March 2019 ~ Easy to understand maths

The differential equation of type

$M\,dx +N \, dy=0$ (where M and N are function of x and y) is called an exact differential equation when
$M\,dx +N \, dy=du$ ( where u is a function of x and y).

Examples :

1. $x\,dy +y \, dx=0$ is an exact differential equation as $x\,dy +y \, dx=d(x,y)$

2. $\frac{ x\,dy -y \, dx }{x^{2}}=0$ is an exact differential equation as $\frac{ x\,dy -y \, dx }{x^{2}}=d\left ( \frac{y}{x} \right )$

3. The differential equation

$\sin x\, \cos y\: dy+\cos x\: \sin y\: dx=0$ is an exact differential equations as $\sin x\, \cos y\: dy+\cos x\: \sin y\: dx=d(\sin x\: \sin y)$

Necessary and sufficient condition

Article. Find the necessary and sufficient condition that the equation $M\,dx +N \, dy=0$ ( where M and N are function of x and y with the condition that $M,N,\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$ are continuous function of x and y) may be exact.

Proof 1. Necessary condition
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

2. Condition is sufficient

$\int_{y\: constant }M\: dx+\int (terms\: in\: N\: not\: containing\: x) dy=c$

Integrating factor

An integrating factor (abbreviatef I.F) of a differential equation is such a factor such that if the equation is multiplied by it, the result equation is exact.

Five rules for finding integrating factor

If   $M\,dx +N \, dy=0$ is not exact and it is difficult to find integrating factor, then following five rules help us in finding integrating factor.

Rule 1. If the equation   $M\,dx +N \, dy=0$ is homogenous in x and y i.e. if M and N are homogenous function of the same degree in x and y, then   $\frac{1}{Mx+Ny}$ is an I.F. provided $M\,dx +N \, dy\neq 0.$

Rule 2. If the equation   $M\,dx +N \, dy=0$ is of the form   $f_{1}(x,y)\: y\: dx + f_{2}(x,y)\: x\: dy=0$ , then   $\frac{1}{Mx-Ny}$ is an I.F. provided   $M\,dx +N \, dy\neq 0.$

Rule 3. If the equation   $M\,dx +N \, dy=0$ , $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$ is a function of x only =f(x) then   $e^{\int f(x)dx}$ is an I.F.

Rule 4. If the equation   $M\,dx +N \, dy=0$ , $\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}$ is a function of y only =f(y) then   $e^{\int f(y)dy}$ is an I.F.

Rule 5. If the equation is

$x^{a}y^{b}(my\: dx +nx\: dy)+x^{a'}y^{b'}(m'y\: dx+n'x\: dy)=0$ , then $x^{h}y^{k}$ is an I.F. when $\frac{a+h+1}{m}=\frac{b+k+1}{n},\frac{a'+h+1}{m'}=\frac{b'+k+1}{n'}$ .

Example

Q1. Solve the differential equation
$(\cos x\cos y-\cot x)dx-(\sin x\sin y)dy=0$ .
Sol. The given differential equation is
$(\cos x\cos y-\cot x)dx-(\sin x\sin y)dy=0$
Comparing it with $M\,dx +N \, dy=0$ , we get
$M=\cos x\cos y-\cot x,\: N=-\sin x\sin y$
$\frac{\partial M}{\partial y}=-\cos x\sin y\: \: and\: \: \frac{\partial N}{\partial x}=-\cos x\sin y$
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Given equation is exact and its solution is

$\int_{y\: constant }M\: dx+\int (terms\: in\: N\: not\: containing\: x) dy=c$

$\int_{y\: constant }(\cos x\cos y-\cot x)\: dx+0=c$

$(\cos y)\left ( \int \cos x\: dx \right )-\int \cot x\: dx=c$

$\cos y\sin x-\log \left | \sin x \right |=c.$

Q2. Solve the following differential equation $(y^{3}-2yx^{2})dx+(2xy^{2}-x^{3})dy =0$
Sol.

