Turning inexact heat transfer differential into an exact one

Consider the first law of thermodynamics,

$$ dU = dq +dw$$

simplfying,

$$ dU + P_{text{ext}} dV = dq$$

Now we can say that $ q $ is a function of $ U$ and $V$

$ dq(U,V) = dU + P_{text{ext}} dV$

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For a differential $dF(x,y) = A, dx + B, dy $ to be exact,

$$ frac{partial A}{partial y} = frac{partial B}{partial x}$$ is a necessary condition.

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Clearly the function $q(U,V)$ doesn’t obey this definition, and hence, let us multiply both sides by an integrating factor $ phi(U,V)$ such that the condition of exact differential is satisfied.

$$ phi(U,V) , dq = phi(U,V) , dU + phi(U,V) P_{text{ext}} , dV$$

For this to be exact,

$$ frac{ partial phi(U,V) P_{text{ext}} }{partial U} = frac{partial phi(U,V) }{partial V}$$

Which leads to:

$$ left( frac{partial P_{text{ext}} }{ partial U} right)_V phi + P_{text{ext}}(frac{partial phi}{partial U})_V =(frac{partial phi}{partial V})_U $$

Now, I’m not sure how to get a general solution for the above partial differential equation…

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My real goal is to derive the expression for entropy at end and prove that $ frac{1}{T}$ is the integrating factor but I’m a bit stuck. I’ve seen the proof that $ frac{1}{T}$ is the integrating factor by people pointing to carnots theorem that the circulation of the differentials over a loop is zero but I wanted to derive it using differential equations.

Now, I’m thinking of how I can include the assumption that the process is reversible because the entropy definition is written used $ dq_{text{rev}}$; also maybe derive the entropy generation term.

Any hints?

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Reference for integrating factors