Question

Find all values of alpha for which all solutions of x^2y'' + (alpha)xy' + (5/2)y = 0 approach 0 as x goes to 0

Answer 1

Mmm so what's the deal here?
We need to power series or something?
:U

Answer 2

I am not positive, but I was considering it, yes. I solved the equation through a non-power series method, but using that solution I couldnt come up with the correct answer.

Answer 3

Solution I came up with was:
\[y=c _{1}x ^{\frac{ 1-\alpha-\sqrt{(1-\alpha)^{2}-10} }{ 2 }}+c _{2}x ^{\frac{ 1-\alpha+\sqrt{(1-\alpha)^{2}-10} }{ 2 }}\]

Answer 4

All of that jazz is in the exponent of each term.

Answer 5

The only reason I didnt jump to power series was only because of the section it was in. Do you think power series is pretty much the route to go, though? Or is there maybe a way to do it through conventional solving?

Answer 6

Ahhh I dunno D: grr

Answer 7

Were you trying something out? D:

Answer 8

Maybe I did the power series set up wrong, but I end up with:
\[\frac{ 5 }{ 2 }\alpha _{0}+\alpha(a _{1})x+\frac{ 5 }{ 2 }a _{1}x+\sum_{n=2}^{\infty}[n(n-1)a _{n}+\alpha(na _{n})+\frac{ 5 }{ 2 }a _{n}]x ^{n}=0\]

In the end, it seems like I get nothing useful out of this. Maybe I did it improperly or just don't know the proper way to finish this off, but this is what I got :/

Answer 9

Look for y = \(\sum_{n=0}^{\infty }a_nx^n\)
\[y'= \sum_{n=0}^{\infty }na_nx^{n-1}\\\alpha xy'= \sum_{n=0}^{\infty }\alpha na_nx^n\]
\[y" =\sum_{n=0}^{\infty }n(n-1)a_nx^{n-2}\\x^2y"=\sum_{n=0}^{\infty }n(n-1)a_nx^n\]
Add them all
\[x^2y" +\alpha xy'+\dfrac{5}{2}y = \sum_{n=0}^{\infty }n(n-1)a_nx^n +\sum_{n=0}^{\infty }\alpha na_nx^n +\sum_{n=0}^{\infty }\dfrac{5}{2}a_nx^n\]
simplify
\[\sum_{n=0}^{\infty }(n(n-1)+\alpha n +5/2)a_nx^n\]
this sum =0 iff \(n(n-1) +\alpha n+5/2)=0 \)
Solve for \(\alpha \) from here I get \(\alpha = n +3/(2n)\)

Answer 10

So that , I make a conclusion as: if alpha = (the above) then y = \(\sum_{n=0}^{\infty }a_nx^n\) is solution of the ODE and limit of this sum approaches 0 when x approaches 0

Answer 11

Since this is an Euler equation, I think your solution is on the right track, but just to check your work, I'll work some of it out. Assume \(y=x^r\) is a solution, then \(y'=rx^{r-1}\) and \(y''=r(r-1)x^{r-2}\).
\[x^2y''+\alpha xy'+\frac{5}{2}y=0\]
Substituting into the equation gives
\[\begin{align*}x^2r(r-1)x^{r-2}+\alpha xrx^{r-1}+\frac{5}{2}x^r&=0\\
r(r-1)x^{r}+\alpha rx^{r}+\frac{5}{2}x^r&=0\\
x^r\left[r(r-1)+\alpha r+\frac{5}{2}\right]&=0\\
r^2+(\alpha-1) r+\frac{5}{2}&=0
\end{align*}\]
and you have roots
\[r=\frac{1-\alpha\pm\sqrt{(\alpha-1)^2-10}}{2}\]
Alright, I'm getting the same solutions.

Note that you must have \((\alpha-1)^2\ge10\), or \(\alpha\ge1+\sqrt{10}\) and \(\alpha\le1-\sqrt{10}\), for the solutions to exist.

Answer 12

Yeah, I figured everything I did was okay, but the answer says \[\alpha < 1\] which gives complex solutions for certain alpha. Attempting the series solution route, I get what was posted above, but I do not know how to turn that into a suitable alpha really. Unless my textbook is wrong (which I wouldn't doubt), I'm not sure how to derive their solution.

Answer 13

Maybe complex solutions are fine since you can use Euler's identity to turn the solution into sines and cosines? Even then, seems fishy x_x

Answer 14

dan, why can't we have sqrt =0, <0 ??
if the sqrt >0 , we have 2 distinct solutions
if it =0 , double root
if it <0, we have complex root
is it not that? woah... My knowledge is.... poor. hihihi

Answer 15

This is the Euler's form, so that the solution will form y = x^r. i don't understand why the root cannot be double one.

Answer 16

Alright, so I figured it out, in case anyone was interested. So, because we end up with a form of y = x^r, the function will approach 0 and x -> 0 given r is not negative or 0. Complex valued solutions are perfectly fine to have because those would be in the form
\[y= |x |^{a}(c _{1}\cos(\beta*lnx)+c _{2}(\sin \beta*lnx))\]
Repeated roots simply give a form of
\[y=(c _{1}+c _{2}\ln|x|)*|x|^{r}\].

All of these solutions, limit as x->0 will approach 0 given that we have a positive (or zero), real exponent of x, regardless of whether or not we have complex or repeated roots. So we just need to simply ensure that in the form of the solution we have
\[r=\frac{ 1-\alpha \pm \sqrt{(\alpha - 1)^{2}-10} }{ 2 }\]
produces some sort of positive (or zero), real exponent for x. This can be done as long as
\[\alpha < 1\].

Hope that makes sense x_X