# Uses of Integrating Factor To Solve Non Exact Differential Equation

Differential Equation / Thursday, September 6th, 2018

In this article we will learn about Integrating Factor and how it is used to solve non exact differential equation. In the previous article we already learned what is exact differential equation and how to solve it.

# Integrating Factor Definition

A function µ(x, y) is said to be an Integrating Factor (I.F.) of the equation Mdx + Ndy = 0 if it is possible to obtain a function u(x, y) such that µ(Mdx + Ndy) = du. In other words, I.F. is a multiplying factor by which the equation can be made exact.
Let us consider the differential equation
$xdy-ydx=0$
We observe that the equation is not exact. But if we multiply the equation using (1/x2) it will become exact.
Then we have
$\frac{xdy-ydx}{{{x}^{2}}}=0$
Which can be put as an exact differential:
$d\left( \frac{y}{x} \right)=0$
$\therefore y=cx$
Is the general solution of equation.
Here, (1/x2) is called an Integrating Factor.

## Rules for finding integrating factor

When the differential equation of the form Mdx + Ndy = 0 is not exact
$i.e.,\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
Then we can find out the I.F. in different ways as given below:
$Rule-1:If,\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=f(x)$
a function of x only, then
${{e}^{\int{f(x)dx}}}$
is and I.F. of the equation.
$Rule-2:If,\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M}=g(y)$
a function of y only, then
${{e}^{-\int{g(y)dy}}}$
is and I.F. of the equation.
$Rule-3:If,Mx+Ny\ne 0$
and M and N are both homogeneous functions in x, y of same degree, then I.F. is
$\frac{1}{Mx+Ny}$
$Rule-4:If-the-equation-of-the-form$
$yf(xy)dx+xg(sy)dy=0,and,Mx-Ny\ne 0$
$then,I.F.=\frac{1}{Mx-Ny}$
$Rule-5:If-the-equation-of-the-form$
${{x}^{a}}{{y}^{b}}\left( mydx+nxdy \right)+{{x}^{a’}}{{y}^{b’}}\left( m’ydx+n’xdy \right)=0$
Where a, b, a’, b’, m, n, m’, n’ are all constant the xhyk is an I.F. of the equation where
$\frac{a+h+1}{m}=\frac{b+k+1}{n},\frac{a’+h+1}{m’}=\frac{b’+k+1}{n’}$

 Example 01

$Solve:\left( x{{y}^{2}}-{{e}^{\frac{1}{{{x}^{3}}}}} \right)dx-{{x}^{2}}ydy=0$
Solution:
$Given,\left( x{{y}^{2}}-{{e}^{\frac{1}{{{x}^{3}}}}} \right)dx-{{x}^{2}}ydy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$We-have:M=x{{y}^{2}}-{{e}^{\frac{1}{{{x}^{3}}}}},and,N=-{{x}^{2}}y$
$\therefore \frac{\partial M}{\partial y}=2xy,and,\frac{\partial N}{\partial x}=-2xy$
$\therefore \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
Therefore the given equation is not exact.
$Now,\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{2xy+2xy}{-{{x}^{2}}y}=\frac{4xy}{-{{x}^{2}}y}=-\frac{4}{x}$
Which is a function of x only,
$\therefore I.F.={{e}^{\int{\left( -\frac{4}{x} \right)dx}}}={{e}^{-4\log x}}={{e}^{\log {{x}^{-4}}}}=\frac{1}{{{x}^{4}}}$
Multiplying the given equation by the I.F., we get
$\frac{1}{{{x}^{4}}}\left( x{{y}^{2}}-{{e}^{\frac{1}{{{x}^{3}}}}} \right)dx-\frac{1}{{{x}^{4}}}.{{x}^{2}}ydy=0$
$\Rightarrow \left( \frac{{{y}^{2}}}{{{x}^{3}}}-\frac{{{e}^{\frac{1}{{{x}^{3}}}}}}{{{x}^{4}}} \right)dx-\frac{y}{{{x}^{2}}}dy=0$
Which is an exact differential equation. Hence the solution of the equation is given by
$\int{\left( \frac{{{y}^{2}}}{{{x}^{3}}}-\frac{{{e}^{\frac{1}{{{x}^{3}}}}}}{{{x}^{4}}} \right)dx=\frac{c}{6}}$
Since there is no term which does not contain x in N.
$\Rightarrow {{y}^{2}}.\left( -\frac{1}{2{{x}^{2}}} \right)+\frac{1}{3}\int{{{e}^{\frac{1}{{{x}^{3}}}}}d\left( \frac{1}{{{x}^{3}}} \right)}=\frac{c}{6}$
$\Rightarrow -\frac{{{y}^{2}}}{2{{x}^{2}}}+\frac{1}{3}{{e}^{\frac{1}{{{x}^{3}}}}}=\frac{c}{6}$
$\therefore 2{{x}^{2}}{{e}^{\frac{1}{{{x}^{3}}}}}-3{{y}^{2}}=c{{x}^{2}}$
is the required solution.

