Question

Y"+x^2y'+xy=0 power series

Answer 1

About \(x=0\)?

Answer 2

Yes

Answer 3

y = ∑(n=0 to ∞) a_n x^n
y' = ∑(n=0 to ∞) a_n n x^(n-1)
y'' = ∑(n=0 to ∞) a_n n(n-1) x^(n-2)

y'' - x^2 y' - xy =
= ∑(n=0 to ∞) a_n n(n-1) x^(n-2) -
- ∑(n=0 to ∞) a_n n x^(n+1) -
- ∑(n=0 to ∞) a_n x^(n+1) =

= ∑(n=-1 to ∞) a_(n+3) (n+3)(n+2) x^(n+1) -
- ∑(n=0 to ∞) a_n n x^(n+1) -
- ∑(n=0 to ∞) a_n x^(n+1) = 0

The first sum starts at n=-1, the other two start at n=0. Then, the term n=-1 of the first sum must be zero:

a_2 ∙ 2 ∙ 1 = 0 => a_2 = 0

For n≥0:

a_(n+3) (n+3)(n+2) - a_n (n+1) = 0
=>
a_(n+3) = a_n (n+1) / [(n+3)(n+2)]

It's right: you can choose a_0 and a_1 arbitrarily (order 2, 2 initial conditions to adjust), and a_2 must be zero. All other a_n are given by the previous recursion.

Answer 4

\[\begin{align*}
y&=\sum_{n=0}^\infty a_nx^n\\[1ex]
y'&=\sum_{n=1}^\infty na_nx^{n-1}\\[1ex]
y''&=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}
\end{align*}\]
Substituting into the ODE:
\[\begin{align*}
0&=\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\\[2ex]

&=\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}\\[2ex]

&=\sum_{n+3=2}^\infty (n+3)(n+3-1)a_{n+3}x^{n+3-2}+\sum_{n=1}^\infty na_nx^{n+1}+\sum_{n=0}^\infty a_nx^{n+1}\\[2ex]

&=\sum_{n=-1}^\infty (n+3)(n+2)a_{n+3}x^{n+1}+\sum_{n=1}^\infty na_nx^{n+1}+\sum_{n=0}^\infty a_nx^{n+1}\\[2ex]

&=\left(2a_2+6a_3x+\sum_{n=1}^\infty (n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n=1}^\infty na_nx^{n+1}\\[1ex]
&\quad\quad+\left(a_0x+\sum_{n=1}^\infty a_nx^{n+1}\right)\\[2ex]

&=2a_2+(6a_3+a_0)x+\sum_{n=1}^\infty \bigg[(n+3)(n+2)a_{n+3}+(n+1)a_n\bigg]x^{n+1}
\end{align*}\]