Find all values of $\alpha$ so that all solutions approach $0$ as $x \to \infty$ Given the equation
$x^2y′′+\alpha xy′+4y=0$
find all values of α so that all solutions approach zero as $x \to \infty$. 
Anyone have advice for this question?
So I get $y = c_1 x^{\frac{1}{2}\sqrt{a^2 -2a -15}-a+1} + c_1 x^{\frac{1}{2}\sqrt{a^2 -2a -15}-a+1}$
I tried solving $\sqrt{a^2 -2a -15}-a+1 < 0$ and got $[5, \infty)$. However, this itnerval is apparently not the answer. What did I do wrong?

As  Dr. Sonnhard Graubner  said:
In
$x^2y′′+\alpha xy′+4y=0$,
set
$y = x^r
$.
Then,
writing $a$ for $\alpha$
because I am lazy,
$y'
=rx^{r-1}
$
and
$y''
=r(r-1)x^{r-2}
$
so
$0
=r(r-1)x^{r}+arx^{r}+4x^r
= x^r(r(r-1)+ar+4)
= x^r(r^2+(a-1)r+4)
$
or
$r^2+(a-1)r+4=0$.
Therefore,
$r
=\dfrac{-(a-1)\pm\sqrt{(a-1)^2-16}}{2}
=\dfrac{1-a\pm\sqrt{a^2-2a-15}}{2}
$.
Call the two roots
$r_1
=\dfrac{1-a+\sqrt{a^2-2a-15}}{2}
$
and
$r_2
=\dfrac{1-a-\sqrt{a^2-2a-15}}{2}
$.
The solutions are therefore
$y
=c_1x^{r_1}
+c_2x^{r_2}
$.
For all solutions
to approach
$\infty$,
both roots
must be positive.
For all solutions
to approach
zero,
both roots
must be negative.
From these,
you can work out
conditions on $a$.
Note that
the roots must be real;
if not,
the solutions will oscillate.
This imposes another condition on $a$.
Also note that
if
$(a-1)^2 = 16$,
the roots are equal,
so you get another type of solution,
which has to be considered.
