The so-called error function is defined as:
$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^{2}}dt$$
show that the function $y(x) = e^{x^2}\operatorname{erf}(x)$ satisfies the differential equation:
$$\frac{dy}{dx}=2xy+\frac{2}{\sqrt{\pi}}$$
I have no idea how to start if someone could show me that would be greatly appreciated.
By using the formula $(uv)'=u'v+uv'$, you just have (prime = diff. with respect to x). $$ \left(e^{x^2}\times\frac{2}{\sqrt{\pi}}\int_0^{{x}}e^{-t^{2}}dt\right)'=2xe^{x^2}\left(\frac{2}{\sqrt{\pi}}\int_0^{{x}}e^{-t^{2}}dt\right)+\left(e^{x^2}\times\frac{2}{\sqrt{\pi}}e^{-x^{2}}\right)=2xy+\frac{2}{\sqrt{\pi}} $$ since $$ \left(\int_0^{{x}}e^{-t^{2}}dt\right)'=e^{-x^{2}}. $$