# Thread: Singular points and finding linearly independent solutions

1. ## Singular points and finding linearly independent solutions

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

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 $y = (x^r)\sum_{n=0}^{\infty} a_nx^{n}$. Using the recurrence relation.

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

Then I found the first and second derivatives.

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

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

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

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

Am I doing this correct so far? I would like to make sure before I continue and would it be easier to take out $x^r$, $x^{r-1}$, 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.

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

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

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

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

Then I found the first and second derivatives.

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

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

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

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

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

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

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

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

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

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

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

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

This recurrence is easy to solve as this is similar to telescoping, but its the ratio that telescopes..

3. Hello !

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

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

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

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

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

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
• $x P(x)=\tfrac{x+3}{2}$ is finite when $x \to 0$.
• $x^2 Q(x)=-\tfrac 12$ is finite when $x \to 0$.

4. 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?

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

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

5. Originally Posted by Kyeong

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

$\lim_{x \to 0} (x - 0)^{2} \frac {(-1)} {(2x^{2})} = 0$
You just have to show these limits are finite.
$\lim_{x\to 0}x\cdot \frac{(x+3)}{2x} = \lim_{x\to 0}\frac{x+3}{2} = \tfrac{3}{2}$
The same with the other one.

6. Originally Posted by ThePerfectHacker
You just have to show these limits are finite.
$\lim_{x\to 0}x\cdot \frac{(x+3)}{2x} = \lim_{x\to 0}\frac{x+3}{2} = \tfrac{3}{2}$
The same with the other one.
Oh, ok. I think I understand it now. So the other one will be $\lim_{x\to 0}x^{2}\cdot \frac{-1}{2x^2} = \lim_{x\to 0}\frac{-1}{2} = \tfrac{-1}{2}$, correct? I think I fully understand this now, I just hate this type of series.

Thanks.

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

Thanks.
Exactly !

But it's not a series it's only limits !

8. 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.

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:

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

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

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

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

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

I then took the $a_0$ 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
$a_0$ values.

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

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

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

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

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.