Answer
(a) We show that $F\left( x \right)$ has infinite radius of convergence.
(b) We show that $y = F\left( x \right)$ is a solution of the differential equation:
$y{\rm{''}} = xy{\rm{''}} + y$, ${\ \ \ \ \ }$ $y\left( 0 \right) = 1$, ${\ \ }$ $y'\left( 0 \right) = 0$
(c) See graph
Work Step by Step
(a) Let ${a_k} = \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{k \to \infty } \left| {\dfrac{{{a_{k + 1}}}}{{{a_k}}}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\dfrac{{{x^{2k + 2}}}}{{{2^{k + 1}}\cdot\left( {k + 1} \right)!}}\dfrac{{{2^k}\cdot k!}}{{{x^{2k}}}}} \right| = \left| {\dfrac{{{x^2}}}{2}} \right|\mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{k + 1}} = 0$
Since $\rho = 0 \lt 1$ for all $x$, by the Ratio Test, the series $F\left( x \right) = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$ converges for all $x$. It means that $F\left( x \right)$ has infinite radius of convergence.
(b) Write
$y = F\left( x \right) = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}} = 1 + \dfrac{{{x^2}}}{{2\cdot 1!}} + \dfrac{{{x^4}}}{{4\cdot 2!}} + \dfrac{{{x^6}}}{{8\cdot 3!}} + \cdot\cdot\cdot$,
where $y\left( 0 \right) = 1$.
Compute the derivatives with respect to $x$:
$y' = \mathop \sum \limits_{k = 1}^\infty \left( {2k} \right)\dfrac{{{x^{2k - 1}}}}{{{2^k}\cdot k!}} = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{x^{2k - 1}}}}{{{2^{k - 1}}\cdot\left( {k - 1} \right)!}} = x + \dfrac{{{x^3}}}{2} + \dfrac{{{x^5}}}{8} + \cdot\cdot\cdot$,
where $y'\left( 0 \right) = 0$.
$y{\rm{''}} = \mathop \sum \limits_{k = 1}^\infty \left( {2k - 1} \right)\dfrac{{{x^{2k - 2}}}}{{{2^{k - 1}}\cdot\left( {k - 1} \right)!}} = 1 + \dfrac{3}{2}{x^2} + \dfrac{5}{8}{x^4} + \cdot\cdot\cdot$
By shifting the index, $y{\rm{''}}$ can be written as
$y{\rm{''}} = \mathop \sum \limits_{k = 0}^\infty \left( {2k + 1} \right)\dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$
$y{\rm{''}} = \mathop \sum \limits_{k = 0}^\infty \left( {2k} \right)\dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}} + \mathop \sum \limits_{k = 0}^\infty \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$
Notice that the first term ($k=0$) of the first sum on the right-hand side is zero, so we can write
$y{\rm{''}} = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{x^{2k}}}}{{{2^{k - 1}}\cdot\left( {k - 1} \right)!}} + \mathop \sum \limits_{k = 0}^\infty \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$
$y{\rm{''}} = x\mathop \sum \limits_{k = 1}^\infty \dfrac{{{x^{2k - 1}}}}{{{2^{k - 1}}\cdot\left( {k - 1} \right)!}} + \mathop \sum \limits_{k = 0}^\infty \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$
Since $y = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$ and $y' = \mathop \sum \limits_{k = 1}^\infty \dfrac{{{x^{2k - 1}}}}{{{2^{k - 1}}\cdot\left( {k - 1} \right)!}}$, so
$y{\rm{''}} = xy{\rm{''}} + y$
(c) Write ${S_N} = \mathop \sum \limits_{k = 0}^N \dfrac{{{x^{2k}}}}{{{2^k}\cdot k!}}$.
Using a computer algebra system, we plot the partial sums ${S_N}$ for $N = 1,3,5,7$ on the same set of axes (please see the figure attached).