The given differential equation is
$(y^{3}-2yx^{2})dx+(2xy^{2}-x^{3})dy =0$

Which is homogenous in x,y.
Comparing with $M\,dx +N \, dy=0$ , we get

$M=y^{3}-2yx^{2},\: \: \: N=2xy^{2}-x^{3}$

$I.F.= \frac{1}{Mx+Ny}=\frac{1}{xy^{3}-2x^{3}y+2xy^{3}-x^{3}y}=\frac{1}{3xy(y^{2}-x^{2})}$
Multiple both side by $\frac{1}{3xy(y^{2}-x^{2})}$ , we get,

$\frac{y^{2}-2x^{2}}{3xy(y^{2}-x^{2})}dx+\frac{(2y^{2}-x^{2})}{ 3xy(y^{2}-x^{2}) }dy=0$

Which is exact and its solution is

$\frac{1}{3}\int_{y\: constant }\frac{y^{2}-2x^{2}}{x(y^{2}-x^{2})}\: dx+\frac{1}{3}\int \frac{1}{y} dy=c$

$\frac{1}{3}\int_{y\: constant }\left ( \frac{1}{x}-\frac{x}{y^{2}-x^{2}} \right )\: dx+\frac{1}{3}\int \log \left | y \right | dy=c$

$\frac{1}{3}\left [ \log|x| +\frac{1}{2}\log|y^{2}-x^{2}|+ \log|y| \right ]=c$

$\log |(x^{2}y^{2})(y^{2}-x^{2})|=6c$

$x^{2}y^{2}(y^{2}-x^{2})=|e^{6c}|=C$

Q3. Solve tge differential equation
$(x^{3}y^{4}+x^{2}y^{3}+xy^{2}+y)dx+(x^{4}y^{3}-x^{3}y^{2}+x^{2}y+x)dy=0$
Sol.
The given differential equation is

$(x^{3}y^{4}+x^{2}y^{3}+xy^{2}+y)dx+(x^{4}y^{3}-x^{3}y^{2}+x^{2}y+x)dy=0$

$(x^{3}y^{3}+x^{2}y^{2}+xy+1)y\: dx+(x^{3}y^{3}-x^{2}y^{2}+xy+1)x\: dy=0$

Which is of form $f_{1}(x,y)y\: dx+f_{2}(x,y)x\: dy=0$

Comparing it with $M\,dx +N \, dy=0$ , we get

$M=y(x^{3}y^{3}+x^{2}y^{2}+xy+1),\: N=x(x^{3}y^{3}-x^{2}y^{2}+xy+1)$

I.F.= $\frac{1}{Mx-Ny}$

= $\frac{1}{ xy(x^{3}y^{3}+x^{2}y^{2}+xy+1)-\: xy(x^{3}y^{3}-x^{2}y^{2}+xy+1) }$

= $\frac{1}{2x^{3}y^{3}}$
Multiple both side by $\frac{1}{2x^{3}y^{3}}$ , we get

$\frac{ x^{3}y^{3}+x^{2}y^{2}+xy+1 }{2x^{3}y^{2}}dx+\frac{ x^{3}y^{3}-x^{2}y^{2}+xy+1 }{2x^{2}y^{3}}dy=0$

Which is exact and its solution is

$\frac{1}{2y^{2}}\int \left ( y^{3}+\frac{y^{2}}{x}+\frac{y}{x^{2}}+\frac{1}{x^{3}} \right )dx+\frac{1}{2}\int \left ( -\frac{1}{y} \right )dy=c$

$\frac{1}{2y^{2}} \left ( xy^{3} +y^{2}\log x-\frac{y}{x}-\frac{1}{2x^{2}}\right )-\frac{1}{2}\log y=c$

Wednesday, 27 March 2019

Exact Differential Equations

Necessary and sufficient condition

Integrating factor

Five rules for finding integrating factor

Example

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