constant evaluation when using differential equations. This is regards to constant evaluation when using differential equations.


*

*A solution is given to be:
$$y=(e^{2x}+e^x )  \ln⁡(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}$$

*A simplified solution in an answer book is given as:
$$y=(e^{2x}+e^x )  \ln⁡(1+e^{-x} )+(c_1 ) e^x+(c_2 ) e^{2x}$$
There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.
Sincerely,
Mary A. Marion

The reason why this makes sense is, the solution is actually a family of solutions, and what you're interested in isn't really any particular parametrisation of this family, but the set of them. So, to be precise, you might write it instead as:
The solutions are
$$\begin{align}
     S & :&&\mathbb{P}(\mathbb{R}\to\mathbb{R}),
  \\ S &=&&\left\{x \mapsto (e^{2x} + e^x)\cdot \ln(1+e^{-x}) - (c_1+1)\cdot e^x + (c_2-1)\cdot e^{2x}\quad\bigr|\quad c_1,c_2 \in \mathbb{R}\right\}
\end{align}$$
and that set of functions is literally the same set as
$$
  S = \left\{x \mapsto (e^{2x} + e^x)\cdot \ln(1+e^{-x}) + c_1\cdot e^x + c_2\cdot e^{2x}\quad\bigr|\quad c_1,c_2 \in \mathbb{R}\right\}
$$
because you're internally quantifying over $c_1$ and $c_2$, i.e. they're bound variables with no meaning outside of the set comprehension. And because they're quantified over all $\mathbb{R}$, it also doesn't make a difference to this whether you apply an affine transformation to them. Note that, if the domain of $c_1$ were constrained to e.g. an interval, this would need to transform.
