How do I know which equations can be treated as differential equations and which can't?

Question

I'm sometimes mystified by the use of differentials in physics. I don't understand which formulas—on which occasions—can be thought of as differential equations and which cannot.
While discussing work done by a piston during an isothermal process, my textbook does not treat $PV=nRT$ as a differential equation. Let me illustrate how $W$ is derived:
$$W=int_{V_1}^{V_2}Pmathrm{d}V=int_{V_1}^{V_2}frac{nRT}{V}mathrm{d}V=nRTlnfrac{V_2}{V_1}$$
My question is, why could I not treat the ideal gas law as a differential equation and say $Pmathrm{d}V=nRmathrm{d}T$? If I could, I'd then say:
$$W=int_{V_1}^{V_2}Pmathrm{d}V=int_{T_1}^{T_1}nRmathrm{d}T=0$$
Since $mathrm{d}T=0$ during an isothermal process. The result is erroneous, but why can I not argue in the following way? Why can't $PV=nRT$ be treated as a differential equation? How do I know which equations can be treated as such and which can't?

Gert · Accepted Answer

$$PV=nRTtag{1}$$
can not be considered a differential equation, simply because it contains no differentials.
Now, you can't just go and differentiate that equation as:
$$Pmathrm{d}V=nRmathrm{d}T$$
because $V=f(n,P,T)$ and $T=g(n,P,V)$ where $f$ and $g$ are multi-variable functions. To differentiate them you would have to use partial differentials ($partial$).
Differentiating $(1)$ you need to apply the product rule:
$$mathrm{d}(PV)=mathrm{d}(nRT)$$
Assuming $n=text{constant}$:
$$Pmathrm{d}V+Vmathrm{d}P=nRmathrm{d}T$$
But to derive the work done by an isothermal expansion/compression we simply use the general definition of work:
$$mathrm{d}W=F(x)mathrm{d}x$$
It's easy to show that for a piston $F(x)mathrm{d}x=Pmathrm{d}V$, so:
$$mathrm{d}W=Pmathrm{d}V$$
Then extract $P$ from the Ideal Gas Law.

Regarding your title question. Differential equations (DEs) typically arise to describe dynamic problems, where change, often (but not exclusively) in time, occurs.
Let's take a simple example. A mass $m$ sits on a rough incline, motionless. This is a static problem and requires no DEs.
Now we apply sufficient force on the mass for it to start moving. Newton's Second Law now states:
$$F_{net}=ma$$
or:
$$F_{net}=mfrac{mathrm{d}v}{mathrm{d}t}$$
This is of course a DE which allows us to calculate the rate of change of the velocity $v$.

GiorgioP · Answer

I'll try to concentrate my answer to your general question, more than concentrating on the example, although I'll touch it as an example of the general answer.
The basic answer is that Thermodynamics deals with functions of more than one variable. Therefore, in general, one has to expect to find partial differential equations. Partial differential equations are related but do not coincide with the differential one can write using thermodynamic laws.
Therefore, the way partial differential equations enter the play is more subtle than just taking the differential of left- and right-hand side of a thermodynamic relation.
Let's look in this perspective the equation of state of the perfect gas.
Le time rewrite the equation as
$$
frac{P}{T}= nfrac{R}{V}.tag{1}
$$
If we recall that
$$
left.frac{partial{S}}{partial{V}}right|_{U,n}=frac{P}{T}(U,V,n),tag{2}
$$
equation $(1)$ correspond to the partial differential equation
$$
left.frac{partial{S}}{partial{V}}right|_{U,n}=nfrac{R}{V}
$$
and in this simple case of the ideal gas it can be integrated to provide
$$
S(U,V,n)=nRlog (V)+ Phi(U,n)
$$
where $Phi(U,n)$ is an arbitrary function of $U$ and $n$ and cannot be determined by equation $(1)$ only. Adding more information, for example the relation between $T$ and $U$, it is possible to obtain the fundamental function $S(U,V,n)$ within an arbitrary constant.
Notice that with a more general equation of state, for example is the right-hand side of equation $(1)$ depends on $V$ and $U$, it would not be possible to avoid to solve a system of partial differential equations.

How do I know which equations can be treated as differential equations and which can't?

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