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.

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