Using termwise (term-by-term) differentiation on an infinite series to satisfy a differential equation. I have a question which asks me to use termwise differentation on the series $$\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(n!)^22^{2n}}$$ to show that it satisfies the differential equation $$x^2y''+xy'+x^2y=0$$
I dont understand what this question is asking me to do. I have found the interval and radius of convergence and the first 3 terms of this in previous questions if that is relevant at all?
Can someone explain this to me or the method etc so that I know how to do complete it?
 A: Excellent question. If you have a power series, you are allowed to differentiate it termwise (within the radius of convergence), just like a polynomial. I.e.
$$
\frac{d}{dx}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty na_nx^{n-1}
$$
Now, you can just differentiate the given series and check that it satisfies the differential equation as you would any other function.
A: Let $$y(x):=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{n!^22^{2n}}.$$
Since a power series can be differentiated term-by-term inside its radius of convergence, we have
$$
\begin{align*}
y'(x)&=\sum_{n=1}^\infty\frac{(-1)^n2nx^{2n-1}}{n!^22^{2n}}\\
y''(x)&=\sum_{n=1}^\infty\frac{(-1)^n2n(2n-1)x^{2n-2}}{n!^22^{2n}}
\end{align*}
$$
and they converges absolutely on the same region.  So
$$
\begin{align*}
&x^2y''(x)+xy'(x)+x^2y\\
&=\sum_{n=1}^\infty\frac{(-1)^n2n(2n-1)x^{2n}}{n!^22^{2n}}
+\sum_{n=1}^\infty\frac{(-1)^n2nx^{2n}}{n!^22^{2n}}
+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{n!^22^{2n}}\\
&=\sum_{n=1}^\infty\frac{(-1)^n2n(2n-1)x^{2n}}{n!^22^{2n}}
+\sum_{n=1}^\infty\frac{(-1)^n2nx^{2n}}{n!^22^{2n}}
+\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{2n}}{(n-1)!^22^{2(n-1)}}\\
&=\sum_{n=1}^\infty\frac{(-1)^n}{n!^2 2^{2n}}[2n(2n-1)
+2n-4n^2]x^{2n}\\
\end{align*}
$$
which is equal to $0$.
