One way to attack this is with the Exponential Shift Theorem.
For any polynomial $P$, constant $k$ and smooth function $u(x)$, $P(D) (\exp(k x) u) = \exp(k x) P(D+k) u$
Here $D$ is the differentiation operator $\dfrac{d}{dx}$. Thus e.g. for the polynomial $P(t) = t^2 + 2 t + 3$, $P(D) u = \dfrac{d^2 u}{dx^2} + 2 \dfrac{du}{dx} + 3 u$.
Now finding one nontrivial solution of the differential equation $P(D) y = 0$ would be easy if the constant term of the polynomial was $0$: $y=1$ would be a solution. But $P(t+k)$ has constant term $0$ (as a polynomial in $t$) if and only if $P(k) = 0$. Thus you look for roots of the polynomial. If $P(k) = 0$, then $P(D+k) 1 = 0$, and by Exponential Shift $P(D) \exp(k x) = 0$. Each distinct root of the polynomial $P$ gives you a solution; it is easy to check that these are linearly independent, and if the polynomial has all distinct roots you have a fundamental set of solutions.
If the roots are not all distinct, Exponential Shift can be applied again. Thus if $k$ is a root of $P$ with multiplicity $m$, the terms of $P(t+k)$ in $t^j$ for $0 \le j \le m-1$ are all $0$. Then $P(D+k) u$ involves only the $m$'th and higher derivatives of $u$, which means that it is $0$ if $u$ is a polynomial of degree less than $m$. So we get solutions $1, x, \ldots, x^{m-1}$ of $P(D+k) u = 0$ and by Exponential Shift solutions $\exp(kx), x \exp(kx), \ldots, x^{m-1} \exp(kx)$ of $P(D) y = 0$.