Respuesta :
[tex]\displaystyle y=\sum_{n\ge0}a_nx^n[/tex]
[tex]\implies\displaystyle y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}[/tex]
[tex]y''-3xy=0[/tex]
[tex]\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0[/tex]
[tex]\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0[/tex]
[tex]\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge3}a_{n-3}x^{n-2}=0[/tex]
[tex]\displaystyle2a_2+\sum_{n\ge3}\bigg(n(n-1)a_n-3a_{n-3}\bigg)x^{n-2}=0[/tex]
This generates the recurrence relation
[tex]\begin{cases}a_0=a_0\\a_1=a_1\\2a_2=0\\n(n-1)a_n-3a_{n-3}=0&\text{for }n\ge3\end{cases}[/tex]
Because you have
[tex]n(n-1)a_n-3a_{n-3}=0\implies a_n=\dfrac3{n(n-1)}a_{n-3}[/tex]
it follows that [tex]a_2=0\implies a_5=a_8=a_{11}=\cdots=a_{n=3k-1}=0[/tex] for all [tex]k\ge1[/tex].
For [tex]n=1,4,7,10,\ldots[/tex], you have
[tex]a_1=a_1[/tex]
[tex]a_4=\dfrac3{4\times3}a_1=\dfrac{3\times2}{4!}a_1[/tex]
[tex]a_7=\dfrac3{7\times6}a_4=\dfrac{3\times5}{7\times6\times5}=\dfrac{3^2\times5\times2}{7!}[/tex]
[tex]a_{10}=\dfrac3{10\times9}a_7=\dfrac{3\times8}{10\times9\times8}a_7=\dfrac{3^3\times8\times5\times2}{10!}a_1[/tex]
so that, in general, for [tex]n=3k-2[/tex], [tex]k\ge1[/tex], you have
[tex]a_{n=3k-2}=\dfrac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}a_1[/tex]
Now, for [tex]n=0,3,6,9,\ldots[/tex], you have
[tex]a_0=a_0[/tex]
[tex]a_3=\dfrac3{3\times2}a_0=\dfrac3{3!}a_0[/tex]
[tex]a_6=\dfrac3{6\times5}a_3=\dfrac{3\times4}{6\times5\times4}a_3=\dfrac{3^2\times4}{6!}a_0[/tex]
[tex]a_9=\dfrac3{9\times8}a_6=\dfrac{3\times7}{9\times8\times7}a_6=\dfrac{3^3\times7\times4}{9!}a_0[/tex]
and so on, with a general pattern for [tex]n=3k[/tex], [tex]k\ge0[/tex], of
[tex]a_{n=3k}=\dfrac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}a_0[/tex]
Putting everything together, we arrive at the solution
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
[tex]y=a_0\underbrace{\displaystyle\sum_{k\ge0}\frac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}x^{3k}}_{n=0,3,6,9,\ldots}+a_1\underbrace{\displaystyle\sum_{k\ge1}\frac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}x^{3k-2}}_{n=1,4,7,10,\ldots}[/tex]
To show this solution is sufficient, I've attached is a plot of the solution taking [tex]y(0)=a_0=1[/tex] and [tex]y'(0)=a_1=0[/tex], with [tex]n=6[/tex]. (I was hoping to be able to attach an animation that shows the series solution (orange) converging rapidly to the exact solution (blue), but no such luck.)
[tex]\implies\displaystyle y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}[/tex]
[tex]y''-3xy=0[/tex]
[tex]\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0[/tex]
[tex]\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0[/tex]
[tex]\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge3}a_{n-3}x^{n-2}=0[/tex]
[tex]\displaystyle2a_2+\sum_{n\ge3}\bigg(n(n-1)a_n-3a_{n-3}\bigg)x^{n-2}=0[/tex]
This generates the recurrence relation
[tex]\begin{cases}a_0=a_0\\a_1=a_1\\2a_2=0\\n(n-1)a_n-3a_{n-3}=0&\text{for }n\ge3\end{cases}[/tex]
Because you have
[tex]n(n-1)a_n-3a_{n-3}=0\implies a_n=\dfrac3{n(n-1)}a_{n-3}[/tex]
it follows that [tex]a_2=0\implies a_5=a_8=a_{11}=\cdots=a_{n=3k-1}=0[/tex] for all [tex]k\ge1[/tex].
For [tex]n=1,4,7,10,\ldots[/tex], you have
[tex]a_1=a_1[/tex]
[tex]a_4=\dfrac3{4\times3}a_1=\dfrac{3\times2}{4!}a_1[/tex]
[tex]a_7=\dfrac3{7\times6}a_4=\dfrac{3\times5}{7\times6\times5}=\dfrac{3^2\times5\times2}{7!}[/tex]
[tex]a_{10}=\dfrac3{10\times9}a_7=\dfrac{3\times8}{10\times9\times8}a_7=\dfrac{3^3\times8\times5\times2}{10!}a_1[/tex]
so that, in general, for [tex]n=3k-2[/tex], [tex]k\ge1[/tex], you have
[tex]a_{n=3k-2}=\dfrac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}a_1[/tex]
Now, for [tex]n=0,3,6,9,\ldots[/tex], you have
[tex]a_0=a_0[/tex]
[tex]a_3=\dfrac3{3\times2}a_0=\dfrac3{3!}a_0[/tex]
[tex]a_6=\dfrac3{6\times5}a_3=\dfrac{3\times4}{6\times5\times4}a_3=\dfrac{3^2\times4}{6!}a_0[/tex]
[tex]a_9=\dfrac3{9\times8}a_6=\dfrac{3\times7}{9\times8\times7}a_6=\dfrac{3^3\times7\times4}{9!}a_0[/tex]
and so on, with a general pattern for [tex]n=3k[/tex], [tex]k\ge0[/tex], of
[tex]a_{n=3k}=\dfrac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}a_0[/tex]
Putting everything together, we arrive at the solution
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
[tex]y=a_0\underbrace{\displaystyle\sum_{k\ge0}\frac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}x^{3k}}_{n=0,3,6,9,\ldots}+a_1\underbrace{\displaystyle\sum_{k\ge1}\frac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}x^{3k-2}}_{n=1,4,7,10,\ldots}[/tex]
To show this solution is sufficient, I've attached is a plot of the solution taking [tex]y(0)=a_0=1[/tex] and [tex]y'(0)=a_1=0[/tex], with [tex]n=6[/tex]. (I was hoping to be able to attach an animation that shows the series solution (orange) converging rapidly to the exact solution (blue), but no such luck.)
