Question

Hallar la solución general de la siguiente ecuación como una serie de potencial alrededor del punto x=0

2y" +xy´ + y =0

Answer 1

$\displaystyle 2y''+xy'+y=0\\\\ y=\sum_{n=0}^{\infty}a_nx^n\\\\ y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\\\\ y''=\sum_{n=2}^{\infty}a_nn(n-1)x^{n-2}$

$\text{Sustituimos}:\\\\ \displaystyle 2\sum_{n=2}^{\infty}a_nn(n-1)x^{n-2}+x\sum_{n=1}^{\infty}a_nnx^{n-1}+\sum_{n=0}^{\infty}a_nx^n=0\\\\\\ \sum_{n=2}^{\infty}2a_nn(n-1)x^{n-2}+\sum_{n=1}^{\infty}a_nnx^{n}+\sum_{n=0}^{\infty}a_nx^n=0\\\\\\ \text{Igualando potencias}:$

$\displaystyle \sum_{n=0}^{\infty}2a_{n+2}(n+2)(n+1)x^{n}+\sum_{n=1}^{\infty}a_nnx^{n}+\sum_{n=0}^{\infty}a_nx^n=0\\\\\\ \text{Igualando sub \'indices}\\ \\ 4a_2+a_0+\sum_{n=1}^{\infty}2a_{n+2}(n+2)(n+1)x^{n}+\sum_{n=1}^{\infty}a_nnx^{n}+\sum_{n=1}^{\infty}a_nx^n=0\\\\\\ 4a_2+a_0+\sum_{n=1}^{\infty}\left[2a_{n+2}(n+2)(n+1)+a_n(n+1)\right]x^n=0$

$\displaystyle \text{Comparando coeficientes:}\\\\ 4a_2+a_0=0\to a_2=-\dfrac{a_0}{4}\\\\ 2a_{n+2}(n+2)(n+1)+a_n(n+1)=0\;,\; \text{para }n\geq1\\\\ a_{n+2}=-\dfrac{a_n}{2(n+2)}\;,\;n\geq1$


$\displaystyle \text{Probemos:}\\ \\ \text{Con }n=1:\;\; a_3=-\dfrac{a_1}{2(3)}\\\\ \text{Con }n=2:\;\; a_4=-\dfrac{a_2}{2(4)}=\dfrac{a_0}{2^3(4)}\\\\ \text{Con }n=3:\;\; a_5=-\dfrac{a_3}{2(5)}=-\dfrac{a_1}{2^2(3\cdot 5)}$


$\displaystyle \text{Con }n=4:\;\; a_6=-\dfrac{a_4}{2(6)}=-\dfrac{a_0}{2^4(4\cdot6)}\\\\ \text{Con }n=5:\;\; a_7=-\dfrac{a_5}{2(7)}=\dfrac{a_1}{2^3(3\cdot5\cdot7)}\\\\ \vdots\\ \\ a_{2n-1}=\dfrac{(-1)^{n+1}a_1}{2^{n}(1\cdot3\cdot5\cdot7\cdots(2n+1))}\\\\ a_{2n-1}=\dfrac{(-1)^{n+1}a_1(2\cdot4\cdot6\cdots(2n))}{2^{n}(2n+1)!}$


$\displaystyle \boxed{a_{2n-1}=\dfrac{(-1)^{n+1}n!}{(2n+1)!}a_1\;,\;n\geq 1}\\\\\\ a_{2n}=\dfrac{(-1)^n}{2^n(2\cdot4\cdot6\cdots(2n))}a_0\\\\ \boxed{a_{2n}=\dfrac{(-1)^{n}}{4^{n}n!}a_0\;,\; n\geq1}\\\\ y=\sum_{n=0}^{\infty}a_nx^n\\\\ y=\sum_{n=0}^{\infty}a_{2n}x^{2n}+\sum_{n=1}^{\infty}a_{2n-1}x^{2n-1}\\\\ \boxed{y=a_0\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{4^{n}n!}x^{2n}+a_1\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}n!}{(2n+1)!}x^{2n-1}}$