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exact differential equation (Definition)

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

$\displaystyle X(x,\,y)+Y(x,\,y)\frac{dy}{dx} = 0$
or
$\displaystyle X(x,\,y)dx+Y(x,\,y)dy = 0$ (1)
is called an exact differential equation, if the condition

$\displaystyle \frac{\partial X}{\partial y} = \frac{\partial Y}{\partial x}$
is true in $R$.

Then there is a function  $f\!: R \to \mathbb{R}$  such that the equation (1) has the form

$\displaystyle d\,f(x,\,y) = 0,$
whence its general integral is

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

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

$\displaystyle f(x,\,y) := \int_{P_0}^P [X(x,\,y)\,dx+Y(x,\,y)\,dy]$ (2)
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

$\displaystyle \frac{2x}{y^3}\,dx+\frac{y^2-3x^2}{y^4}\,dy = 0.$
This equation is exact, since

$\displaystyle \frac{\partial}{\partial y}\frac{2x}{y^3} = -\frac{6x}{y^4} = \frac{\partial}{\partial x}\frac{y^2-3x^2}{y^4}.$
If we use as the integrating way the broken line from  $(0,\,1)$  to  $(x,\,1)$  and from this to  $(x,\,y)$,  the integral (2) is simply

$\displaystyle \int_0^x\frac{2x}{1^3}\,dx+\!\int_1^y\frac{y^2-3x^2}{y^4}\,dy = \... ...}{y}+1 = x^2-\frac{1}{y}+\frac{x^2}{y^3}+1-x^2 = \frac{x^2}{y^3}-\frac{1}{y}+1.$
Thus we have the general integral

$\displaystyle \frac{x^2}{y^3}-\frac{1}{y} = C$
of the given differential equation.



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Cross-references: differential equation, functions

This is version 1 of exact differential equation, born on 2009-04-18.
Object id is 657, canonical name is ExactDifferentialEquation.
Accessed 248 times total.

Classification:
Physics Classification: 02.30.Hq (Ordinary differential equations)
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