Form of solutions of linear ODEs with constant coefficients I am studying differential equations, and there is something I cannot accept. 
For homogeneous linear ODEs with constant coefficients, the textbook always tests candidate solutions of the form $ e^{cx} $. However, I wonder whether there exist solutions of another form, not based on the exponential function. 
Must the solution of such ODEs be some linear combination of $x^k e^{cx}$? Can no other form be solution? Why should the general solution be of such form?
 A: 'homogeneous linear ODE' implies that some linear combination of various derivatives of your initial function sums to zero.
The remarkable thing about the exponential function is that it is its own derivative. More generally, the exponential function is the eigenfunction of the differential operator and will remain unchanged up to a multiplicative constant upon differentiation, whereas for example polynomials will change their order.
A: I will try to address some of your questions making an initial core assumption, whose importance has already been pointed out in the previous answer.
Assume that we have at hands a function, which we call $u(\cdot)$, such that
$$ (*) \qquad
\begin{cases}
u'(t)=u(t), \qquad t \in \mathbb{R} \\
u(0) = 1.
\end{cases}
 $$
It is easy to see that such a function plays a role in findind the solution of linear, homogeneous ODEs with constant coefficients, at least for first and second order where computations are rather clean. Indeed for example if $c \in \mathbb{R}$ and for all $x$,
$$ (u(cx))'=u'(cx)c = c u(cx),  \qquad u(c \cdot 0)=u(0)=1,$$
which means that $y_0u(cx)$ is a solution of the Cauchy problem $y'=cy, \ y(0)=y_0$. The same goes on for second order equations, for all $x$:
$$ (u(-x))''=( - u'(-x))'=(-u(-x))'= u'(-x) = u(-x), $$
which tells us that the functions $u(x), u(-x), u(x)\pm u(-x)$ are all general integrals for the equation $y''=y$ (here, we also used the linearity property). 
To conclude this first part, let us use some Taylor expansions. Consider the problem
$$ 
\begin{cases}
y'' - 2y' + y = 0 \\
y(0)=0, \\
y'(0)=1. \\
\end{cases}
$$
Differentiating the equation and using the initial data, we find that if $y(\cdot)$ is a solution of this problem and it analytic (read: it is infinitely differentiable and it is the sum of its Taylor series), then for $n \ge 0$, 
$$ y^{(n)}(0)=n \implies y(x)= \sum_{n \ge 1} \frac{n x^n}{n!} = \sum_{n \ge 1} \frac{x^n}{(n-1)!} = x \sum_{n \ge 1 } \frac{x^{n-1}}{(n-1)!} = x \sum_{n \ge 0} \frac{x^n}{n!} = x u(x); $$
where the last equality holds if we assume that $u(\cdot)$ is also analytic (notice that owing to the definition, $u^{(n)}(0)=1 \ \forall n$). Similar manipulations can be done for higher order homogeneous, linear equations.
Now, $(*)$ is precisely one possible way of defining the exponential function (see the interesting https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function). In some sense, the exponential function is intrinsically involved in linear homogeneous ODEs because of its very definition(s). Note, for instance, that a function satisfying $(*)$ immediately turns out to have derivatives of every possible order! 
The question why can't there be solutions of other type? is more tricky. My perspective is the following: one proves/observes that some kind of functions involving our $u(\cdot)$, or $\exp(\cdot)$, are solutions to a given problem (see above), or a class of problems; then proves separately, and perhaps with different techniques, that the solution is unique or almost unique (i.e, the set of solutions in a finite dimensional vector space...), closing the loop.
