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 $0$.


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 $g=\exp\frac{x^2}{2}$.
