$$f(t)=\sum\limits_{k=0}^{\infty}\frac{ (-1)^k\,(t/2)^{2k}}{(k!)^2}$$ show $t\, f''(t)+f'(t)+t\, f(t)=0$
By differentiating it, I got
$$f'(t)= \sum\limits_{k=0}^{\infty}\frac {(-1)^k\,(2k)(t)^{2k-1}}{(k!)^2 \,2^{2k}}$$
and $$f''(t)=\sum\limits_{k=0}^{\infty} \frac{(-1)^k(2k)(2k-1)(t)^{2k-2}}{(k!)^2 \, 2^{2k}}$$
when I plug these into the equation I then have LHS of the equation
$$\sum\limits_{k=0}^{\infty} \frac{(2k)^2+t^2}{(k!)^2 \, 2 ^{2k}}$$
Where did I make my mistake? I couldn't end up with the $LHS=0$.