 Example 02

$Solve:\left( {{y}^{4}}+2y \right)dx+\left( x{{y}^{3}}+2{{y}^{4}}-4x \right)dy=0$
Solution:
$Given,\left( {{y}^{4}}+2y \right)dx+\left( x{{y}^{3}}+2{{y}^{4}}-4x \right)dy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$We-have:M={{y}^{4}}+2y,and,N=x{{y}^{3}}+2{{y}^{4}}-4x$
$\therefore \frac{\partial M}{\partial y}=4{{y}^{3}}+2,and,\frac{\partial N}{\partial x}={{y}^{3}}-4$
$\therefore \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
Therefore the given equation is not exact.
$Now,\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=4{{y}^{3}}+2-{{y}^{3}}+4=3\left( {{y}^{3}}+2 \right)$
$\therefore \frac{1}{M}\left( \frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} \right)=\frac{3\left( {{y}^{3}}+2 \right)}{y\left( {{y}^{3}}+2 \right)}=\frac{3}{y}$
Which is a function of y only,
$\therefore I.F.={{e}^{-\int{\frac{3}{y}dy}}}={{e}^{-3\log y}}={{e}^{\log {{y}^{-3}}}}=\frac{1}{{{y}^{3}}}$
Multiplying the given equation by the I.F., we get
$\frac{1}{{{y}^{3}}}\left( {{y}^{4}}+2y \right)dx+\frac{1}{{{y}^{3}}}\left( x{{y}^{3}}+2{{y}^{4}}-4x \right)dy=0$
$\Rightarrow \left( y+\frac{2}{{{y}^{2}}} \right)dx+\left( x+2y-\frac{4x}{{{y}^{3}}} \right)dy=0$
Which is an exact differential equation. Hence the solution of the equation is given by
$\int{\left( y+\frac{2}{{{y}^{2}}} \right)dx+\int{2ydy=c}}$
$\Rightarrow \left( y+\frac{2}{{{y}^{2}}} \right)x+{{y}^{2}}=c$
is the required solution.

