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I'm rather new to differential equations and so am having some trouble finding a general solution to the equation:

y'' = xy ; with initial values y(0) = 1, y'(0) = 0

I've gone through the routine series differentiation, index shift and taking the n=0 term out, to get,

--> $$2a_2+\sum_{n=1}^\infty \left[ a_{n+2}{(n+2)(n+1)}-a_{n-1}\right]*x^n=0$$

and found the recurrence relation

a(n+2) = a(n-1)/(n+2)(n+1)

Using this, I've found the terms

$$a_3=\frac{a_0}{3 \cdot 2}=\frac{1}{6}$$ $$a_4=\frac{a_1}{4 \cdot 3}=0$$ $$a_5=\frac{a_2}{5 \cdot 4}=0$$ $$a_6=\frac{a_3}{6 \cdot 5 \cdot 3 \cdot 2}=\frac{1}{180}$$ $$a_7=\frac{a_4}{7 \cdot 6}=0$$ $$a_8=\frac{a_5}{8 \cdot 7}=0$$ $$a_9=\frac{a_6}{9 \cdot 8 \cdot 6\cdot 5 \cdot 3 \cdot 2}=\frac{1}{12960}$$ ...

The trouble I'm having, is finding the general $$a_n$$ term.

Considering the initial values, and y'(0)=0, I've come up with something along the lines of

$$a_{3n}=\frac{a_0}{(3n)!}$$ --> $$a_0\left[1+\sum_{n=1}^\infty\frac{x^{3n}}{(3n)!}\right]$$

But am unsure if this is correct.

Any help or tips appreciated!

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  • $\begingroup$ You may want to make more explicit the nature of the denominator there: You specifically want the factors $3n$ through $1$ but with $1,4,7,\ldots,3n-2$ omitted. Perhaps a clearer way to write this is as $$a_{3n}=\frac{a_0}{(3n)!}(3n-2)(3n-5)\cdots(4)(1).$$ $\endgroup$ – Semiclassical May 15 '17 at 6:10
  • $\begingroup$ Ah yeah. And if I wanted to find the radius of convergence for this, how could I do that using the ratio test? The (3n-2)(3n-5)...(4)(1) part confuses me. $\endgroup$ – theGreatWhatever May 15 '17 at 6:25
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    $\begingroup$ Well, the ratio test isn't very useful if applied directly here due to how many coefficients vanish. But this is remedied by introducing $u=x^3$, so that the relevant sequence to be summed is $\{a_{3n} u^n\}$. Hence one considers the limit $$\lim_{n\to\infty}\frac{a_{3n+3}u^{n+1}}{a_{3n}u^n}=\lim_{n\to\infty}u\frac{(3n)(3n-1)(3n-3)(3n-4)\cdots (3)(2)}{(3n+3)(3n+2)(3n)(3n-1)\cdots (3)(2)}=\lim_{n\to\infty} \frac{u}{(3n+3)(3n+2)}.$$ (The version of $a_n$ you wrote above is convenient for this purpose.) $\endgroup$ – Semiclassical May 15 '17 at 6:34
  • $\begingroup$ Gotcha. So if I take the limit of (x^3)/(3n+3)(3n+2), I'm getting that the radius of convergence would be |x| < sqrt(3). Would this be correct? $\endgroup$ – theGreatWhatever May 15 '17 at 7:18
  • $\begingroup$ The relevant is with $x$ held fixed and $n\to\infty$. So no, that's not the radius of convergence... $\endgroup$ – Semiclassical May 15 '17 at 7:19
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This too long for a comment.

  1. Given the condition, you should use $a_0=1$
  2. If you look at the denominators of the non-zero coefficients you already obtained, they correspond to sequence $A176730$ at $OEIS$ (this could be of some interest, I hope)
  3. Totally off-topic (just given for your curiosity), the non-zero coefficients are given by $$a_{3n}=\frac{ 1}{3^{2n} n!}\frac{ \Gamma \left(\frac{2}{3}\right)}{\Gamma \left(n+\frac{2}{3}\right) }$$ where appears the gamma function. From this last result, you would easily obtain (as already given by Semiclassical) $$\frac{a_{3n+3}}{a_{3n}}=\frac{1}{(3n+3)(3n+2)}$$
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