Show that the following Euler Cauchy differential equation is oscillatory and disconjugate Question: Show that the Euler Cauchy differential equation 
$t^2x''+2tx'+bx=0 $
is oscillatory on $[1,\infty)$ if $b>\frac{1}{4}$ and disconjugate on $[1,\infty)$ if $b\leq\frac{1}{4}$.
So, we've got the indical equation $t^2r(r-1)+2tr+b=0$ which has roots $r= \frac{-(2t-t^2)\pm\sqrt{(2t-t^2)^2-4t^2b}}{2t^2}$
I looked at a similar problem, which showed that the solutions are oscillatory when $Im(r)\neq 0$ (but I'm unsure of why this is the case).
How would I solve the problem from this point on?
 A: HINT :
The equation is not $t^2r(r-1)+2tr+b=0$ , but is :
$$r(r-1)+2r+b=0$$
And the roots are not $r= \frac{-(2t-t^2)\pm\sqrt{(2t-t^2)^2-4t^2b}}{2t^2}$ , but are :
$$r_1= \frac{-1+\sqrt{1-4b}}{2}\quad\text{and}\quad r_2= \frac{-1-\sqrt{1-4b}}{2}$$
The general solution is :
$$x(t)=c_1t^{r_1}+c_2t^{r_2}\tag 1$$
Sinusoidal terms appear when $r_1$ and $r_2$ are not real $(b>1/4)$. 
NOTE : Case $b>\frac14$ 
$$r_1=\alpha+i\beta \qquad\begin{cases}
\alpha=-\frac12 \\ 
\beta=\frac{\sqrt{4b-1}}{2}
\end{cases}$$
$$t^{r_1}=t^{\alpha+i\beta }=t^{\alpha}e^{i\beta\ln(t) }=t^{\alpha}\left(\cos(\beta\ln(t))+i\sin(\beta\ln(t))\right)$$
See the Euler formula : http://mathworld.wolfram.com/EulerFormula.html
Similarly for $\quad t^{r_2}=t^{\alpha}\left(\cos(\beta\ln(t))-i\sin(\beta\ln(t))\right)$
Let $C_1=c_1+c_2$ and $C_2=i(c_1-c_2)$. The form of the above general solution is :
$$x(t)=t^{-1/2}\bigg(C_1\cos(\beta\ln(t))+C_2\sin(\beta\ln(t)) \bigg) \tag 2$$
In the most general case, the coefficients $c_1,c_2,C_1,C_2$ can be complex.
But if you restrict the solutions to be only real, just take $c_1, c_2$ real in Eq.$(1)$ case $b<\frac14$ , and take $C_1,C_2$ real in Eq.$(2)$ case $b>\frac14$.
A: Assuming $x = C_0 t^\alpha$ and substituting into the differential equation we have
$$
C_0t^\alpha(\alpha^2+\alpha+b) = 0
$$
so the solutions are for $\alpha$ such that
$$
\alpha = (-1\pm\sqrt{1-4b})/2
$$
Those solutions are complex for $b >\frac{1}{4}$ in which case the solution will be periodic.
In this case, calling $\beta = \sqrt{1-4b}/2$ we have as general solution
$$
x(t) = t^{-1/2}(C_1 \sin(\beta\log t)+C_2\cos(\beta\log t))
$$
