how to solve the equation $$ (z-a) (z-b) = 0 $$ well, we would have two solutions $z=a$ and $z=b$ (assuming $a \neq b$).
how to solve equation ( $a \neq b$) $$ (d/dx - a) (d/dx - b) f(x) = 0? $$ Instead of saying we have two solutions, we say the solution space is a two dimensional vector space. it is easy to check that $e^{ax}$ and $e^{bx}$ solves the equations, and so are their linear combinations, thus we can write down the general solution as $$ f(x) = c_1 e^{ax} + c_2 e^{bx} $$
wait, how do you know you have found ALL the solutions? how do you know you didn't miss any? There is a theorem saying that, a constant coeff ODE of order $n$ will have a $n$-dimensional solution space.
How about the case where $a=b$? $$ (d/dx - a) (d/dx - a) f(x) = 0? $$ Here we claim that the general solution is $$ f(x) = (c_0 + c_1 x) e^{ax} $$
Proof of the claim: suppose $f(x)$ solves the equation, then we can always write $f(x) = g(x) e^{ax}$ (since $e^{ax}$ is never 0), and figure out the equation that $g(x)$ satisfies. We see $$ (d/dx - a)[ g(x) e^{ax} ] = d/dx [ g(x) e^{ax} ] - ag(x) e^{ax} = g'(x) e^{ax} + a e^{ax} g(x) - a g(x) e^{ax} = e^{ax} (d/dx) (g(x)) $$ thus, we have $$ e^{ax} (d/dx)^2 (g(x)) = 0 $$ That means $g(x) = c_0 + c_1 x$. Hence we get the claimed general solution
Initial condition / Boundary condition.