Question

Find all values of a for which all solutions of x^2*y"+axy'+(5/2)y=0 approach zero as x-->0. (I used y=x^r, y'=rx^r-1, y"=r(r-1)x^r-2 and got r^2-r+ar+5/2=0. What's next?)

Answer 1

How do you do that?

Answer 2

But how do you solve for r from there?

Answer 3

But the answer is alpha<1, how do I get there?

Answer 4

I typed a instead of alpha, that's why.

Answer 5

Does that have anything to do with this?

Answer 6

@SithsAndGiggles help me, please

Answer 7

HAHAHA

Answer 8

Yes, you will. Because you need to learn and get benefits, too. Hehehe.

Answer 9

I don't mind.

Answer 10

I'll come back in 5 minutes.

Answer 11

\[x^2y''+\alpha xy'+\frac{5}{2}y=0\]
As a guess you use \(y=x^r\), so that \(y'=rx^{r-1}\) and \(y''=r(r-1)x^{r-2}\).
\[x^2\left(r(r-1)x^{r-2}\right)+\alpha x\left(rx^{r-1}\right)+\frac{5}{2}x^r=0\\
r(r-1)x^r+\alpha rx^r+\frac{5}{2}x^r=0\\
r^2-r+\alpha r+\frac{5}{2}=0\\
r^2+(\alpha-1)r+\frac{5}{2}=0\]
Now I'm all caught up to where you are.

So you want to find at least one solution of the form \(x^r\). This solution will approach 0 as \(x\to0\) for \(r>0\) only.
(If r is negative, then you have \(1/x^\text{(some power)}\), and as \(x\to0\), this term approaches positive infinity. If r is 0, you have the indeterminate form \(0^0\). So r must be positive.)

So now all you have to do is find \(\alpha\) such that the roots of the equation are positive. Applying the quadratic formula, you have as your root(s)
\[r=\frac{-(\alpha-1)\pm\sqrt{(\alpha-1)^2-10}}{2}\]
\(r\) must be real and positive. The simplest case for this is when the discriminant is zero, leaving you with
\[r=\frac{1-\alpha}{2}\]
This value is only positive for \(\alpha<1\).

I really hope I'm not missing anything here, but I got what you said is the answer.

Answer 12

Thanks SithsAndGiggles. you save my face, hehehe. at least I went on the right track. just not get the neat logic like you. hihihi...

Answer 13

But how did you get r=(1-alpha)/2?

Answer 14

I assumed the discriminant \(\left((\alpha-1)^2-10)\right)\) was 0, leaving me with
\[r=\frac{-(\alpha-1)\pm0}{2}\\
r=\frac{-\alpha+1}{2}\\
r=\frac{1-\alpha}{2}
\]

Answer 15

But how do you know that alpha<1 when r=1-alpha/2?

Answer 16

We have the restriction that \(r\) must be real and positive. \(r\) depends on \(\alpha\), and \(\alpha<1\) allows for \(r\) to be real and positive. Reread my first post, I had a little paragraph with a quick explanation.

Answer 17

Thanks for the help.

Answer 18

You're welcome!