According to the question,

$\frac{dy}{dx}=x+xy$

$\Rightarrow \frac{{\displaystyle dy}}{{\displaystyle dx}}-xy=x$

The linear differential equation can be written as $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

So, here $P\left(x\right)=-x \& Q\left(x\right)=x$

The integrating factor of the equation is 

$e^{-\int xdx}=e^{-\frac{x^2}{2}}$

So, the solution of differential equation is

$y\times e^{\frac{-x^2}{2}}=\int x\times e^{\frac{-x^2}{2}}dx+c ...\left(1\right)$

Assume 

$-\frac{x^2}{2}=t$

$\Rightarrow -xdx=dt$

So, from $\left(1\right)$, we can say that

$y\times e^{\frac{-x^2}{2}}=-\int e^{t}dt+c$

$\Rightarrow y\times e^{\frac{-x^2}{2}}=-e^{\frac{-x^2}{2}}+c$

Given, $y\left(0\right)=1$

So, we can say that

$e^0=-e^0+c$

$\Rightarrow c=2$

So, the equation of the curve is

$y\times e^{\frac{-x^2}{2}}=-e^{\frac{-x^2}{2}}+2$

$\Rightarrow y=2e^{\frac{x^2}{2}}-1$