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

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

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$e^{2e}$

Vedclass · Answer

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

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

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