If $$y = y(x)$$ is the solution of the differential equation $$x\frac{dy}{dx} + 2y = xe^x, y(1) = 0$$ then the local maximum value of the function $$z(x) = x^2y(x) - e^x, x \in R$$ is

We need to solve $$x\frac{dy}{dx} + 2y = xe^x$$ with $$y(1) = 0$$, then find the local maximum of $$z(x) = x^2y(x) - e^x$$.

First, we rewrite the differential equation as $$\frac{dy}{dx} + \frac{2}{x}y = e^x$$ and identify the integrating factor as $$\mu = e^{\int \frac{2}{x}dx} = x^2$$.

Next, multiplying through by this integrating factor gives $$\frac{d}{dx}(x^2 y) = x^2 e^x$$, and integrating both sides yields $$x^2 y = \int x^2 e^x\,dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C$$.

Now applying the initial condition $$y(1) = 0$$ leads to $$1\cdot 0 = e^1(1 - 2 + 2) + C \implies 0 = e + C \implies C = -e$$, so that $$x^2 y = e^x(x^2 - 2x + 2) - e$$.

Since $$z(x) = x^2 y - e^x$$, substituting the expression for $$x^2 y$$ gives $$z(x) = e^x(x^2 - 2x + 2) - e - e^x = e^x(x^2 - 2x + 1) - e = e^x(x-1)^2 - e$$.

Next, differentiating yields $$z'(x) = e^x(x-1)^2 + e^x \cdot 2(x-1) = e^x(x-1)[(x-1) + 2] = e^x(x-1)(x+1)$$, so the critical points are $$x = 1$$ and $$x = -1$$.

Performing a sign analysis of $$z'(x) = e^x(x-1)(x+1)$$ shows that for $$x < -1$$ both factors $$(x-1)$$ and $$(x+1)$$ are negative, yielding $$z' > 0$$, while for $$-1 < x < 1$$ the factor $$(x-1)$$ is negative and $$(x+1)$$ is positive, giving $$z' < 0$$. Therefore, $$x = -1$$ is a local maximum.

Finally, evaluating at this point gives $$z(-1) = e^{-1}(-1-1)^2 - e = \frac{4}{e} - e$$. Hence the correct answer is Option D: $$\dfrac{4}{e} - e$$.