Given that $\frac{dy}{dx} = ye^{x}$ such that x = 0, y = e. The value of y (y > 0) when x = 1 will be :

e $\frac{1}{e}$ $e^c$ $e^2$

$e^c$

$\frac{d y}{y}=e^{x} dx \Rightarrow \ln y=e^{x}+c$ At x = 0, y = 1 ; c = 0 $\ln y=e^x$ Therefore At x = 1, $y=e^c$ Hence (3) is the correct answer