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I have been solving a thermodynamics problem where I have two representations:

$(\dfrac{∂v}{∂p})_{θ}$ = $vg(p)e^{-θ}$; $(\dfrac{∂v}{∂θ})_{p}$ = $vp^{2}e^{-θ}$

I want to find $v(p,θ)$, so I am using the fact that $dv$ is an exact differential:

$dv = (\dfrac{∂v}{∂p})_{θ} dp + (\dfrac{∂v}{∂θ})_{p} dθ$;

So $(\dfrac{∂M}{∂θ})_{p} = (\dfrac{∂N}{∂p})_{θ}$, where $M = (\dfrac{∂v}{∂p})_{θ}$ and $N = (\dfrac{∂v}{∂θ})_{p}$

$-vg(p)e^{-θ} = 2vpe^{-θ}$

So for this equality I found $g(p) = -2p$

Substituting I get this expression

$dv(p,θ) = -2vpe^{-θ}dp + vp^{2}e^{-θ}dθ$

But now I'm stuck on this part, cause I dont know how to solve an exact differential equation with two independent variables.

Rubén Guridi
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1 Answers1

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We have \begin{align} \frac{\partial v}{\partial p}&=-2vpe^{-\theta}, \tag{1a}\\ \frac{\partial v}{\partial \theta}&=vp^2e^{-\theta}. \tag{1b} \end{align} We first solve $(1\text{a})$ as a conventional ODE, treating $\theta$ as a constant: $$ (1\text{a})\implies \int\frac{dv}{v}=-e^{-\theta}\int 2p\,dp \implies \ln |v| = -e^{-\theta}p^2+f(\theta) $$ $$ \implies v=\exp\left(-e^{-\theta}p^2+f(\theta)\right). \tag{2} $$ Now we plug $(2)$ into $(1\text{b})$ and solve the resulting ODE for $f(\theta)$: $$ \left(e^{-\theta}p^2+f'(\theta)\right)\exp\left(-e^{-\theta}p^2+f(\theta)\right) =p^2e^{-\theta}\exp\left(-e^{-\theta}p^2+f(\theta)\right) $$ $$ \implies f'(\theta)=0 \implies f(\theta)=c. \tag{3} $$ We conclude, therefore, that $$ v(p,\theta)=C\exp\left(-e^{-\theta}p^2\right). \tag{4} $$

Gonçalo
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