 Example 03

$Solve:\left( 3{{x}^{2}}{{y}^{4}}+2xy \right)dx+\left( 2{{x}^{3}}{{y}^{3}}-{{x}^{2}} \right)dy=0$
Solution:
$Given,\left( 3{{x}^{2}}{{y}^{4}}+2xy \right)dx+\left( 2{{x}^{3}}{{y}^{3}}-{{x}^{2}} \right)dy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$We-have:M=3{{x}^{2}}{{y}^{4}}+2xy,and,N=2{{x}^{3}}{{y}^{3}}-{{x}^{2}}$
$\therefore \frac{\partial M}{\partial y}=12{{x}^{2}}{{y}^{3}}+2x,and,\frac{\partial N}{\partial x}=6{{x}^{2}}{{y}^{3}}-2x$
$\therefore \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
Therefore the given equation is not exact.
$Now,\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=12{{x}^{2}}{{y}^{3}}+2x-6{{x}^{2}}{{y}^{3}}+2x=6{{x}^{2}}{{y}^{3}}+4x$
$\therefore \frac{1}{M}\left( \frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} \right)=\frac{6{{x}^{2}}{{y}^{3}}+4x}{3{{x}^{2}}{{y}^{4}}+2xy}=\frac{2}{y}$
Which is a function of y only,
$\therefore I.F.={{e}^{-\int{\frac{2}{y}dy}}}={{e}^{-2\log y}}={{e}^{\log {{y}^{-2}}}}=\frac{1}{{{y}^{2}}}$
Multiplying the given equation by the I.F., we get
$\frac{1}{{{y}^{2}}}\left( 3{{x}^{2}}{{y}^{4}}+2xy \right)dx+\frac{1}{{{y}^{2}}}\left( 2{{x}^{3}}{{y}^{3}}-{{x}^{2}} \right)dy=0$
$\Rightarrow \left( 3{{x}^{2}}{{y}^{2}}+\frac{2x}{y} \right)dx+\left( 2{{x}^{3}}y-\frac{{{x}^{2}}}{{{y}^{2}}} \right)dy=0$
Which is an exact differential equation. Hence the solution of the equation is given by
$\int{\left( 3{{x}^{2}}{{y}^{2}}+\frac{2x}{y} \right)dx=c}$
$\Rightarrow {{x}^{3}}{{y}^{2}}+\frac{{{x}^{2}}}{y}=c$
is the required solution.

 Example 04

$Solve:\left( {{x}^{4}}+{{y}^{4}} \right)dx-x{{y}^{3}}dy=0$
Solution:
$Given,\left( {{x}^{4}}+{{y}^{4}} \right)dx-x{{y}^{3}}dy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$We-have:M={{x}^{4}}+{{y}^{4}},and,N=-x{{y}^{3}}$
$\therefore \frac{\partial M}{\partial y}=4{{y}^{3}},and,\frac{\partial N}{\partial x}=-{{y}^{3}}$
$\therefore \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
Therefore the given equation is not exact.
$Now,Mx+Ny=x\left( {{x}^{4}}+{{y}^{4}} \right)+y\left( -x{{y}^{3}} \right)={{x}^{5}}\ne 0$
$\therefore I.F.=\frac{1}{Mx+Ny}=\frac{1}{{{x}^{5}}}$
Multiplying the given equation by the I.F., we get
$\frac{1}{{{x}^{5}}}\left( {{x}^{4}}+{{y}^{4}} \right)dx-\frac{1}{{{x}^{5}}}.x{{y}^{3}}dy=0$
$\Rightarrow \left( \frac{1}{x}+\frac{{{y}^{4}}}{{{x}^{5}}} \right)dx-\frac{{{y}^{3}}}{{{x}^{4}}}dy=0$
Which is an exact differential equation. Hence the solution of the equation is given by
$\int{\left( \frac{1}{x}+\frac{{{y}^{4}}}{{{x}^{5}}} \right)dx=\frac{c}{4}}$
$\Rightarrow \log x-\frac{{{y}^{4}}}{4{{x}^{4}}}=\frac{c}{4}$
$\therefore 4{{x}^{4}}\log x-{{y}^{4}}=c{{x}^{4}}$
is the required solution.

