Given that $\frac{dy}{dx} = ye^{x}$ such that x = 0, y = e. The value of y (y > 0) when x = 1 will be :
Answer & explanation
Correct answer: option 3
$\frac{d y}{y}=e^{x} dx \Rightarrow \ln y=e^{x}+c$
At x = 0, y = 1 ; c = 0
$\ln y=e^x$
Therefore At x = 1, $y=e^c$
Hence (3) is the correct answer