Nice expressions for dS and dH individually
in terms of the height and slope of the melting curve at
any and every temperature:

dH =  T^2 R m'(t) ({m+1 \over m( 1-m) } )


dS =  - R ( \ln {(1-m)^2 [DNA] \over m })
- T m'(t) ({m+1 \over m( 1-m) } )

(sorry for the math typesetting format, but I think
you can see the gist.)

Both come from differentiating the equation in the equal initial
concentrations [DNA], in either form (here p is the coiled proportion,
m=1-p)

dH -T dS = - R T \ln {1-p \over p^2[DNA]}

or

T^{-1}dH - dS = - R \ln {1-p \over p^2[DNA]}

In the first, dH drops out and we can solve for dS
and in the second, the opposite occurs.


The full formulation for the melting curve, m=1-p,
as an initial value problem with a differential equation give
in terms of either dH or dS alone, and an initial condition
which incorporates the other:

p'=f(T,p,dH) p(T_m)=1/2

- specifically, given dH, p(T) satisfies the differential equation

p'(T) = dH p(1-p) / ( RT^2 (p-2))

and the point condition which involves dS as well,

p(  dH / ( dS + R ln [DNA] /2 )) = 1/2


or alternatively, given dS,

p' = dS p(1-p)/(RT(p-2)) +  p(1-p)/((p-2)T) ln ( p^2[DNA]/(1-p)) )

and the same point condition, which involves dH as well.

Setting the differential equations equal and multiplying
through by (p-2)RT^2/(p(1-p))  recovers

dH  = T dS +  RT ln ( p^2[DNA]/(1-p)) )

from which these differential equations were derived in the first place,
and which also leads to the quadratic formula form of their
solution.


Exact solution of DE by separation and partial fractions


p'(T) = dH p(1-p) / ( RT^2 (p-2))

p' 0 when p=0,1, p never 2 so no singularity there,
p' 0  at T= infty, infty at 0 but p=0 so cancels?

separable?

dH R^{-1} T^{-2} dT = (p-2)/(p(1-p)) dp

integrate:

The left:

-dH/(RT) =

The right:

Partial fractions!

C/p + D/(1-p)  =  (p-2)/(p(1-p))

cross multiplying gives

D-C = 1 C=-2 So D=-1

So the right becomes:

-2/p - 1/(1-p) dp

Integrating: (p and 1-p are positive so no absolute values) u=1-p du=-dp

-2 ln p + ln (1-p)  = ln ((1-p)/p^2) + C

(C=dS+ concentration term, of course!)

in other words,

dH = -RT ln ((1-p)/p^2) + C

Gets us back where we started from which we solved using the
quadratic formula!