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.