 Example 05

$Solve:({{x}^{3}}{{y}^{2}}+x)dy+({{x}^{2}}{{y}^{3}}-y)dx=0$
Solution:
$Given:({{x}^{3}}{{y}^{2}}+x)dy+({{x}^{2}}{{y}^{3}}-y)dx=0$
$\Rightarrow y({{x}^{2}}{{y}^{2}}-1)dx+x({{x}^{2}}{{y}^{2}}+1)dy=0$
The given equation is of the form
$yf(xy)dx+xg(xy)dy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$We-have:M={{x}^{2}}{{y}^{3}}-y,and,N={{x}^{3}}{{y}^{2}}+x$
$Now,Mx-Ny=x({{x}^{2}}{{y}^{3}}-y)-y({{x}^{3}}{{y}^{2}}+x)$
$\Rightarrow Mx-Ny={{x}^{3}}{{y}^{3}}-xy-{{x}^{3}}{{y}^{3}}-xy=-2xy\ne 0$
$\therefore I.F.=\frac{1}{Mx-Ny}=\frac{1}{-2xy}$
Multiplying the given equation by the I.F., we get
$\frac{1}{-2xy}y({{x}^{2}}{{y}^{2}}-1)dx+\frac{1}{-2xy}x({{x}^{2}}{{y}^{2}}+1)dy=0$
$\Rightarrow -\frac{1}{2}\left( x{{y}^{2}}-\frac{1}{x} \right)dx-\frac{1}{2}\left( {{x}^{2}}y+\frac{1}{y} \right)dy=0$
Which is an exact differential equation. Hence the solution of the equation is given by
$-\frac{1}{2}\int{\left( x{{y}^{2}}-\frac{1}{x} \right)dx}-\frac{1}{2}\int{\frac{1}{y}dy=-\frac{c}{2}}$
$\Rightarrow \frac{1}{2}{{x}^{2}}{{y}^{2}}-\log x+\log y=c$
$\therefore \frac{1}{2}{{x}^{2}}{{y}^{2}}+\log \left( \frac{y}{x} \right)=c$
is the required solution.

 Example 06

$Solve:3ydx-2xdy+\frac{{{x}^{2}}}{y}\left( 10ydx-6xdy \right)=0$
Solution:
$Given,3ydx-2xdy+\frac{{{x}^{2}}}{y}\left( 10ydx-6xdy \right)=0$
$\Rightarrow \left( 3ydx-2xdy \right)+{{x}^{2}}{{y}^{-1}}\left( 10ydx-6xdy \right)=0$
Comparing the given equation with
${{x}^{a}}{{y}^{b}}\left( mydx+nxdy \right)+{{x}^{a’}}{{y}^{b’}}\left( m’ydx+n’xdy \right)=0$
$we-have,a=0,b=0,m=3,n=-2$
$and,a’=2,b’=-1,m’=10,n’=-6$
Let xhyk be an integrating factor. Then
$\frac{a+h+1}{m}=\frac{b+k+1}{n}\Rightarrow \frac{0+h+1}{3}=\frac{0+k+1}{-2}$
$\therefore 2h+3k=-5……….(1)$
Again
$\frac{a’+h+1}{m’}=\frac{b’+k+1}{n’}\Rightarrow \frac{2+h+1}{10}=\frac{-1+k+1}{-6}$
$\therefore 3h+5k=-9……….(2)$
Solving (1) and (2) we get, h = 2, k = -3
$\therefore I.F.={{x}^{h}}{{y}^{k}}={{x}^{2}}{{y}^{-3}}$
Multiplying the given equation by the I.F., we get
${{x}^{2}}{{y}^{-3}}\left( 3ydx-2xdy \right)+{{x}^{2}}{{y}^{-3}}.{{x}^{2}}{{y}^{-1}}\left( 10ydx-6xdy \right)=0$
$\Rightarrow \left( 3{{x}^{2}}{{y}^{-2}}dx-2{{x}^{3}}{{y}^{-3}}dy \right)+\left( 10{{x}^{4}}{{y}^{-3}}dx-6{{x}^{5}}{{y}^{-4}}dy \right)=0$
$\Rightarrow \left( 3{{x}^{2}}{{y}^{-2}}+10{{x}^{4}}{{y}^{-3}} \right)dx-\left( 2{{x}^{3}}{{y}^{-3}}+6{{x}^{5}}{{y}^{-4}} \right)dy=0$
Which is an exact differential equation. Hence the solution of the equation is given by
$\int{\left( 3{{x}^{2}}{{y}^{-2}}+10{{x}^{4}}{{y}^{-3}} \right)dx=c}$
$\therefore {{x}^{3}}{{y}^{-2}}+2{{x}^{5}}{{y}^{-3}}=c$
is the required solution.

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