Fundamental set of solutions of $y'' + \frac{a_1}{t} y' + \frac{a_2}{t^2} y = 0$ Consider $y'' + \frac{a_1}{t} y' + \frac{a_2}{t^2} y = 0$ where $0<t<\infty$
I began by assuming that the solution of this ODE was of the form $e^{rt}$, then when I solved the characteristic equation for $r$, I got that $y_{1,2}$ = $e^{- \frac{1}{2} a_1 \pm \frac{1}{2} \sqrt{a_1^2 - 4a_2} } $
But (1) this doesn't appear to actually be a solution AND (2) even if it were it's a constant so it can't possibly be both elements of the fundamental set.
Questions:
(a) is this enough to know that the solution is not of the form $e^{rt}$?
(b) if $e^{rt}$ is out for all $r$, then that should also rule out trig functions as well, correct?
(c) my current guess looking at other problems for known $a_1$, $a_2$ is that I need a $t^r$ type function.  I have an idea of using initial conditions $y(1) = a_1$, $y'(1)=-a_2$ to find the fundamental set.  Will this work, or is there a better way forward?  
(d) Do I need to lay down some restrictions on $a_1$, $a_2$ in order to ensure a solution exists?
 A: The characteristic equation is usually a tool for solving equations with constant coefficients, of which this is not an example. The answer to your question (a) is hence yes.
For (b), not necessarily. After all, why not $\tan t$? That's not easily expressible as an exponential. (It doesn't work, but that's beside the point).
For (c), you are correct: The solutions to this are power functions whose powers depend on the roots of the indicial polynomial
$$r(r - 1) + a_1 r + a_2 = 0$$
The initial data will not tell you anything about the powers.
For (d), no. There are solutions for all choices of $a_1$ and $a_2$, but the form of the solution will be radically different depending on whether the indicial polynomial has two real roots, two complex roots, or one real root.
For more, this is a well-known type of equation.
A: I think that just looking at the equation gives the idea that solutions of the equation $$y'' + \frac{a_1}{t} y' + \frac{a_2}{t^2} y = 0$$ are of the type $y=c t^k$. Replacing and simplifying leads to $$ct^{k-2}\big(k^2+(a_1-1)k+a_2\big)=0$$ Then, the two possible values of $k$ $$k_\pm=\frac 12\Big(a_1-1\pm \sqrt{(a_1-1)^2-4a_2} \Big)$$
