# All Things You Must Know About Exact Differential Equation

Differential Equation / Wednesday, September 5th, 2018
(Last Updated On: September 5, 2018)

We are going learn about Exact Differential Equation in the following 2 Part article series:
1. How to solve Exact Differential Equation.
2. How to solve non Exact Differential Equation using Integrating Factor.

I suggest you to first check my article on Cheat Sheet of First Order First Degree Differential Equation Solution Methods for better understanding of Differential Equation.

# Exact Differential Equation Definition

If the first order first degree differential equation Mdx + Ndy = 0, without being multiplied by any factor, can be expressed in the form du = 0, where u is a function of x and y, then the equation Mdx + Ndy = 0 is said to be an exact differential equation and the general solution of this equation is given by u(x, y) =c, where c is arbitrary constant.

## Condition for a differential equation to be exact

### Theorem

The necessary condition and sufficient condition for the ordinary differential equation Mdx + Ndy = 0 to be exact is
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Here it is assumed that the functions M and N have continuous partial derivatives.
Proof:
To prove that the condition is necessary.
If the equation
$Mdx+Ndy=0……….(1)$
is exact, then there must exist a function u of x and y, such that Mdx + Ndy is equal to the total differential du of u.
$Since,du=\frac{\partial u}{\partial x}.dx+\frac{\partial u}{\partial y}.dy……….(2)$
Where x and y are variables independent of each other,
$Mdx+Ndy=\frac{\partial u}{\partial x}.dx+\frac{\partial u}{\partial y}.dy$
$So,M=\frac{\partial u}{\partial x},and,N=\frac{\partial u}{\partial y}$
$Then,\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left( \frac{\partial u}{\partial x} \right)=\frac{{{\partial }^{2}}u}{\partial y\partial x}$
$and,\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left( \frac{\partial u}{\partial y} \right)=\frac{{{\partial }^{2}}u}{\partial x\partial y}$
If the partial derivatives of M and N are continuous,
$\frac{{{\partial }^{2}}u}{\partial y\partial x}=\frac{{{\partial }^{2}}u}{\partial x\partial y}$
$\therefore \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Thus it is proved that the condition is necessary.

To prove that the condition is sufficient.
Suppose that the condition ,
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
holds.
$Put,F=\int{Mdx,}$
where the integration is performed on the supposition that y is constant.
$Then,\frac{\partial F}{\partial x}=M$
$\Rightarrow \frac{{{\partial }^{2}}F}{\partial x\partial y}=\frac{{{\partial }^{2}}F}{\partial y\partial x}=\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
$\therefore \frac{\partial }{\partial x}\left( N-\frac{\partial F}{\partial y} \right)=0$
$\therefore \frac{\partial }{\partial x}\left( N-\frac{\partial F}{\partial y} \right)$
Is constant so far as x is concerned, that is, a function of
$y=\phi \left( y \right),say.$
$Hence,N=\phi \left( y \right)+\frac{\partial F}{\partial y}$
We next take function u(x, y) so that
$u=F+\int{\phi \left( y \right)dy}\Rightarrow \frac{\partial u}{\partial y}=N$
$Also,M=\frac{\partial F}{\partial x}=\frac{\partial u}{\partial y}\left[ \because u=F+\int{\phi \left( y \right)dy} \right]$
Thus,
$Mdx+Ndy=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=du$
A perfect differential.
So, the equation is exact and condition is sufficient.

## Some exact differentials

Sometimes some differential equation can be solved by inspection, by re-arranging the terms suitably to form a exact differential and then integrating it. In those cases it will be convenient to remember the following exact differentials:
$(i)dx\pm dy=d\left( x\pm y \right)$
$(ii)xdy+ydx=d\left( xy \right)$
$(iii)xdx\pm ydy=\frac{1}{2}d\left( {{x}^{2}}\pm {{y}^{2}} \right)$
$(iv)\frac{ydx-xdy}{{{y}^{2}}}=d\left( \frac{x}{y} \right)$
$i.e.,ydx-xdy={{y}^{2}}d\left( \frac{x}{y} \right)$
$(v)\frac{xdy-ydx}{{{x}^{2}}}=d\left( \frac{y}{x} \right)$
$i.e.,xdy-ydx={{x}^{2}}d\left( \frac{y}{x} \right)$
$(vi)\frac{ydx-xdy}{{{x}^{2}}+{{y}^{2}}}=d\left( {{\tan }^{-1}}\frac{x}{y} \right)$
$(vii)\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=d\left( {{\tan }^{-1}}\frac{y}{x} \right)$
$(viii)\frac{ydx-xdy}{xy}=d\left( \log \frac{x}{y} \right)$
$(ix)\frac{xdy-ydx}{xy}=d\left( \log \frac{y}{x} \right)$
$(x)\frac{2xydx-{{x}^{2}}dy}{{{y}^{2}}}=d\left( \frac{{{x}^{2}}}{y} \right)$
$(xi)\frac{2xydy-{{y}^{2}}dx}{{{x}^{2}}}=d\left( \frac{{{y}^{2}}}{x} \right)$

