# Singular points and finding linearly independent solutions

• Jul 8th 2008, 09:14 PM
Kyeong
Singular points and finding linearly independent solutions
The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$

1. How would I go proving a singular point that is regular?

I have the textbook in front of me and it doesn't help me at all. I was able to find x=0, but the question asks me to show it. So I did by making y'' equaling to 1 and divided the variable to the rest. I'm just confused if that's good enough. In my end, I don't think it is because they say show that it's a regular singluar point. I'm just a bit confused if I should show more or if what I just mentioned is ok.

2. Find two linearly independent solutions of the form $\displaystyle y = (x^r)\sum_{n=0}^{\infty} a_nx^{n}$. Using the recurrence relation.

Ok so I changed the given y to $\displaystyle y = \sum_{n=0}^{\infty} a_nx^{n+r}$

Then I found the first and second derivatives.

$\displaystyle f'(x) = \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1}$

$\displaystyle f''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2}$

When I plug it in back into the original equation, I get:

$\displaystyle 2 \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r}$ + $\displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+r+1}$ + $\displaystyle 3\sum_{n=0}^{\infty} (n+r)a_nx^{n+r}$ - $\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r}$

Am I doing this correct so far? I would like to make sure before I continue and would it be easier to take out $\displaystyle x^r$,$\displaystyle x^{r-1}$, or does it not matter?

Thanks. I'll come back after seeing if I'm doing the first part correct and post my solution to see if y'all agree.
• Jul 8th 2008, 11:00 PM
Isomorphism
Quote:

Originally Posted by Kyeong
The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$

1. How would I go proving a singular point that is regular?

I have the textbook in front of me and it doesn't help me at all. I was able to find x=0, but the question asks me to show it. So I did by making y'' equaling to 1 and divided the variable to the rest. I'm just confused if that's good enough. In my end, I don't think it is because they say show that it's a regular singluar point. I'm just a bit confused if I should show more or if what I just mentioned is ok.

Sorry, I dont know what is a regular point (Wondering)

Quote:

Originally Posted by Kyeong
2. Find two linearly independent solutions of the form $\displaystyle y = (x^r)\sum_{n=0}^{\infty} a_nx^{n}$. Using the recurrence relation.

Ok so I changed the given y to $\displaystyle y = \sum_{n=0}^{\infty} a_nx^{n+r}$

Then I found the first and second derivatives.

$\displaystyle f'(x) = \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1}$

$\displaystyle f''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2}$

When I plug it in back into the original equation, I get:

$\displaystyle 2 \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r}$ + $\displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+r+1}$ + $\displaystyle 3\sum_{n=0}^{\infty} (n+r)a_nx^{n+r}$ - $\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r}$

Am I doing this correct so far? I would like to make sure before I continue and would it be easier to take out $\displaystyle x^r$,$\displaystyle x^{r-1}$, or does it not matter?

Yes you are right...

From the expression, by comparing the power of $\displaystyle x^{n+r}$ on both sides, we can get the following:

$\displaystyle 2(n+r)(n+r-1)a_n +(n-1+r)a_{n-1}+3(n+r)a_n - a_n = 0$

$\displaystyle (2(n+r)(n+r-1)+3(n+r) -1)a_n +(n-1+r)a_{n-1}= 0$

$\displaystyle (2(n+r)^2+(n+r) -1)a_n +(n-1+r)a_{n-1}= 0$

$\displaystyle (2(n+r)+1) (n+r -1)a_n +(n-1+r)a_{n-1}= 0$

$\displaystyle (n-1+r)((2n+2r+1)a_n + a_{n-1})= 0$

$\displaystyle a_{n-1} = -(2n+2r+1)a_n$

This recurrence is easy to solve as this is similar to telescoping, but its the ratio that telescopes..
• Jul 9th 2008, 07:16 AM
Moo
Hello !

Quote:

The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$

1. How would I go proving a singular point that is regular?

I have the textbook in front of me and it doesn't help me at all. I was able to find x=0, but the question asks me to show it. So I did by making y'' equaling to 1 and divided the variable to the rest. I'm just confused if that's good enough. In my end, I don't think it is because they say show that it's a regular singluar point. I'm just a bit confused if I should show more or if what I just mentioned is ok.
Read this : Regular Singular Point -- from Wolfram MathWorld

$\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0 \implies y''+\frac{x+3}{2x} y'-\frac{y}{2x^2}=0$

$\displaystyle P(x)=\frac{x+3}{2x}$

$\displaystyle Q(x)=\frac{-1}{2x^2}$

You found 0 to be a singular point. It's ok.

