# Find an expression for S(T,x) from tension and specific heat

I'm working on a problem from a Statistical Mechanics lecture series online, and on the homework, I hit a bump in this problem.

Short version, we are given that the force to move the filament is $J=ax-bT+cTx$ and that the specific heat for a constant displacement is $C_x=A(x)T$, and in part be it is shown that $A$ is just a constant.
My first thought was to write the entropy as follows $$S(T,x)=\Delta S+S_0=S_0+\int_{(0,0)}^{(T,x)}\frac{\delta Q_{rev}}{T}.$$ But I get stuck around trying to find a reliable expression for $\delta Q_{rev}$.
I also tried using Maxwell Relations to find $\frac{\partial S}{\partial x}$ and $\frac{\partial S}{\partial T}$ and integrate the two to find $S(T,x)$. That's about all I can think of for solution methods. Any guidance would be appreciated!
The route of finding $\frac{\partial S}{\partial x}$ and $\frac{\partial S}{\partial T}$ is the more practical route. $\frac{\partial S}{\partial x}$ can be found using a simple Maxwell relation for the Helmholtz free energy: $$dA=dU-d(TS)=Jdx-SdT$$ $$\left. \frac{\partial A}{\partial x}\right|_T = J \, \, \mathrm{and} \, \, \left. \frac{\partial A}{\partial T}\right|_x = -S.$$ So then $-\frac{\partial J}{\partial T} = \frac{\partial S}{\partial x}=b-cx$. The other relation is obtained from the equation for specific heat: $$C=\frac{\delta Q}{\partial T}=C_x=AT$$ $$TdS= \delta Q = ATdT$$ $$\frac{\partial S}{\partial T} = A.$$ Integrating both expressions gives $S(T,x) = bx -\frac{cx^2}{2} + f(T)=AT+g(x)=AT+bx-\frac{cx^2}{2}+C$. Lastly, you simply put in $(T,x)=(0,0)$ and get that $C=S_0$.