I was messing with some partial derivative thing (probably incorrectly) and ended up with $$y=\frac {\sinh(ax)} x -a$$ somehow. (note: $a$ is any constant independent of $x$)
I went ahead and graphed it because why not, and found that it is (or at least appears to be) a polynomial with its vertex (right?) directly on $(0,0)$.
This lead me to two questions: Firstly, I noticed that as a decreases, the graph started to "become" the $x$-axis, which I assume has something to do with euler's identity as $\frac {e^{ax}-e^{-ax}} 2$ gets closer and closer to zero as $a$ (and/or $x$ I guess) gets smaller. On the other hand, as $a$ grows larger (not even that large) the graph seems to "become" the positive $y$-axis. This got me thinking about the dirac-$\delta$ function, as it has a "somewhat" similar in it's behavior, and I was wondering if there's any link between the two functions.
Secondly, I also found that $y=cosh(ax)-a$ seems to also be a positive polynomial centered at $(0,0)$. Then I noticed that if I take $\frac {sinh(ax)} x -a$ and multiply/divide it by an odd polynomial (I used $$\sum_{n=0}^{\infty}(x^{2n+1})$$ and just chose a random number as my upper limit for each time I tested it) I get an even polynomial. The opposite is true for the $cosh(ax)$ function (where I used $$ \sum_{n=0}^{\infty}(x^{2n})$$ instead, still just choosing a random number to stop on, and the $a$ subtracted in both cases is the lowest degree term of the series, a 1 for $cosh(ax)$ and whatever I decided was my lowest term in $sinh(ax)$ which if dividing by a polynomial, would be raised to a negative power if I recall correctly).
I assume this is because sinh is an odd function and cosh is even, but if I skip terms in the series, the graph becomes wildly different, seemingly able to take on any shape. So what I wanted to ask was can you represent all (or at least a significant portion) possible solutions to PDEs (at least 2-D solutions) as just $sinh(ax)$ or $cosh(ax)$ multiplied by some function derived from a PDE $\pm$ some constant (either independent or dependent, but with $a$ still remaining an independent constant)? That sounds a little outlandish to be true, but the question has been bugging me and reading papers on PDEs makes my eyes glaze over.