 Example 01

Examine whether
$\left( \cos y+y\cos x \right)dx+\left( \sin x-x\sin y \right)dy=o$
Is an exact differential equation.

Solution:
$Given,\left( \cos y+y\cos x \right)dx+\left( \sin x-x\sin y \right)dy=o$
Comparing the given equation with Mdx +Ndy = 0, we have
$M=\cos y+y\cos x,N=\sin x-x\sin y$
$\therefore \frac{\partial M}{\partial y}=-\sin y+\cos x,and,\frac{\partial N}{\partial x}=\cos x-\sin y$
$Here,\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
So the given equation is exact.

 Example 02

$Solve:ydx-xdy=xydx$
Solution:
$Given,ydx-xdy=xydx$
$\Rightarrow \frac{ydx-xdy}{{{y}^{2}}}=\frac{x}{y}dx$
$\Rightarrow d\left( \frac{x}{y} \right)=\left( \frac{x}{y} \right)dx$
$\Rightarrow \frac{d\left( \frac{x}{y} \right)}{\left( \frac{x}{y} \right)}=dx$
Integrating we get,
$\log \left( \frac{x}{y} \right)=x+c$
Where c is arbitrary constant, is the general solution of the given equation.

 Example 03

$Solve:xdx+ydy+\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=0$
Solution:
$Given,xdx+ydy+\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=0$
$\Rightarrow \frac{1}{2}d\left( {{x}^{2}}+{{y}^{2}} \right)+\frac{\frac{xdy-ydx}{{{x}^{2}}}}{\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}}}=0$
$\Rightarrow \frac{1}{2}d\left( {{x}^{2}}+{{y}^{2}} \right)+\frac{d\left( \frac{y}{x} \right)}{1+{{\left( \frac{y}{x} \right)}^{2}}}=0$
Integrating we get,
$\frac{1}{2}\left( {{x}^{2}}+{{y}^{2}} \right)+{{\tan }^{-1}}\left( \frac{y}{x} \right)=c$
$\Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)+2{{\tan }^{-1}}\left( \frac{y}{x} \right)=k,[where,k=2c]$

 Example 04

$Solve:\left( x{{y}^{2}}+2{{x}^{2}}{{y}^{3}} \right)dx+\left( {{x}^{2}}y-{{x}^{3}}{{y}^{2}} \right)dy=0$
Solution:
$Given,\left( x{{y}^{2}}+2{{x}^{2}}{{y}^{3}} \right)dx+\left( {{x}^{2}}y-{{x}^{3}}{{y}^{2}} \right)dy=0$
$\Rightarrow xydx+2{{x}^{2}}{{y}^{3}}dx+{{x}^{2}}ydy-{{x}^{3}}{{y}^{2}}dy=0$
$\Rightarrow xy\left( ydx+xdy \right)+{{x}^{2}}{{y}^{2}}\left( 2ydx-xdy \right)=0$
$\Rightarrow d\left( xy \right)+{{x}^{2}}{{y}^{2}}\left( 2\frac{dx}{x}-\frac{dy}{y} \right)=0$
$\Rightarrow \frac{d\left( xy \right)}{{{x}^{2}}{{y}^{2}}}+2\frac{dx}{x}-\frac{dy}{y}=0$
Integrating we get,
$-\frac{1}{xy}+2\log x-\log y=\log c$
$\Rightarrow \log {{x}^{2}}-\log y-\log c=\frac{1}{xy}$
$\Rightarrow \log \left( \frac{{{x}^{2}}}{cy} \right)=\frac{1}{xy}$
$\therefore {{x}^{2}}=cy{{e}^{\frac{1}{xy}}}$
is the required general solution.

