If the curve y = y(x) passes through the poin | vedclass

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(C) Given the differential equation: $\frac{dy}{y} = (2 + \log_e x) dx$.Integrating both sides: $\int \frac{dy}{y} = \int (2 + \log_e x) dx$.$\log_e y

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$e^{2e}$

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(C) Given the differential equation: $\frac{dy}{y} = (2 + \log_e x) dx$.Integrating both sides: $\int \frac{dy}{y} = \int (2 + \log_e x) dx$.$\log_e y = 2x + (x \log_e x - x) + C = x \log_e x + x + C$.Since the curve passes through $(1, e)$, substitute $x = 1$ and $y = e$:$\log_e e = 1 \cdot \log_e 1 + 1 + C$.$1 = 0 + 1 + C \Rightarrow C = 0$.Thus, the equation of the curve is $\log_e y = x \log_e x + x$.To find $y(e)$, substitute $x = e$:$\log_e y = e \log_e e + e = e(1) + e = 2e$.Therefore, $y = e^{2e}$.

If the curve $y = y(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx$, $x > 0$, then $y(e)$ is equal to:

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