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,

$n(n-1)a_n+2a_{n-2}=0\implies a_n=-\dfrac2{n(n-1)}a_{n-2}$

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

$k=0\implies n=0\implies a_0=a_0$

$k=1\implies n=2\implies a_2=-\dfrac2{2\cdot1}a_0$

$k=2\implies n=4\implies a_4=-\dfrac2{4\cdot3}a_2=\dfrac{(-2)^2}{4!}a_0$

$k=3\implies n=6\implies a_6=-\dfrac2{6\cdot5}a_4=\dfrac{(-2)^3}{6!}a_1$

and so on, with

$a_{2k}=\dfrac{(-2)^k}{(2k)!}a_0$

• If $n=2k+1$, then

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

$k=1\implies n=3\implies a_3=-\dfrac2{3\cdot2}a_1$

$k=2\implies n=5\implies a_5=-\dfrac2{5\cdot4}a_3=\dfrac{(-2)^2}{5!}a_1$

and so on, with

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

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

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