If the curve $y = f(x)$ passes through the point ( $1,e$ ) and satisfies the differential equation $

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

Accepted Answer

Sol. [Key Concept: Separate variables and integrate $\int(2 + ln\ x)dx = x + xln\ x$; apply the point to fix $C$.]$$\frac{dy}{dx} = y\left( 2 + \mathcal{l}nx \right) $$$$\int\frac{dy}{y} = \int\left( 2 + \mathcal{l}nx \right)dx $$$$\mathcal{l}ny = 2x + x\mathcal{l}nx - x + C $$$$\mathcal{l}ny = x + x\mathcal{l}nx + C $$Since it passes through $(1,e)$$$1 = 1 + 0 + C \Rightarrow C = 0 $$$${\mathcal{l}ny = x + x\mathcal{l}nx }f$$$$(x) = y = e^{x + {xln}x} $$$$\Rightarrow f(e) = e^{ee} = e^{2e} $$

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