Finding the kernel, eigenvalues, and eigenvectors of the operator $L(x) := x'' + 3 x' + 4 x$ I want to find the kernel, eigenvalues and eigenvectors of the differential operator:
$$L(x)=x''+3x'-4x$$
on the $\Bbb C \space \space \text{vectorspace} \space \space  C^{\infty}(\Bbb R)$ as well as the solution the the homogenous differential equation:
$$x''+3x'-4x=0$$
First question: I have only seen differential operators in the from of $\frac{d}{dx}$. Is there something special about $L(x)$? Does it have a special meaning in linear algebra?
Second question: As far as I know, eigenvalues and eigenvectors are only defined for square matrices. However the coefficient matrix of this differential equation is not square. Do I have to turn this second order diff. equation into a system of first order equations in order to find the eigenvalues and eigenvectors?
Third question: If the answer to the second question is yes. How do I turn $L(x)=x''+3x'-4x$ into a system of first order differential equations? I looked up how to convert an $n^{th}$ order diff. equation into a system of lower order differential equation (LINK) but the author always has equations that are equal to zero. Here my equation is equal to $L(x)$. How do I deal with that?
 A: First answer:
Your operator can be written in terms of the usual derivative operator.
We have $L=(\frac{d}{dx})^2+3\frac{d}{dx}-4$, where $4$ stands for multiplication by the constant.
From the point of view of linear algebra, $L$ is a linear operator/mapping/function from a vector space to itself.
Second answer:
A square matrix as we usually understand it corresponds to a linear map from a Euclidean space to itself.
The operator $L$ is also square in the sense that it goes from a space to itself.
The space is infinite dimensional, but still.
Eigenfunctions and eigenvalues are defined as usual: $Lf=\lambda f$ for some $\lambda\in\mathbb C$ and $f\in C^\infty(\mathbb R)$.
This has nothing to do with the fact that $L$ contains derivatives of order two; it could have order 17 and eigenthings would be just as reasonable.
There are many ways to find eigenfunctions, and converting to a first order system is one.
Third answer:
Fix $\lambda$ and let us try to find a nonzero $f$ so that $Lf=\lambda f$.
We get
$$
f''(x)+3f'(x)-(4+\lambda)f(x)=0.
$$
You can reduce this to a first order system by denoting $g=f'$.
Then you get the first order system
$$
\begin{cases}
f'=g\\
g'=-3g+(4+\lambda)f.
\end{cases}
$$
If the operator $L$ had order 17, you would end up with a first order system of 17 dimensions.
Here is how I would find eigenfunctions:
Let $p_\lambda(y)=y^2+3y-(4+\lambda)$.
We can formally write a polynomial of the operator $\frac{d}{dx}$, and an eigenfunction should satisfy $p_\lambda(\frac{d}{dx})f=0$.
Find the roots $r_{\lambda,1}$ and $r_{\lambda,2}$ of $p_\lambda$.
Now check that the functions $f_i(x)=e^{r_{\lambda,i}x}$ for $i=1,2$ are eigenfunctions.
(To see why something like this should happen, you can factor the polynomial into $p_\lambda(y)=(y-r_{\lambda,1})(y-r_{\lambda,2})$ and observe that $(\frac{d}{dx}-r_{\lambda,i})f_i=0$.)
The equation $Lf=\lambda f$ is a second order equation so it should have two linearly independent solutions.
You have found them.
A: (1) No, there is nothing uniquely special about this $L$.
(2) The notions of eigenvalues and eigenvectors are defined for any linear map from a vector space $\Bbb V$ to itself: A (nonzero) vector $v$ is an eigenvector, of eigenvalue $\lambda$, for the linear transformation $T: \Bbb V \to \Bbb V$ iff $$T(v) = \lambda v.$$ If $\Bbb V$ is finite-dimensional, then for any basis $(E_a)$ of $\Bbb V$ (e.g., the standard basis of $\Bbb F^n$, where $\Bbb F$ is the underlying field), then we can encode $T$ as a matrix $[T]$ with entries in $\Bbb F$ and speak of eigenvalues and eigenvectors of any square matrix, in which case the above definition specializes to the usual ones for matrices.
In our case, however, though it's unspecified, we're probably regarding $L$ as a linear operator on the vector space of $C^{\infty}$ (i.e., infinitely differentiable) functions, so there is no (finite) matrix representation. You need not encode the equation as a first-order system.
Now, by definition, an eigenvector of $L$ is a solution to
$$L(x) = \lambda x$$
for some $\lambda$ in the appropriate field (probably $\Bbb R$, but possibly also $\Bbb C$, depending on context. Rearranging gives that this is a solution to the homogeneous d.e.
$$x'' + 3 x' - (4 + \lambda) x = 0.$$
