1. Given differential equation

We have $y\frac{dx}{dy}=2x+y^3(y+1)e^y,$ with initial condition $x(1)=0.$

We need to find $x(e)$.

2. Rewrite in standard linear form

Divide both sides by $y$ (assuming $y\neq 0$, and here we work near $y=1$ to $y=e$): $\frac{dx}{dy}-\frac{2}{y}x=y^2(y+1)e^y.$

So this is a linear differential equation in $x$ as a function of $y$: $\frac{dx}{dy}+P(y)x=Q(y),$ where $P(y)=-\frac{2}{y}, \qquad Q(y)=y^2(y+1)e^y.$

3. Find the integrating factor

The integrating factor is $\mathrm{I.F.}=e^{\int P(y)dy}=e^{\int -\frac{2}{y}dy}=e^{-2\ln y}=y^{-2}.$

4. Multiply the equation by the integrating factor

Multiplying throughout by $y^{-2}$: $y^{-2}\frac{dx}{dy}-\frac{2}{y}y^{-2}x=(y+1)e^y.$

The left-hand side becomes $\frac{d}{dy}(x y^{-2}).$ Hence, $\frac{d}{dy}(x y^{-2})=(y+1)e^y.$

5. Integrate both sides

Integrate with respect to $y$: $x y^{-2}=\int (y+1)e^y\,dy + C.$

Now, $\int (y+1)e^y\,dy = ye^y,$ because $\frac{d}{dy}(ye^y)=e^y+ye^y=(y+1)e^y.$

Therefore, $x y^{-2}=ye^y+C.$ So, $x=y^2(ye^y+C).$

6. Use the initial condition

Given $x(1)=0$: $0=1^2(1\cdot e^1+C)=e+C.$ So, $C=-e.$

Thus, $x=y^2(ye^y-e).$

7. Evaluate at $y=e$

$x(e)=e^2\big(e\cdot e^e-e\big)=e^2\cdot e\,(e^e-1)=e^3(e^e-1).$

8. Compare with the options

We obtained $x(e)=e^3(e^e-1).$ This matches:

Option A: $e^3(e^e-1).$