To see if it's regular at x=0, you have to see if
• $\displaystyle x P(x)=\tfrac{x+3}{2}$ is finite when $\displaystyle x \to 0$.
• $\displaystyle x^2 Q(x)=-\tfrac 12$ is finite when $\displaystyle x \to 0$.

(Wink)
• Jul 9th 2008, 07:45 AM
Kyeong
About the first part. I read the definition and still confused. I wished there were examples I can work with because that's just how it works for me to understand things.

Would I set it up like this?

$\displaystyle \lim_{x \to 0} (x - 0) \frac {(x+3)} {(2x)} = 1$

$\displaystyle \lim_{x \to 0} (x - 0)^{2} \frac {(-1)} {(2x^{2})} = 0$
• Jul 9th 2008, 07:56 AM
ThePerfectHacker
Quote:

Originally Posted by Kyeong

$\displaystyle \lim_{x \to 0} (x - 0) \frac {(x+3)} {(2x)} = 1$

$\displaystyle \lim_{x \to 0} (x - 0)^{2} \frac {(-1)} {(2x^{2})} = 0$

You just have to show these limits are finite.
$\displaystyle \lim_{x\to 0}x\cdot \frac{(x+3)}{2x} = \lim_{x\to 0}\frac{x+3}{2} = \tfrac{3}{2}$
The same with the other one.
• Jul 9th 2008, 08:44 AM
Kyeong
Quote:

Originally Posted by ThePerfectHacker
You just have to show these limits are finite.
$\displaystyle \lim_{x\to 0}x\cdot \frac{(x+3)}{2x} = \lim_{x\to 0}\frac{x+3}{2} = \tfrac{3}{2}$
The same with the other one.

Oh, ok. I think I understand it now. So the other one will be $\displaystyle \lim_{x\to 0}x^{2}\cdot \frac{-1}{2x^2} = \lim_{x\to 0}\frac{-1}{2} = \tfrac{-1}{2}$, correct? I think I fully understand this now, I just hate this type of series.

Thanks.
• Jul 9th 2008, 08:52 AM
Moo
Quote:

Originally Posted by Kyeong
Oh, ok. I think I understand it now. So the other one will be $\displaystyle \lim_{x\to 0}x^{2}\cdot \frac{-1}{2x^2} = \lim_{x\to 0}\frac{-1}{2} = \tfrac{-1}{2}$, correct? I think I fully understand this now, I just hate this type of series.

Thanks.

Exactly !

But it's not a series :eek: it's only limits !
• Jul 9th 2008, 10:25 AM
Kyeong
Lol, yea it's limits. I meant power series in general. It just confuses me, but now I got the hang of it. Thanks a lot everyone!

Now onto the second part and I'll come back with my work.
• Jul 9th 2008, 12:09 PM
Kyeong

What Isomorphism posted seemed different than how my professor did it and I was a bit confused also. So I tried it similarly to what my professor has been showing us.

I got:

$\displaystyle 2 \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r}$ + $\displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+r}$ + $\displaystyle 2 \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r}$ + $\displaystyle 3\sum_{n=0}^{\infty} (n+r)a_nx^{n+r}$ - $\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r}$

After that, I got a bit confused but was able to get through. This is what I got in the end.

$\displaystyle x^r$$\displaystyle [\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_nx^{n}$ + $\displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+1}$ + $\displaystyle \sum_{n=0}^{\infty} 3(n+r)a_nx^{n}$ -$\displaystyle \sum_{n=0}^{\infty} a_nx^{n}]$

Then I tried to get $\displaystyle a_0$ and changed some of the n values to $\displaystyle j$and $\displaystyle j+1$ like my professor did.

$\displaystyle x^r [(2r(r-1)+3-1)a_0$ + $\displaystyle \sum_{n=1}^{/infty}$
$\displaystyle 2(j+r)(j+r-1)a_j + 3(j+r)a_j - a_j - (j-1+r)a_{j-1}]x^j$

I then took the $\displaystyle a_0$ values and set it equaled to zero. I got r = 1/2 and r = -1.

Then I set r = 1/2 and plugged it into the second part of the last equation after the
$\displaystyle a_0$ values.

I got
$\displaystyle 2(j+ {1/2})(j+{1/2})a_j + 3(j+{1/2})a_j - a_j - (j+{1/2})a_{j-1}$

Then $\displaystyle a_j = \frac {-(j+{1/2})a_{j-1}} {2j^2 + 5j + 1}$

I did the same thing for $\displaystyle r = -1$ to get the second function and the equation I got was:

$\displaystyle a_j = \frac {-(j+{1/2})a_{j-1}} {2j^2 + j - 2}$

So have I done it correctly?

After all this I'm suppose to plug in values for each function and get the first 4 non zero terms. I understand this last part, but getting the equation is just confusing me.