Solve Differential Equation to Find Function Value at E

To find the value of $f(e)$, we first solve the given differential equation: $d y = y\left(2+\log _e x\right) d x, \quad x>0$

Separating the variables $x$ and $y$, we get: $\frac{d y}{y} = \left(2+\log _e x\right) d x$

Integrating both sides with respect to their respective variables: $\int \frac{1}{y} d y = \int \left(2+\log _e x\right) d x$

Using the integration by parts formula for $\int \log_e x \, d x = x \log_e x - x$, the right-hand side becomes: $\log_e y = 2x + \left(x \log_e x - x\right) + C$ $\log_e y = x \log_e x + x + C$

We are given that the curve $y = f(x)$ passes through the point $(1, e)$, which means $y = e$ when $x = 1$. Substituting these values to find the constant of integration $C$: $\log_e e = 1 \cdot \log_e 1 + 1 + C$ $1 = 0 + 1 + C \implies C = 0$

Substituting $C = 0$ back into the equation yields: $\log_e y = x \log_e x + x$ $y = f(x) = e^{x \log_e x + x} = e^{x(1 + \log_e x)}$

Now, to find $f(e)$, we substitute $x = e$ into the expression for $f(x)$: $f(e) = e^{e\left(1 + \log_e e\right)}$

Since $\log_e e = 1$: $f(e) = e^{e(1 + 1)} = e^{2e}$

Hence, the correct option is C.