Solution of a linear differential equation Find the general solution of the equation $$ \frac{dy}{dx}=1+xy $$
My attempt: Arranging it in standard linear equation form $\frac{dy}{dx}+Py=Q$, we get
$$\frac{dy}{dx}+(-x)y=1$$
Hence, the integrating factor(I.F.) = $e^{\int-xdx}=e^{-x^2/2}$
Hence, the solution is $$y(e^{-x^2/2})=\int{e^{-x^2/2}dx}+c$$
How do I solve this integral now? 
 A: As a commenter mentions, the solution is:
$$y(x)=c_1 e^{\frac{x^2}{2}}+\sqrt{\frac{\pi }{2}} e^{\frac{x^2}{2}}    \text{erf}\left(\frac{x}{\sqrt{2}}\right)$$
One thing to note here is that the erf(x) function is sometimes not all that well known. According to wolfram alpha, https://www.wolframalpha.com/input/?i=erf(), the erf() function is defined as:
$$\text{erf(x)}=\frac{2}{\sqrt{\pi}} \int_{0}^{x}{e^{-t^2}dt}$$
Now when you have:
$$\int{e^{-x^2/2}dx}$$
It can be rewritten as:
$$\int{e^{-(x/\sqrt{2})^2}dx}$$
Let's replace $x/\sqrt{2}=t$. This tells us that $dx=\sqrt{2}dt$.
Making the substitution:
$$\sqrt{2}\int{e^{-t^2}dt}$$
Since limits are need to be inserted, it can be shown that:
$$\int_0^t{e^{-z^2}dz}=\int{e^{-t^2}dt}+C$$
Combining all of that:
$$\sqrt{2}(\frac{\sqrt{\pi}}{2}\text{erf(t)}+C)=\sqrt{2}\int{e^{-t^2}dt}$$
This means that:
$$y(e^{-x^2/2})=\sqrt{2}(\frac{\sqrt{\pi}}{2}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+C)$$
I trust you can simplify from there.
A: The incomplete equation is
$$\frac {y'}{y}=x $$
thus
$$\ln (\frac {y}{\lambda})=\frac {x^2}{2} $$
and
$$y_h=\lambda e^{\frac {x^2}{2}} $$
the constante variation method gives
$$\lambda'(x)=e^{\frac {-x^2}{2}} $$
the general solution is
$$y_g=\Bigl(\lambda+\int e^{\frac {-x^2}{2}}dx\Bigr)e^{\frac {x^2}{2}}$$
