PlanetPhysics/Exact Differential Equation

Let $R$ be a region in ${ℝ}^2$ and let the [[../Bijective/|functions\,]] $X:R\to ℝ$,\, $Y:R\to ℝ$ have continuous partial derivatives in $R$.\, The first order [[../DifferentialEquations/|differential equation]] $X(x,y)+Y(x,y)\frac{dy}{dx}=0$ or

$\begin{array}{c}X(x,y)dx+Y(x,y)dy=0\end{array}$

is called an 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\, $f:R\to ℝ$\, such that the equation (1) has the form $df(x,y)=0,$ whence its general integral is $f(x,y)=C.$

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

$\begin{array}{c}f(x,y):=\underset{P_0}{\overset{P}{\int }}[X(x,y)dx+Y(x,y)dy]\end{array}$

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 (2) is simply $\underset{0}{\overset{x}{\int }}\frac{2x}{1^3}dx+\underset{1}{\overset{y}{\int }}\frac{y^2-3x^2}{y^4}dy=\frac{x^2}{y^3}-\frac{1}{y}+1=x^2-\frac{1}{y}+\frac{x^2}{y^3}+1-x^2=\frac{x^2}{y^3}-\frac{1}{y}+1.$ Thus we have the general integral $\frac{x^2}{y^3}-\frac{1}{y}=C$ of the given differential equation.