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$$y''+x^2y'+xy=0$$

I'm a bit confused.

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Hello Paul, it seems that you are new to this website. It would help if you explain what you have tried so far and it would also be appreciated if you invest more time into formulating your question. You can use $y''+x^2y'+xy=0$ to use Latex, the result would be $y''+x^2y'+xy=0$ which looks nicer than the formatting you used. –  sonystarmap Apr 29 '13 at 18:33
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Suppose there is a solution $y(x) = \sum_n a_n x^n$. Then figure out what $y',y''$ are (in terms of the coefficients of $x^n$), and then write the above equation using these expressions for $y,y',y''$. Then equate all the $x^n$ coefficients to zero, and figure out the values of the coefficients. –  copper.hat Apr 29 '13 at 18:34
    
Sorry about that, I thought it was easy enough to read without Latex. I tried to do what you described, copper.hat, but I couldn't figure out the recurrence relation. –  Paul Apr 29 '13 at 18:38
    
Can anyone help with this, though? I don't understand how I should make each x have the same power for all of the expressions in the final summation that is equal to 0. –  Paul Apr 29 '13 at 19:26

1 Answer 1

Hints, start off with:

$\displaystyle y = \sum_{n=0}^{\infty} a_nx^n$, so:

$\displaystyle y' = \sum_{n=1}^{\infty} na_nx^{n-1}$

$\displaystyle y'' = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}$

Substitute these back into the DEQ:

$ 0 = y''+x^2y'+xy$

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

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

Now, we want all terms to just be $x^n$, so write them out so it is so.

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

Next, we want to get the sum to start at a common $n$, so choose $n = 2$, since that is where the highest sum is already starting. So we have,

$0 = 2a_2 +6a_3x + a_0x + \sum_{n=2}^{\infty}\left( (n+2)(n+1)a_{n+2}+ na_{n-1}\right)x^n$

From this equation, we get three things:

  • $(1)$ $2a_2 = 0 \rightarrow a_2 = 0$
  • $(2)$ $\displaystyle 6a_3 + a_0 = 0 \rightarrow a_3 = -\frac{a_0}{6}$
  • $(3)$ The recursion relation, for $n \ge 2$:

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

Now, we need to figure out the general recursion (i am going to kick start you, but you have to work some of the problem).

  • $n = 2$, $\displaystyle a_4 = \frac{2}{4 \times 3}a_1$, where $a_1$ is arbitrary
  • $n = 3$, $\displaystyle a_5 = \frac{3}{5 \times 4}a_2 = 0$, since $a_2 =0$ from earlier
  • $n = 4$, $\displaystyle a_6 = \frac{4}{6 \times 5}a_3 = \frac{4}{6 \times 5}-\frac{a_0}{6}$, where $a+0$ is an arbitrary constant
  • $n = 5$, $\displaystyle a_7 = \frac{5}{7 \times 6}a_4$
  • $n = 6$, $\displaystyle a_8 = \frac{6}{8 \times 7}a_5 = 0$, since $a_5 = 0$

Do you see which terms will be zero?

Do you see how to write this as a general series?

Once you have that, you can write out the general solution for $y$

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I know how to proceed in general, but I'm not quite sure on this one. How should I make the first summation have an $\displaystyle x^{n+1}$ term? –  Paul Apr 29 '13 at 21:07
    
@Paul: see updates and see if you can take it from there. –  Amzoti Apr 29 '13 at 21:16
    
+1 Very nice hints in precisely the right direction. –  amWhy Apr 30 '13 at 0:36
    
@amWhy: It would be great to hear back and see if s/he got it! Thanks! –  Amzoti Apr 30 '13 at 0:36
    
Hm, I would think one way to go would be to evaluate the sum that starts at two to make it start at 0, but I don't know what I would do with a separated term of $\displaystyle a_{2}x^2$. Sorry for all the questions, I'm just really bad at these. The help is amazing, by the way. This community is so good. –  Paul Apr 30 '13 at 3:51

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