Talk:PlanetPhysics/Exact Differential Equation

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: exact differential equation %%% Primary Category Code: 02.30.Hq %%% Filename: ExactDifferentialEquation.tex %%% Version: 1 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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Let $R$ be a region in $\mathbb{R}^2$ and let the \htmladdnormallink{functions\,}{http://planetphysics.us/encyclopedia/Bijective.html} $X\!: R \to \mathbb{R}$,\, $Y\!: R \to \mathbb{R}$ have continuous partial derivatives in $R$.\, The first order \htmladdnormallink{differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} $$X(x,\,y)+Y(x,\,y)\frac{dy}{dx} = 0$$

or \begin{align} X(x,\,y)dx+Y(x,\,y)dy = 0 \end{align} is called an {\em exact differential equation}, if the condition $$\frac{\partial X}{\partial y} = \frac{\partial Y}{\partial x}$$ is true in $R$.

Then there is a function\, $f\!: R \to \mathbb{R}$\, such that the equation (1) has the form $$d\,f(x,\,y) = 0,$$ whence its general integral is $$f(x,\,y) = C.$$

The solution function $f$ can be calculated as the line integral \begin{align} f(x,\,y) := \int_{P_0}^P [X(x,\,y)\,dx+Y(x,\,y)\,dy] \end{align} 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 $\equiv 1$).\\

\textbf{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 (2) 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.

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