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


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*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
 A: As $c_1$ and $c_2$ are constants, we could define two other constants $\tilde{c_1}=-(c_1+1)$ and $\tilde{c_2}=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.
The answer key shows that 
$$y=(e^{2x}+e^x )  \ln⁡(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}\tag{1}$$
can be simplified to
$$y=(e^{2x}+e^x )  \ln⁡(1+e^{-x} )+(c_1) e^x+(c_2 ) e^{2x}\tag{2}$$
where $(2)$ could be rewritten as
$$y=(e^{2x}+e^x )  \ln⁡(1+e^{-x} )+(\tilde{c_1} ) e^x+(\tilde{c_2} ) e^{2x}\tag{3}$$
provided that
$$\tilde{c_1}=-(c_1+1)$$
$$\tilde{c_2}=c_2-1$$
The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.
A: Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $\epsilon=\frac{\epsilon}{2},$ for example, since $\epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.
Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.
A: Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$
Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants. 
A: 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.
