PlanetPhysics/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

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

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\, $f\!:R\to \mathbb {R}$\, 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

${\begin{matrix}f(x,\,y):=\int _{P_{0}}^{P}[X(x,\,y)\,dx+Y(x,\,y)\,dy]\end{matrix}}$

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 $\int _{0}^{x}{\frac {2x}{1^{3}}}\,dx+\!\int _{1}^{y}{\frac {y^{2}-3x^{2}}{y^{4}}}\,dy={\frac {x^{2}}{y^{3}}}-{\frac {1}{y}}+1=x^{2}-{\frac {1}{y}}+{\frac {x^{2}}{y^{3}}}+1-x^{2}={\frac {x^{2}}{y^{3}}}-{\frac {1}{y}}+1.$ Thus we have the general integral ${\frac {x^{2}}{y^{3}}}-{\frac {1}{y}}=C$ of the given differential equation.