Answer to Question #205866 in Differential Equations for sania khan

Question

solve power series solution

y'' - 2xy' + y = 0

Answer 1

Assuming a power series solution like this:

$y=\displaystyle\sum_{n=0}^{\infin}a_nx^n$

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

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

we get

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

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

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

$n=0:2a_2+a_0=0$

$=>a_2=-\dfrac{1}{2}a_0=-\dfrac{1}{(2(1))!}a_0$

$n\geq1: (n+2)(n+1)a_{n+2}-2na_n+a_n=0$

$a_{n+2}=\dfrac{2n-1}{(n+2)(n+1)}a_n, n=0,1,2,...$

$a_{3}=\dfrac{2(1)-1}{(1+2)(1+1)}a_1=\dfrac{1}{(2(1)+1)!}a_1$

$a_{4}=\dfrac{2(2)-1}{(2+2)(2+1)}a_2=\dfrac{3}{12}a_2=-\dfrac{3}{(2(2))!}a_0$

$a_{5}=\dfrac{2(3)-1}{(3+2)(3+1)}a_3=\dfrac{5}{20}a_3=\dfrac{1\cdot5}{(2(2)+1)!}a_1$

$a_{6}=\dfrac{2(4)-1}{(4+2)(4+1)}a_4=\dfrac{7}{30}a_4=-\dfrac{3\cdot7}{(2(3))!}a_0$

$a_{2k}=-\dfrac{3\cdot7\cdot...\cdot(4k-5)}{(2k)!}a_0, k=2,3,...$

$a_{2k+1}=\dfrac{1\cdot5\cdot...\cdot(4k-3)}{(2k+1)!}a_1, k=1,2,3,...$

$y=a_0\bigg(1-\dfrac{x^2}{2!}-\displaystyle\sum_{n=2}^{\infin}\dfrac{3\cdot7\cdot...\cdot(4n-5)}{(2n)!}x^{n}\bigg)$

$+a_1\bigg(x+\displaystyle\sum_{n=2}^{\infin}\dfrac{1\cdot5\cdot...\cdot(4n-3)}{(2n+1)!}x^{n+1}\bigg)$