Question

Find a power series solution to the differential equation at the point x0. (2+x2)y′′-xy′+4y=0

Answer 1

`(x^2+2)y''-xy'+4y=0`

I'll assume you are looking for a series centered at
`x=0`, which is an ordinary point for the ODE. Substituting


`y=\displaystyle\sum_(n\ge0)a_nx^n`

into the ODE, we can rewrite it as


`\displaystyle\sum_(n\ge2)n(n-1)a_nx^n+2\sum_(n\ge2)n(n-1)a_nx^(n-2)`

`\,\,\,\,\,\,\,\,-\displaystyle\sum_(n\ge1)na_nx^n+4\sum_(n\ge0)a_nx^n=0`


`\displaystyle\sum_(n\ge2)n(n-1)a_nx^n+2\sum_(n\ge0)(n+2)(n+1)a_(n+2)x^n`

`\displaystyle\,\,\,\,\,\,\,\,-\sum_(n\ge1)na_nx^n+4\sum_(n\ge0)a_nx^n=0`


`\displaystyle4(a_0+a_2)+3(a_1+4a_3)x+\sum_(n\ge2)\bigg[2(n+2)(n+1)a_(n+2)+(n^2-2n+4)a_n\bigg]x^n=0`

so the coefficients of the power series are defined by the recurrence


`\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=-((n-3)^2+3)/(2n(n-1))a_(n-2)&\text{for }n\ge2\end{cases}`

For even
`n`, i.e.
`n=2k`
`(k\ge0)`, we have


`a_0=a_0`

`a_2=-a_0`

`a_4=-\frac4{4*3*2}a_0=\frac4{4!}a_0`

`a_6=-(12)/(2*6*5)a_4=-(12*4)/(2*6!)a_0`

`a_8=-(28)/(2*8*7)a_6=(28*12*4)/(2^2*8!)a_0`

and so on. The pattern in the denominator is pretty clear, but we can also find a compact form for the numerator. When
`k=5`, we can write it as


`4*12*28*52=4^4(1*3*7*13)`

`=4^4\bigg(1*(1+2)*(1+2+4)*(1+2+4+6)\bigg)`

`=4^4\bigg(1*(1+2(1))*(1+2(1+2))*(1+2(1+2+3)\bigg)`

`=4^4\displaystyle\prod_(i=1)^3\left(1+2\sum_(j=1)^(i-1)j\right)`

`=4^4\displaystyle\prod_(i=1)^3(i^2-i+1)`

So in general we have


`a_(2k)=((-1)^k2^k\displaystyle\prod_(i=1)^(k-1)(i^2-i+1))/((2k)!)`

We can treat the odd-indexed terms similarly. For
`n=2k-1`
`(k\ge1)` we have


`a_1=a_1`

`a_3=-\frac14a_1=-\frac3{2*3*2}a_1=-\frac3{2*3!}a_1`

`a_5=-\frac7{2*5*4}a_3=(7*3)/(2^2*5!)a_1`

`a_7=-(19)/(2*7*6)a_5=-(19*7*3)/(2^3*7!)a_1`

and so on. Again, the pattern in the denominator is simple. For
`k=5`, we would get a numerator of


`3*7*19*39=3*(3+4)*(3+4+12)*(3+4+12+20)`

`=3*(3+4(1))*(3+4(1+3))*(3+4(1+3+5))`

`=\displaystyle\prod_(i=1)^4\left(3+4\sum_(j=1)^(i-1)(2j-1)\right)`

`=\displaystyle\prod_(i=1)^4(4i^2-8i+7)`

and in general we'd have


`a_(2k-1)=((-1)^(k+1)\displaystyle\prod_(i=1)^(k-1)(4i^2-8i+7))/(2^(k-1)(2k-1)!)`

Thus the power series solution to this ODE is


`y(x)=\displaystyle\sum_(k\ge0)a_(2k)x^(2k)+\sum_(k\ge1)a_(2k-1)x^(2k-1)`

Attached below is a plot of a numerical solution (blue) compared to the first 9 terms
`(0\le n\le8)` and first 21 terms
`(0\le n\le21)` of the series solution over the interval
`|x|\le3`, assuming initial values of
`y(0)=y'(0)=1`
`(a_0=a_1=1)`.