Exact and inexact differentials

An exact differential is a differential equation $du$ , for instance of two variables, of the form $du=M(x,y)dx+N(x,y)dy$ , where $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

Consider the function $u=xy+c$ , where $c$ is a constant. Its total differential is $du=\left(\frac{\partial u}{\partial x}\right)_ydx+\left(\frac{\partial u}{\partial y}\right)_xdy$ . Comparing with the general form of an exact differential, we have $M(x,y)=\left(\frac{\partial u}{\partial x}\right)_y$ and $N(x,y)=\left(\frac{\partial u}{\partial y}\right)_x$ . Since $\frac{\partial M}{\partial y}=\frac{\partial^2u}{\partial y\partial x}=1=\frac{\partial^2u}{\partial x\partial y}=\frac{\partial N}{\partial x}$ , $du$ is an exact differential. The equality of the mixed partials $\frac{\partial^2u}{\partial y\partial x}=\frac{\partial^2u}{\partial x\partial y}$ implies that the change in $u$ is independent of the path taken.

Conversely, an inexact differential is a differential equation $du$ of the form $du=M(x,y)dx+N(x,y)dy$ , where $\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$ . The change in $u$ , in this case, is dependent on the path taken.

Question

Show that $du=3ydx+xdy$ is an inexact differential.

Answer

We have $\frac{\partial M}{\partial y}=3$ and $\frac{\partial N}{\partial x}=1$ . So, $\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$ .

Another difference between an exact differential and an inexact differential is that an exact differential integrates directly to give the function $u$ , whereas an inexact differential does not.

Question

Using the integral criterion, show that $du=xdy+ydx$ is an exact differential, while $du=3ydx+xdy$ is an inexact differential.

Answer

The integrated form of the first differential is evidently $u=xy+c$ . The detailed analysis involves two steps. First, integrating $M(x,y)=\left(\frac{\partial u}{\partial x}\right)_y$ with respect to $x$ , treating $y$ as a constant, yields

$u=xy+f(y)\;\;\;\;\;\;\;\;17a$

The function $f(y)$ accounts for terms in $u$ involving $y$ or constants, which differentiate to zero when differentiating $u$ with respect to $x$ . Secondly, integrating $N(x,y)=\left(\frac{\partial u}{\partial y}\right)_x$ with respect to $y$ , with $x$ as a constant, gives

$u=xy+g(x)\;\;\;\;\;\;\;\;17b$

Comparing eq17a and eq17b, we find $u=xy+c$ , where $f(x)=cx^0=cy^0=f(y)=c$ . In other words, $u=xy+c$ integrates directly from $du=xdy+ydx$ .

For the second differential, integrating with respect to $x$ yields $u=3xy+g(y)$ , and integrating with respect to $y$ gives $u=xy+h(x)$ . Clearly, $3xy+g(y)\neq xy+h(x)$ , for all $g(y)$ and $h(x)$ , indicating that the second differential does not directly integrate to give a function.

The fact that an exact differential integrates directly to give the function $u$ but an inexact differential does not, implies that $M(x,y)$ and $N(x,y)$ for a differentiable function must be of the appropriate forms of $\left(\frac{\partial u}{\partial x}\right)_y$ and $\left(\frac{\partial u}{\partial y}\right)_x$ , respectively. In other words, the total differential of a differentiable function must be an exact differential.