Suppose I have a set containing $n$ elements and I consider the number of partitions of it in which all the constituent classes are even. I let $R_n$ denote this number ($R_0=1$) and proceed as follows to obtain the exponential generating function for $R_n$.
Since $$R_n=\sum_{k=1}^{\lfloor n/2\rfloor}(\text{No. of such partitions where n lies in a class containing 2k elements})$$ so $R_n=\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n-1}{2k-1}R_{n-2k}$. Now let $r(x)$ be the exponential generating function of $R_n$. Then
$\begin{align}r'(x)&=\sum_{n=1}^\infty\frac{R_nx^{n-1}}{(n-1)!}\\&=\sum_{n=1}^\infty\sum_{k=1}^{\lfloor n/2\rfloor}\frac{R_{n-2k}x^{n-1}}{(n-2k)!(2k-1)!}\\&=\sum_{k=1}^\infty\sum_{n=2k}^\infty \frac{R_{n-2k}x^{n-1}}{(n-2k)!(2k-1)!}\\&=\sum_{k=1}^\infty\frac{x^{2k-1}}{(2k-1)!}\sum_{n=2k}^\infty \frac{x^{n-2k}R_{n-2k}}{(n-2k)!}\\&=\sum_{k=1}^\infty\frac{x^{2k-1}}{(2k-1)!}\sum_{n=0}^\infty \frac{x^nR_n}{n!}\\&=r(x)\sum_{k=1}^\infty \frac{x^{2k-1}}{(2k-1)!}\end{align}$.
This shows that $r'(x)=r(x)\sinh x$. This differential equation has the solution $r(x)=e^{\cosh x+c}$. The boundary condition yields $r(x)=e^{\cosh x}$.
My doubt is that in my book the answer is provided as $r(x)=e^{\cosh x-1}$. I simply want to know whether it is a misprint in the book or if I am wrong, where have I gone wrong.
The reference cited is : Alfred Renyi, New methods and results in combinatorial analysis, I, II, MTA III. Osztaly Kozlemenyei, 16(1966), 77-105, 159-177. English summary on p.177 (paper is in Hungarian), note esp. pp.78-91. I do not have access to it.
It is easiest to approach this problem using symbolic combinatorics as formalized by Flajolet and Sedgewick. Let $\mathcal{R}$ be the combinatorial class in question. Then we have straightforwardly the following combinatorial class equation $$\mathcal{R} = \mathfrak{P}(\mathfrak{P}_2(\mathcal{Z}) + \mathfrak{P}_4(\mathcal{Z}) +\mathfrak{P}_6(\mathcal{Z})+\cdots),$$ where $\mathcal{Z}$ is the singleton class.
Translating this into generating functions we obtain for the exponential generating function $R(z)$ that $$ R(z) = \exp\left(\frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \cdots\right) = \exp\left(-1 + \frac{z^0}{0!} + \frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \cdots\right)\\= \exp\left(-1 + \frac{1}{2}\exp(z) + \frac{1}{2}\exp(-z)\right) = \exp(-1 + \cosh(z)).$$ This shows from a modern perspective that there is no misprint in the book.