exact differential equation

Let $R$ be a region in ${ℝ}^2$ and let the functions  $X:R\to ℝ$,  $Y:R\to ℝ$ have continuous partial derivatives in $R$.  The first order differential equation

$$X(x,y)+Y(x,y)\frac{dy}{dx}=\mathrm{\hspace{0.33em}0}$$

or

$X(x,y)dx+Y(x,y)dy=\mathrm{\hspace{0.33em}0}$(1)

is called an exact differential equation, if the condition

${\displaystyle \frac{\partial X}{\partial y}}={\displaystyle \frac{\partial Y}{\partial x}}$(2)

is true in $R$.

By (2), the left hand side of (1) is the total differential of a function, there is a function  $f:R\to ℝ$  such that the equation (1) reads

$$df(x,y)=\mathrm{\hspace{0.33em}0},$$

whence its general integral is

$$f(x,y)=C.$$

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

$f(x,y):={\displaystyle {\int }_{P_0}^P}[X(x,y)dx+Y(x,y)dy]$(3)

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

$$\frac{2x}{y^3}dx+\frac{y^2-3x^2}{y^4}dy=\mathrm{\hspace{0.33em}0}.$$

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  $(0,\mathrm{\hspace{0.17em}1})$  to  $(x,\mathrm{\hspace{0.17em}1})$  and from this to  $(x,y)$,  the integral (3) 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.