Solving the differential equation $f''=f'+f$, where $f$ is an exponential function meant to give a term of Fibonacci series I'm currently working through the H.S. Wilf's book generatingfunctionology, which is about generating functions, and have reached page 40 (or page 45 on the pdf) where the book is starting to deal with  calculus in exponential generating functions. Looking at Example 1 in the image below, I understand how the equation $f''= f' + f$ is created, but what I don't understand is how to solve it, which the author skims over. How would you solve equations like these?

 A: Let's discuss how to solve linear differential equations
The simplest exaple is:
$\frac {dy}{dt} = y$
This is a separable diff eq.
$\int \frac 1y\ dy = \int\ dt\\
\ln y = t + C\\
y = C e^{t}$
Upping the degree of difficulty...
$\frac {dy}{dt} = ay\\
y = C e^{at}$
Note that: $\frac {dy}{dt} - ay = 0$ is the same problem.
And on to second-order diff eq.
$a\frac {d^2y}{dt^2} + b\frac {dy}{dt} + c = 0\\
(\frac {d}{dt} - \frac {-b + \sqrt{b^2-4ac}}{2a})(\frac {d}{dt} - \frac {-b - \sqrt{b^2-4ac}}{2a})y = 0$
If you think I am playing fast and loose with the differential operators, perhaps, I am.  But, this method holds up.  You can factor these like they are polynomials.
$y = C_1 e^{\frac {-b + \sqrt{b^2-4ac}}{2a}t} + C_2  e^{\frac {-b - \sqrt{b^2-4ac}}{2a}t}$
And if you don't believe me, start with the solution and see if it does not indeed solve the original diff eq.
$f'' - f' - f = 0\\
f = C_1 e^{\frac{1 + \sqrt 5}{2} t} + C_2e^{\frac{1 - \sqrt 5}{2} t}$
A: This is a second-order linear differential equation.  Any textbook on differential equations will deal with these.