 Example 05

$Solve:y\left( 2xy+1 \right)dx+x\left( 1+2xy+{{x}^{2}}{{y}^{2}} \right)dy=0$
Solution:
$We-have,y\left( 2xy+1 \right)dx+x\left( 1+2xy+{{x}^{2}}{{y}^{2}} \right)dy=0$
$\Rightarrow \left( 2xy+1 \right)\left( ydx+xdy \right)+{{x}^{3}}{{y}^{2}}dy=0$
$\Rightarrow \left( 2xy+1 \right)d\left( xy \right)=-{{x}^{3}}{{y}^{2}}dy$
$\Rightarrow \frac{\left( 2xy+1 \right)}{{{x}^{3}}{{y}^{3}}}d\left( xy \right)=-\frac{dy}{y}$
$\Rightarrow \left[ \frac{2}{{{\left( xy \right)}^{2}}}+\frac{1}{{{\left( xy \right)}^{3}}} \right]d\left( xy \right)=-\frac{dy}{y}$
Integrating we get,
$-\frac{2}{xy}-\frac{1}{2}.\frac{1}{{{\left( xy \right)}^{2}}}=-\log y-\log c$
$\Rightarrow \frac{2}{xy}+\frac{1}{2{{\left( xy \right)}^{2}}}=\log yc$
is the required general solution.

 Example 06

$Solve:xdy-ydx=\sqrt{{{x}^{2}}+{{y}^{2}}}dx$
Solution:
$Given,xdy-ydx=\sqrt{{{x}^{2}}+{{y}^{2}}}dx$
$\Rightarrow \frac{xdy-ydx}{{{x}^{2}}}=\frac{1}{{{x}^{2}}}.x\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}}dx$
$\Rightarrow d\left( \frac{y}{x} \right)=\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\left\{ \frac{dx}{x} \right\}$
$\Rightarrow \frac{d\left( \frac{y}{x} \right)}{\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}}}=\frac{dx}{x}$
$Putting,u=\frac{y}{x}\Rightarrow \frac{du}{\sqrt{1+{{u}^{2}}}}=\frac{dx}{x}$
Integrating we get,
$\log \left( u+\sqrt{1+{{u}^{2}}} \right)=\log x+\log c$
$\Rightarrow u+\sqrt{1+{{u}^{2}}}=cx$
$\Rightarrow \frac{y}{x}+\sqrt{1+\frac{{{y}^{2}}}{{{x}^{2}}}}=cx$
$\therefore y+\sqrt{{{x}^{2}}+{{y}^{2}}}=c{{x}^{2}}$
is the required general solution.

 Example 07

$Solve:\left\{ xy\sin \left( xy \right)+\cos \left( xy \right) \right\}ydx+\left\{ xy\sin \left( xy \right)-\cos \left( xy \right) \right\}xdy=0$
Solution:
$Given,\left\{ xy\sin \left( xy \right)+\cos \left( xy \right) \right\}ydx+\left\{ xy\sin \left( xy \right)-\cos \left( xy \right) \right\}xdy=0$
$\Rightarrow xy\sin \left( xy \right).ydx+\cos \left( xy \right).ydx+xy\sin \left( xy \right).xdy-\cos \left( xy \right).xdy=0$
$\Rightarrow xy\sin \left( xy \right)\left\{ ydx+xdy \right\}+\cos \left( xy \right)\left\{ ydx-xdy \right\}=0$
$\Rightarrow \frac{\sin \left( xy \right)}{\cos \left( xy \right)}d\left( xy \right)+\frac{dx}{x}-\frac{dy}{y}=0$
Integrating we get,
$-\log \left| \cos \left( xy \right) \right|+\log \left| x \right|-\log \left| y \right|=\log \left| c \right|$
$\Rightarrow \log \left| x \right|=\log \left| c \right|+\log \left| y \right|+\log \left| \cos \left( xy \right) \right|$
$\Rightarrow \log \left| x \right|=\log \left| cy\cos \left( xy \right) \right|$
$\therefore x=cy\cos \left( xy \right)$
is the required general solution.

 Example 08

$Solve:\left( {{y}^{2}}{{e}^{x{{y}^{2}}}}+4{{x}^{3}} \right)dx+\left( 2xy{{e}^{x{{y}^{2}}}}-3{{y}^{2}} \right)dy=0$
Solution:
Comparing the given equation with
$Mdx+Ndy=0$
$M={{y}^{2}}{{e}^{x{{y}^{2}}}}+4{{x}^{3}},and,N=2xy{{e}^{x{{y}^{2}}}}-3{{y}^{2}}$
$\therefore \frac{\partial M}{\partial y}=2y{{e}^{x{{y}^{2}}}}+2x{{y}^{3}}{{e}^{x{{y}^{2}}}}$
$and,\frac{\partial N}{\partial x}=2y{{e}^{x{{y}^{2}}}}+2x{{y}^{3}}{{e}^{x{{y}^{2}}}}$
$\therefore \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
So the given equation is an exact equation.
Hence the solution of the equation given by
$\int{Mdx+\int{N\left( without,x-terms \right)dy=c}}$
$\therefore \int{\left( {{y}^{2}}{{e}^{x{{y}^{2}}}}+4{{x}^{3}} \right)dx+\int{\left( -3{{y}^{2}} \right)}}dy=c$
$\Rightarrow {{y}^{2}}.\frac{{{e}^{x{{y}^{2}}}}}{{{y}^{2}}}+4.\frac{{{x}^{4}}}{4}-3.\frac{{{y}^{3}}}{3}=c$
$\therefore {{e}^{x{{y}^{2}}}}+{{x}^{4}}-{{y}^{3}}=c$
is the required general solution.

