Question

Begin with the series solution y-?~anz" to derive the recursive relation (n +2)(n +1)an+ -an-2, n2 2 for the differential equation y+2y0 and write out the first six non-zero terms.

Answer 1

I guess the ODE is supposed to be

`y''+2y=0`

So if

`y=\displaystyle\sum_(n\ge0)a_nz^n\implies y''=\sum_(n\ge0)(n+2)(n+1)a_(n+2)z^n`

then

`\displaystyle\sum_(n\ge0)(n+2)(n+1)a_(n+2)z^n+2\sum_(n\ge0)a_nz^n=0`

`\displaystyle\sum_(n\ge0)\bigg[(n+2)(n+1)a_(n+2)+2a_n\bigg]z^n=0`

so that

`(n+2)(n+1)a_(n+2)+2a_n=0`

for
`n\ge0`, or equivalently,

`n(n-1)a_n+2a_(n-2)=0\implies a_n=-\frac2{n(n-1)}a_(n-2)`

for
`n\ge2`. Note the dependency between every other coefficient - this means we can consider two cases:

• If
  `n=2k`, where
  `k\ge0` is an integer, then

`k=0\implies n=0\implies a_0=a_0`

`k=1\implies n=2\implies a_2=-\frac2{2\cdot1}a_0`

`k=2\implies n=4\implies a_4=-\frac2{4\cdot3}a_2=((-2)^2)/(4!)a_0`

`k=3\implies n=6\implies a_6=-\frac2{6\cdot5}a_4=((-2)^3)/(6!)a_1`

and so on, with

`a_(2k)=((-2)^k)/((2k)!)a_0`

• If
  `n=2k+1`, then

`k=0\implies n=1\implies a_1=a_1`

`k=1\implies n=3\implies a_3=-\frac2{3\cdot2}a_1`

`k=2\implies n=5\implies a_5=-\frac2{5\cdot4}a_3=((-2)^2)/(5!)a_1`

and so on, with

`a_(2k+1)=((-2)^k)/((2k+1)!)a_1`

Then the ODE has solution

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

and the first six non-zero terms occur for
`0\le n\le5`, for which we get

`\boxed{a_0\left(1-x^2+\frac{x^4}6\right)+a_1\left(x-\frac{x^3}3\right)}`