7.5: Determining Whether an Expression is an Exact Differential

Since exact differentials have these important characteristics, it is valuable to know whether a given differential expression is exact or not. That is, given a differential expression of the form

\[df=M\left(x,y\right)dx+\ N\left(x,y\right)dy, \label{eq1} \]

we would like to be able to determine whether \(df\) is exact or inexact. It turns out that there is a simple test for exactness:

test for exactness

The differential in the form of Equation \ref{eq1} is exact if and only if

\[\dfrac{\partial M}{ \partial y} = \dfrac{\partial N}{ \partial x}. \label{eq2} \]

That is, this condition is necessary and sufficient for the existence of a function, \(f\left(x,y\right)\), for which \(M\left(x,y\right)=f_x\left(x,y\right)\) and \(N\left(x,y\right)=f_y\left(x,y\right)\).

In §4 we demonstrate that the condition is necessary. Now we want to show that it is sufficient. That is, we want to demonstrate: If Equation \ref{eq2} hold, then there exists a \(f\left(x,y\right)\), such that \(M\left(x,y\right)=f_x\left(x,y\right)\) and \(N\left(x,y\right)=f_y\left(x,y\right)\). To do this, we show how to find a function, \(f\left(x,y\right)\), that satisfies the given differential relationship. If we integrate \(M\left(x,y\right)\) with respect to \(x\), we have \[f\left(x,y\right)=\int{M\left(x,y\right)dx+h\left(y\right)} \nonumber \]

where \(h\left(y\right)\) is a function only of \(y\); it is the arbitrary constant in the integration with respect to \(x\), which we carry out with \(y\) held constant.

To complete the proof, we must find a function \(h\left(y\right)\) such that this \(f\left(x,y\right)\) satisfies the conditions:

\[\begin{align} \label{GrindEQ_1} M\left(x,y\right) &=f_x\left(x,y\right)\Leftrightarrow \\[4pt] &=\frac{\partial }{\partial x}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right] \end{align} \]

\[\begin{align}\label{GrindEQ_2} N\left(x,y\right) &=f_y\left(x,y\right) \Leftrightarrow \\[4pt] &=\frac{\partial }{\partial y}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right] \end{align} \]

The validity of condition in Equation \ref{GrindEQ_1} follows immediately from the facts that the order of differentiation and integration can be interchanged for a continuous function and that \(h\left(y\right)\) is a function only of \(y\), so that \({\partial h}/{\partial x=0}\).

To find \(h\left(y\right)\) such that condition in Equation \ref{GrindEQ_2} is satisfied, we observe that

\[\frac{\partial }{\partial y}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right]=\int{\left(\frac{\partial M\left(x,y\right)}{\partial y}\right)}dx+\frac{dh\left(y\right)}{dy} \nonumber \]

But since \[\frac{\partial M\left(x,y\right)}{\partial y} = \frac{\partial N\left(x,y\right)}{\partial x} \nonumber \]

this becomes

\[\frac{\partial }{\partial y}\left[\int{M\left(x,y\right)dx+h\left(y\right)}\right] = \int{\left(\frac{\partial N\left(x,y\right)}{\partial x}\right)}dx+\frac{dh\left(y\right)}dy =N\left(x,y\right)+\frac{dh\left(y\right)}{dy} \nonumber \]

Hence, condition in Equation \ref{GrindEQ_2} is satisfied if and only if \({dh\left(y\right)}/{dy}=0\), so that \(h\left(y\right)\) is simply an arbitrary constant.