Question

Assume a series solution x(t) =summation ^ infinity _ n = 0 a_n t ^n for the equation x" + X = 0. Show that the recursion formula for the coefficients is a_m + 2= -a _m/(m + l1(m + 2), m = 0, 1, 2,... and that this leads to the general solution x (t) = a_0 (1 - t^2/2! + t^4/4! + ...) +a_1 (t - t^3/3! + t^5/5! + ... = a_0 cos(t) + a_1 sin(t)

Answer 1

Assume

`x=\displaystyle\sum_(n\ge0)a_nt^n`

`\implies x''=\displaystyle\sum_(n\ge0)(n+1)(n+2)a_(n+2)t^n`

Substituting these series into the ODE gives

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

from which we get

`(n+1)(n+2)a_(n+2)+a_n=0\implies a_(n+2)=-(a_n)/((n+1)(n+2))`

for
`n\ge0`, given
`x(0)=a_0` and
`x'(0)=a_1`. Now,

• if
  `n=2k` for integer
  `k\ge0`, then

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

`k=1\implies n=2\implies a_2=-(a_0)/(1\cdot2)=(-1)^1(a_0)/(2!)`

`k=2\implies n=4\implies a_4=-(a_2)/(3\cdot4)=(-1)^2(a_0)/(4!)`

`k=3\implies n=6\implies a_6=-(a_4)/(5\cdot6)=(-1)^3(a_0)/(6!)`

and so on, with the general rule

`a_(2k)=(-1)^k(a_0)/((2k)!)`

• and if
  `n=2k+1`, then

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

`k=1\implies n=3\implies a_3=-(a_1)/(2\cdot3)=(-1)^1(a_1)/(3!)`

`k=2\implies n=5\implies a_5=-(a_3)/(4\cdot5)=(-1)^2(a_1)/(5!)`

`k=3\implies n=7\implies a_7=-(a_5)/(6\cdot7)=(-1)^3(a_1)/(7!)`

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

So the series solution is

`\displaystyle x(t)=\sum_(k\ge0)\bigg(a_(2k)t^(2k)+a_(2k+1)t^(2k+1)\bigg)`

`x(t)=\displaystyle a_0\sum_(k\ge0)((-1)^kt^(2k))/((2k)!)+a_1\sum_(k\ge0)((-1)^kt^(2k+1))/((2k+1)!)`

`\implies x(t)=a_0\cot t+a_1\sin t`