Show that $\int\limits_0^x \exp\left(-\frac{t^2}2\right)dt = \frac{f(x)}{g(x)}$ 
Let $f(x) = x+\dfrac{x^3}{1\cdot 3} + \dfrac{x^5}{1\cdot 3 \cdot 5} + \dfrac{x^7}{1\cdot 3\cdot 5\cdot 7}+\dots$ and let $g(x) = 1+\dfrac{x^2}{2} + \dfrac{x^4}{2\cdot 4} + \dfrac{x^6}{2\cdot 4\cdot 6} + \dots$ Show that $\displaystyle\int_0^x \exp\left(-\frac{t^2}2\right)dt = \frac{f(x)}{g(x)}.$

I was thinking of finding a closed form for $f(x)$ and $g(x)$ but I couldn't do that (I did get that $\sinh(x) = \displaystyle\sum_{i=0}^\infty\dfrac{x^{2i+1}}{(2i+1)!}$ and $\cosh(x) = \displaystyle\sum_{i=0}^\infty \dfrac{x^{2i}}{(2i)!}$). 

Edit: Apparently I need to differentiate both sides and show that they vanish at $0$.

 A: For completeness, here is the naive/systematic approach.
Let
$$n!!=\begin{cases}n\cdot (n-2)\cdot (n-4)\cdot\ldots\cdot 1 & \textrm{if } n>0 \\ 1 & \textrm{otherwise.}\end{cases}$$
Then we have
$$f(x)=\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!!}, \qquad f'(x)=\sum_{n=0}^\infty \frac{x^{2n}}{(2n-1)!!}$$
$$g(x)=\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!!}, \qquad g'(x)=\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-2)!!}$$
The cross-products with the derivatives are
$$f'(x)g(x)=\sum_{n=0}^\infty\sum_{m=0}^\infty \frac{x^{(2n+2m)}}{(2n-1)!!(2m)!!}=\sum_{i=0}^\infty x^{2i}\left(\sum_{n=0}^i \frac{1}{(2n-1)!!(2i-2n)!!}\right)$$
$$g'(x)f(x)=\sum_{k=1}^\infty\sum_{m=0}^\infty \frac{x^{(2k+2m)}}{(2k-2)!!(2m+1)!!}=\sum_{i=0}^\infty x^{2i}\left(\sum_{k=1}^i \frac{1}{(2k-2)!!(2i-2k+1)!!}\right).$$
where I made the changes of variables $i=n+m$ (first line) and $i=k+m$ (second line). Note that, for a fixed $i$, the $n$-th term of the inner sum in $f'(x)g(x)$ is equal to the term with $k=i-n+1$ in $g'(x)f(x)$. This means that, when computing $f'(x)g(x)-g'(x)f(x)$, all terms in the inner sums will cancel except the term with $n=0$ in $f'(x)g(x)$ (which requires $k=i+1$, which is not in $g'(x)f(x)$). Therefore,
$$f'(x)g(x)-g'(x)f(x)=\sum_{i=0}^\infty x^{2i}\frac{1}{(2i)!!}=g(x)$$
which turns out to be equal to $g(x)$. Then,
$$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}=\frac{1}{g(x)}.$$
On the other hand, it is easy to see that $(2i)!!=i! 2^i$, and therefore
$$g(x)=\sum_{i=0}^{\infty}\frac{x^{2i}}{2^i}\frac{1}{i!}=e^{\frac{x^2}{2}}.$$
This means that 
$$\frac{\mathrm{d}}{\mathrm{d}x}\int_0^x e^{-t^2/2}=e^{-x^2/2}=\frac{1}{g(x)}=\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}.$$
As pointed out by MoonLightSyzygy in the comments, the equality of the derivatives, together with the fact that both sides are equal at $x=0$, proves that both functions are equal:
$$\int_0^x e^{-t^2/2}=\frac{f(x)}{g(x)}.$$
A: Since $f^\prime+xf=1$, integration factors tell us a constant $c$ exists for which $f=\exp-\frac{x^2}{2}\cdot\int_c^x\exp\frac{t^2}{2}dt$. Since $f(0)=0$, $c=0$. The desired claim then follows from $g=\exp\frac{x^2}{2}$.
