# exact differential equation

Let $R$ be a region in $\mathbb{R}^{2}$ and let the functions  $X\!:R\to\mathbb{R}$,  $Y\!:R\to\mathbb{R}$ have continuous partial derivatives in $R$.  The first order differential equation

 $X(x,\,y)+Y(x,\,y)\frac{dy}{dx}\;=\;0$

or

 $\displaystyle X(x,\,y)dx+Y(x,\,y)dy\;=\;0$ (1)

is called an exact differential equation, if the condition

 $\displaystyle\frac{\partial X}{\partial y}\;=\;\frac{\partial Y}{\partial x}$ (2)

is true in $R$.

By (2), the left hand side of (1) is the total differential of a function, there is a function  $f\!:R\to\mathbb{R}$  such that the equation (1) reads

 $d\,f(x,\,y)\;=\;0,$

whence its general integral is

 $f(x,\,y)\;=\;C.$

The solution function $f$ can be calculated as the line integral

 $\displaystyle f(x,\,y)\;:=\;\int_{P_{0}}^{P}[X(x,\,y)\,dx+Y(x,\,y)\,dy]$ (3)

along any curve $\gamma$ connecting an arbitrarily chosen point  $P_{0}=(x_{0},\,y_{0})$  and the point  $P=(x,\,y)$  in the region $R$ (the integrating factor is now $\equiv 1$).

Example.  Solve the differential equation

 $\frac{2x}{y^{3}}\,dx+\frac{y^{2}-3x^{2}}{y^{4}}\,dy\;=\;0.$

This equation is exact, since

 $\frac{\partial}{\partial y}\frac{2x}{y^{3}}\;=\;-\frac{6x}{y^{4}}\;=\;\frac{% \partial}{\partial x}\frac{y^{2}-3x^{2}}{y^{4}}.$

If we use as the integrating way the broken line from  $(0,\,1)$  to  $(x,\,1)$  and from this to  $(x,\,y)$,  the integral (3) is simply

 $\int_{0}^{x}\frac{2x}{1^{3}}\,dx+\!\int_{1}^{y}\frac{y^{2}-3x^{2}}{y^{4}}\,dy% \;=\;\frac{x^{2}}{y^{3}}-\frac{1}{y}+1\;=\;x^{2}-\frac{1}{y}+\frac{x^{2}}{y^{3% }}+1-x^{2}=\frac{x^{2}}{y^{3}}-\frac{1}{y}+1.$

Thus we have the general integral

 $\frac{x^{2}}{y^{3}}-\frac{1}{y}\;=\;C$

of the given differential equation.

Title exact differential equation ExactDifferentialEquation 2013-03-22 18:06:17 2013-03-22 18:06:17 pahio (2872) pahio (2872) 10 pahio (2872) Result msc 34A05 HarmonicConjugateFunction Differential TotalDifferential exact differential equation