How are eigenfunctions found? Let's stick to the derivative operator and any other operators built up from it.  For instance, if $D = \frac{d}{dt}$, then the eigenfunction of $D$ is known to be $e^{at}$, associated with eigenvalue $a$.  However, what about for a more complicated operator?  Even something as simple as $1 + D$.
Is there a general method that works for any such operator?  For that matter, how was the eigenfunction for $D$ itself found?
 A: Eigenfunction of $D$
$$\begin{array}{rcl}
Df &=& af \\
\dfrac{\mathrm df}{\mathrm dt} &=& af \\
\dfrac{\mathrm dt}{\mathrm df} &=& \dfrac1{af} \\
t &=& \displaystyle \int \dfrac1{af} \ \mathrm df \\
t &=& \dfrac1a\ln f + C \\
f &=& ke^{at}
\end{array}$$

Eigenfunction of $1+D$
$$\begin{array}{rcl}
1+Df &=& af \\
Df &=& af-1 \\
\dfrac{\mathrm df}{\mathrm dt} &=& af-1 \\
\dfrac{\mathrm dt}{\mathrm df} &=& \dfrac1{af-1} \\
t &=& \displaystyle \int \dfrac1{af-1} \ \mathrm df \\
t &=& \dfrac1a\ln (af-1) + C \\
at-b &=& \ln(af-1) \\
f &=& \dfrac{1+e^{at-b}}a \\
\end{array}$$

In general
Apart from directly solving the equation, there are some general approaches.
For certain problems, one can use matrix to find eigenfunctions.
First, we choose a basis. A common basis is $\{1,x,x^2,x^3,\cdots\}$ (the polynomials). Then, we use a matrix to represent the operator, and then find the eigenvalue and the eigenvector of the matrix. The eigenvector represents the eigenfunction.
Limitations: Only works when the operator is linear, and obviously the basis cannot represent all functions.
A: For a minimal example an eigenfunction to the operator ${\bf T = D}^2+2{\bf D}$ on the space of $\{\sin(x),\cos(x)\}$.
The operator $D$ on this space is a matrix: $ {\bf D} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$
Therefore $${\bf T} = {{\bf D}^2+2{\bf D}} = \left[\begin{array}{cc}-1&2\\-2&-1\end{array}\right]$$
We can now solve $\det({\bf T}-\lambda {\bf I})$
$\lambda = -1\pm 2i$ $e = [\pm \sqrt{1/2} i , \sqrt{2}]^T$
So if we allow complex coefficients, apparently second derivative plus 2 times derivative of $$\left(\frac{\partial^2}{\partial x^2} +2 \frac{\partial}{\partial x}\right) \{i\sin(x)+\cos(x)\} = (-1+2i)(i\sin(x)+\cos(x))$$
And we can check it on Wolfram Alpha.