 Example 09

$Solve:\left( {{x}^{3}}-3x{{y}^{2}} \right)dx+\left( {{y}^{3}}-3{{x}^{2}}y \right)dy=0,when,y(0)=1$
Solution:
Comparing the given equation with
$Mdx+Ndy=0$
$M={{x}^{3}}-3x{{y}^{2}},and,N={{y}^{3}}-3{{x}^{2}}y$
$\therefore \frac{\partial M}{\partial y}=-6xy,and,\frac{\partial N}{\partial x}=-6xy$
$\therefore \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
So the given equation is an exact equation.
Hence the solution of the equation given by
$\int{Mdx+\int{N\left( without,x-terms \right)dy=c}}$
$\Rightarrow \int{\left( {{x}^{3}}-3x{{y}^{2}} \right)dx+\int{{{y}^{3}}dy=c}}$
$\Rightarrow \frac{{{x}^{4}}}{4}-\frac{3{{x}^{2}}{{y}^{2}}}{2}+\frac{{{y}^{4}}}{4}=c$
$\therefore {{x}^{4}}-6{{x}^{2}}{{y}^{2}}+{{y}^{4}}=4c$
Given y = 1 when x = 0
$\therefore 0-6.0.1+1=4c\Rightarrow 4c=1$
Hence the required solution is
${{x}^{4}}-6{{x}^{2}}{{y}^{2}}+{{y}^{4}}=1$

 Example 10

$Solve:y+{{e}^{x}}+x\frac{dy}{dx}=0$
Solution:
$Given,y+{{e}^{x}}+x\frac{dy}{dx}=0$
$\Rightarrow \left( y+{{e}^{x}} \right)dx+xdy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$\therefore M=y+{{e}^{x}},and,N=x$
$\therefore \frac{\partial M}{\partial y}=1,and,\frac{\partial N}{\partial x}=1$
$\therefore \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
So the given equation is an exact equation.
Hence the solution of the equation given by
$\int{Mdx+\int{N\left( without,x-terms \right)dy=c}}$
$\Rightarrow \int{\left( y+{{e}^{x}} \right)dx+\int{0dy=c}}$
$\therefore xy+{{e}^{x}}=c$
is the required general solution.

 Example 11

Find the value of ‘a’, for which the differential equation
$\left( x{{y}^{2}}+a{{x}^{2}}y \right)dx+\left( x+y \right){{x}^{2}}dy=0$ is exact and solve it for this value of ‘a’.

Solution:
$We-have,\left( x{{y}^{2}}+a{{x}^{2}}y \right)dx+\left( x+y \right){{x}^{2}}dy=0$
Comparing the given equation with
$Mdx+Ndy=0$
$\therefore M=x{{y}^{2}}+a{{x}^{2}}y,and,N=\left( x+y \right){{x}^{2}}$
$\therefore \frac{\partial M}{\partial y}=2xy+a{{x}^{2}},and,\frac{\partial M}{\partial y}=3{{x}^{2}}+2xy$
Now the given equation will be exact if
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
$\Rightarrow 2xy+a{{x}^{2}}=3{{x}^{2}}+2xy$
$\therefore a=3$
Hence the given equation is exact for a = 3.
Thus the solution of the equation for a = 3 is given by
$\int{Mdx+\int{N\left( without,x-terms \right)dy=c}}$
$\Rightarrow \int{\left( x{{y}^{2}}+3{{x}^{2}}y \right)dx+\int{0dy=c}}$
$\Rightarrow \frac{1}{2}{{x}^{2}}{{y}^{2}}+3.\frac{{{x}^{3}}y}{3}=c$
$\therefore {{x}^{2}}{{y}^{2}}+2{{x}^{3}}y=k,[where,k=2c]$
is the required general solution.