If $$x = x(y)$$ is the solution of the differential equation $$y\frac{dx}{dy} = 2x + y^3(y+1)e^y, x(1) = 0$$; then $$x(e)$$ is equal to

We are given the differential equation $$y\frac{dx}{dy} = 2x + y^3(y+1)e^y$$, with $$x(1) = 0$$.

Rewrite as a first-order linear ODE in x.

$$\frac{dx}{dy} - \frac{2x}{y} = y^2(y+1)e^y$$

Find the integrating factor.

$$\text{I.F.} = e^{\int -\frac{2}{y}\,dy} = e^{-2\ln y} = \frac{1}{y^2}$$

Multiply through and integrate.

$$\frac{d}{dy}\left(\frac{x}{y^2}\right) = (y+1)e^y = ye^y + e^y$$

$$\frac{x}{y^2} = \int (ye^y + e^y)\,dy$$

$$= ye^y - e^y + e^y + C = ye^y + C$$

Apply the initial condition x(1) = 0.

$$\frac{0}{1} = 1 \cdot e + C \implies C = -e$$

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

$$x = y^2(ye^y - e) = y^2 \cdot e(y \cdot e^{y-1} - 1)$$

Evaluate at y = e.

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

Answer: Option B