Day: June 1, 2022

Posted on

202206011159 Problem 5E (Q25)

Test each of the following differentials to see whether they are exact, using two methods for each:

(a) -y\sin x\,\mathrm{d}x+\cos x\,\mathrm{d}y,

(b) y\,\mathrm{d}x+x\,\mathrm{d}y,
(c) yx^3e^x\,\mathrm{d}x+x^3e^x\,\mathrm{d}y,
(d) (1+x)ye^x\,\mathrm{d}x+xe^x\,\mathrm{d}y,
(e) 4x^3y^{-2}\,\mathrm{d}x-2x^4y^{-3}\,\mathrm{d}y.

K. S. Stowe. (2007). An Introduction to Thermodynamics and Statistical Mechanics

Revision. (exact differentials)

The differential of a function is given by Eq. (5.8):

\mathrm{d}F=\displaystyle{\frac{\partial F}{\partial x}\,\mathrm{d}x+\frac{\partial F}{\partial y}\,\mathrm{d}y}.

Therefore, one way to determine whether a differential given by Eq. (5.9):

\mathrm{d}\Phi =g(x,y)\,\mathrm{d}x+h(x,y)\,\mathrm{d}y

is exact is to see whether we can find some function F(x,y) such that

\displaystyle{\frac{\partial F}{\partial x}=g(x,y)}\qquad \textrm{and}\qquad\displaystyle{\frac{\partial F}{\partial y}=h(x,y)}.

If we can, the differential is exact, and if we can’t, it is inexact. Alternatively, we can use the identity

\displaystyle{\frac{\partial^2F}{\partial y\partial x}=\frac{\partial^2F}{\partial x\partial y}}.

Combining this with equations (5.8) and (5.9), we can see that for exact differentials, Eq. (5.10):

\displaystyle{\frac{\partial g}{\partial y}=\frac{\partial h}{\partial x}}.

Text on pg. 89, Sec. E, Ch. 5

Roughwork.

(a)

We can see that this is indeed an exact differential of the function

F=y\cos x+\textrm{constant},

because

\displaystyle{\frac{\partial F}{\partial x}=-y\sin x}\qquad\textrm{and}\qquad\displaystyle{\frac{\partial F}{\partial y}=\cos x}.

Or we can use Eq. (5.10). For this example g=-y\sin x and h=\cos x, so

\displaystyle{\frac{\partial g}{\partial y}=-\sin x}\qquad\textrm{and}\qquad\displaystyle{\frac{\partial h}{\partial x}=-\sin x}.

The two are the same, so the differential is exact.

Parts (b) to (e) are left to the reader as an exercise. Cf. analytic functions et Cauchy-Riemann equations.