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?
 A: 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.
A: For this problem you found only part of it.  You essentially need to make your roots negative as x approaches infinity.  So you need the roots to be negative for these three cases, distinct real roots, equal real roots, and imaginary roots.  You found the interval for the distinct real roots.  Now find it for the other two. (Hint: the imaginary roots will help you find the rest of your solution.)
A: We are working with our old friend the Euler equation again.  This time, let's look at how the solutions differ for the essentially different cases determined by the discriminant of the quadratic equation that results when we substitute the postulated form of the solution, y = xr, into the equation; namely, the discriminant of
r(r - 1) + αr + (5/2) = 0,
whose roots we will call r1 and r2.
The discriminant, d = (α - 1)2 - 10, tells us what the solution will look like, which, by now, you are probably already familiar with:
For d > 0, we get y = c1 |x|r1 + c2 |x|r2.
For d = 0, we get y = (c1 + c2 ln |x|) |x|r1.
For d < 0, we get y = |x|a( c1 cos(b ln |x|) + c2 sin(b ln |x|) ), where r1, r2 = a ± bi, respectively.
Now, looking over these cases, we have to untangle under what (if any) conditions each of the above solutions can approach zero as x tends to zero.
In the first case, d > 0, this is only possible if both r1 and r2 are positive, since otherwise, the solution would become unbounded near zero.  But this is only possible if both the sum of the roots and the product of the roots is positive, i.e. if r1 + r2 > 0 and r1r2 > 0, which in turn is true only if (1 - α) > 0, which is tantamount to α < 1.
In the second case, d = 0, the solution only approaches zero as x approaches zero if the root is positive, since  the presence of the logarithmic term dooms the solution to be unbounded near zero otherwise.  But this is the same as having (1 - α)/2 > 0, which again is tantamount to α < 1.
In the last case, the complex case, the presence of the term |x|a will take the solution towards zero near zero, as long as a > 0.  But a > 0 only when (1 - α)/2 > 0, which again is tantamount to α < 1.
