Part of my assignment is to show that
$$ J(t) = \sum_{n=0}^\infty \frac{(-1)^n\cdot (t/2)^{2n}}{(n!)^2} $$ solves the ODE $$ t\cdot J''(t) + J'(t) + t\cdot J(t) = 0. $$ By differentiating within the radius of convergence and then substitution I'm left with a very long solution that I doesn't look like it will go to zero. I was wondering if there were any short cuts rather than brute force substitution.
$$t\cdot J''(t) + J'(t) + t\cdot J(t) = 0.$$ $$t^2\cdot J''(t) + tJ'(t) + t^2\cdot J(t) = 0.$$ So we have : $$t^2J(t) = \sum_{n=0}^\infty \frac{(-1)^n\cdot (t)^{2(n+1)}}{(n!)^22^{2n}}$$ $$tJ'(t) = \sum_{n=0}^\infty 2n\frac{(-1)^n\cdot (t)^{2n}}{(n!)^22^{2n}}$$ $$t^2J''(t) = \sum_{n=0}^\infty 2n(2n-1)\frac{(-1)^n\cdot (t)^{2n}}{(n!)^22^{2n}}$$
You need to rewrite properly the three series. note that the second and third serie reduces to: $$\sum_{n=0}^\infty 4n^2\frac{(-1)^n\cdot (t)^{2n}}{(n!)^22^{2n}}+\sum_{n=0}^\infty \frac{(-1)^n\cdot (t)^{2(n+1)}}{(n!)^22^{2n}}=0$$
Change the indice of the second sum: $$\sum_{n=0}^\infty 4n^2\frac{(-1)^{n}\cdot (t)^{2n}}{(n!)^22^{2n}}-\sum_{m=1}^\infty \frac{(-1)^m\cdot (t)^{2m}}{((m-1)!)^22^{2(m-1)}}=0$$ The first serie starts at $n=1$ because the first term is zero: $$\sum_{n=1}^\infty 4n^2\frac{(-1)^n\cdot (t)^{2n}}{(n!)^22^{2n}}-\sum_{m=1}^\infty \frac{(-1)^m4m^2(t)^{2m}}{(m!)^22^{2m}}=0$$ It's zero. Both series are the same with opposite sign.