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Math Help - Singular points and finding linearly independent solutions

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Kyeong View Post
    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

    Quote 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..
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  3. #3
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    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.


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  4. #4
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    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
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  5. #5
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    Quote Originally Posted by Kyeong View Post

    \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.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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  7. #7
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    Quote Originally Posted by Kyeong View Post
    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 !
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  8. #8
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    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.
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  9. #9
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    About the second part.

    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.
    Last edited by Kyeong; July 9th 2008 at 06:18 PM.
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