write the solution.
Summary
- Write the equation in the form
\[
M(x,y)dx+N(x,y)dy=0
\]
\[
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
\]
- If the equation is not exact then evaluate
\[
\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}
\]
If the result is a function of only \(x\) then
\[
u(x) = \exp\left(\int \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}dx\right)
\]
otherwise evaluate
\[
\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}
\]
If the result is a function of only \(y\) then
\[
u(y)=\exp\left(\int\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}dy\right)
\]
- Test if the equation is
homogeneus and
\[
xM+yN\ne0
\]
then
\[
u=\frac{1}{xM+yN}
\]
- Test if the equation is of the form
\[
yf(xy)dx+xg(xy)dy=0
\]
and
\[
xM-yN\ne0
\]
then
\[
u=\frac{1}{xM-yN}
\]
- Multiply the old equation by \(u\) to get a new one.
- Solve the new equation.
Example
\[
\frac{dy}{dx}=-\frac{3xy+y^{2}}{x^{2}+xy}
\]
Step 1
Rewriting the equation
\[
(3xy+y^{2})dx+(x^{2}+xy)dy=0
\]
\[
M(x,y)=3xy+y^{2}
\]
\[
N(x,y)=x^{2}+xy
\]
Step 2
\[
\frac{\partial M}{\partial y}=3x+2y, \quad \frac{\partial N}{\partial x}=2x+y
\]
\[
\therefore \quad \frac{\partial M}{\partial y}\ne\frac{\partial N}{\partial x}
\]
Step 3
\[
\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{1}{x}
\]
Step 4
\[
u(x)=e^{\int\frac{1}{x}dx}=e^{ln(x)}=x
\]
Step 5
\[
(3x^{2}y+xy^{2})dx+(x^{3}+x^{2}y)dy=0
\]
Step 6
\[
\frac{\partial M}{\partial y}=3x^{2}+2xy=\frac{\partial N}{\partial x}
\]
Step 7
\[
\begin{cases}
\frac{\partial F}{\partial x}=3x^{2}y+xy^{2}\\
\frac{\partial F}{\partial y}=x^{3}+x^{2}y
\end{cases}
\]
Step 8
\[
F(x,y)=x^{3}y+\frac{x^{2}}{2}y^{2}+\theta(y)
\]
Step 9
\[
\frac{\partial F}{\partial y}=x^{3}+x^{2}y+\theta'(y)=x^{3}+x^{2}y
\]
Because \(\theta^\prime(y)\) is independent of \(x\).
\[\theta^\prime = 0\]
Step 10
Integration with respect to \(y\), we get
\[
F(x,y)=x^{3}y+\frac{x^{2}}{2}y^{2}
\]
Step 11
\[\because F(x, y) = C\]
\[
x^{3}y+\frac{x^{2}y^{2}}{2}=C
\]
Note
It is possible that the integrating factors may not be unique.
References
Read more about notations and symbols.