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I want to calculate the integration factor $\mu(E,V)$ for the differential $\mathrm{d}\sigma = \mu(E,V) \delta Q = \mu(E,V) \mathrm{d}E + \mu(E,V)p(u) \mathrm{d}V$. $p$ is a function of $u$ with $u = \frac{E}{V}$.

For a general $p$ I did the following

$\begin{align} \frac{\partial\mu}{\partial V} &= \frac{\partial (\mu \cdot p)}{\partial E} \\ &= p \cdot \frac{\partial \mu}{\partial E} + \mu \cdot \frac{\partial p}{\partial E} \\ &= p \cdot \frac{\partial \mu}{\partial E} + \mu \cdot \frac{\partial p}{\partial u} \frac{\partial u}{\partial E} \\ &= p \cdot \frac{\partial \mu}{\partial E} + \frac{\mu}{V} \cdot\frac{\partial p}{\partial u} \end{align}$

Can I solve this even further for a general $p$?

Now, I tried calculating $\mu$ for a specific $p = c \cdot u$. When I inserted this into the above equation I get

$\begin{align} \frac{\partial \mu}{\partial V} &= c \cdot \frac{E}{V} \cdot \frac{\partial \mu}{\partial E} + \frac{\mu}{V} \cdot c = \frac{c}{V} \left( E \cdot \frac{\partial \mu}{\partial E} + \mu \right) \end{align}$

How do I go on from here for getting $\mu$?

1 Answers1

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Write equation $\begin{align} \frac{\partial \mu}{\partial V} &= c \cdot \frac{E}{V} \cdot \frac{\partial \mu}{\partial E} + \frac{\mu}{V} \cdot c = \frac{c}{V} \left( E \cdot \frac{\partial \mu}{\partial E} + \mu \right) \end{align}$ as $\begin{align} \frac{V}{\mu} \frac{\partial \mu}{\partial V} &= {c} \left(1+ \frac{E}{\mu} \cdot \frac{\partial \mu}{\partial E} \right) \end{align}$ and introduce new variables $\begin{align} \textrm{ln}E = \epsilon, \textrm{ln}V = \nu, \textrm{ln}\mu(E,V) = M(\epsilon, \nu) \end{align}$ and upon substitution you get $$\begin{align} \frac{\partial M}{\partial \nu} &= {c} \left(1+ \frac{\partial M}{\partial \epsilon} \right) \end{align}$$ A trial solution to this is $M(\epsilon, \nu) = f(\epsilon) + g(\nu)$, substituting you get $$g'(\nu) = c(1+f'(\epsilon)) = K,$$ where $K$ is independent of $\epsilon, \nu$ hence we can integrate: $g'(\nu) = \nu_0 +K\nu$ and $f(\epsilon)=\epsilon_0 + \epsilon(K/c-1)$. Now going back to the original variables the integrating factor will be $$\mu(E,V) = \textrm{exp}[M(\epsilon, \nu)]\\ =\textrm{exp}[f(\epsilon)+g(\nu)]\\ =\textrm{exp}[\epsilon_0 + \epsilon(K/c-1)]\textrm{exp}[\nu_0 +K\nu] =E_0\textrm{exp}[\textrm{ln} E (K/c-1)]V_0\textrm{exp}[K\textrm{ln}V] =V_0E_0E^{(K/c-1)}V^K $$

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