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Math Help - Trying to express a solution as a series

  1. #1
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    Trying to express a solution as a series

    Hello everyone!

    I'm trying to practice on the numerous steps one has to follow to obtain a power series solution so I tried to solve the simple ODE
    (E): y''-2y'+y=0 whos solutions are very well known to be y_1=e^x and y_2=x\cdot e^x.

    Happily, I laid down the buliding blocks substituting in (E), \ y=\sum_{n=0}^\infty c_n \dot x^n and, after reindexing and grouping of terms, got the following recurrence relation:
    c_{k+2}=\frac{2(k+1) c_{k+1}-c_k}{(k+2)(k+1)}.
    Okay, I tried to continue by assuming once, that c_0 = 0 and again c_1=0 but I got no pattern at all, I was expecting something familiar - a taylor series expansion at 0.

    Question: (1) When to consider 2 cases, a null c_0 and a real c_1 and the opposite.
    (2) Why didn't I get a pattern?

    Thanks!
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  2. #2
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    Quote Originally Posted by rebghb View Post
    Hello everyone!

    I'm trying to practice on the numerous steps one has to follow to obtain a power series solution so I tried to solve the simple ODE
    (E): y''-2y'+y=0 whos solutions are very well known to be y_1=e^x and y_2=x\cdot e^x.

    Happily, I laid down the buliding blocks substituting in (E), \ y=\sum_{n=0}^\infty c_n \dot x^n and, after reindexing and grouping of terms, got the following recurrence relation:
    c_{k+2}=\frac{2(k+1) c_{k+1}-c_k}{(k+2)(k+1)}.
    Okay, I tried to continue by assuming once, that c_0 = 0 and again c_1=0 but I got no pattern at all, I was expecting something familiar - a taylor series expansion at 0.

    Question: (1) When to consider 2 cases, a null c_0 and a real c_1 and the opposite.
    (2) Why didn't I get a pattern?

    Thanks!
    As for a pattern, you actually do. Let me write out the first few terms
     <br />
\begin{aligned}<br />
c_2 &= c_1 - \frac{c_0}{2},\\<br />
c_3 &= \frac{c_1}{2} - \frac{c_0}{3},\\<br />
c_4 &= \frac{c_1}{3!} - \frac{c_0}{8},\\<br />
c_5 &= \frac{c_1}{4!} - \frac{c_0}{30}.<br />
\end{aligned}<br />

    Setting c_0 = 0 and c_1 = 1 gives

     <br />
x + \frac{x^2}{1} + \frac{x^3}{2!} + \frac{x^4}{3!} + \cdots<br /> <br />

    or

     <br />
x\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \cdots \right)<br />
(look familar?)

    Setting c_0 = 1 and c_1 = 0 gives

     <br />
1 - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{8} - \frac{x^5}{30} - \cdots<br />

    which, our course, doesn't look familar. So, what happened? Remember in this case, your series solution should recover

     <br />
(ax+b)e^x<br />

    for suitable choices of a and b. For example, try a Taylor series for

    (1-x)e^x and see what you get.
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  3. #3
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    Oh dear God how one earth did you get that, I was searching for something like that (duh) but I got nowhere ... Thanks!!
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