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

(1)

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

(2)

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.