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

\(e\) \(\frac{1}{e}\) \(e^{e}\) None of these

\(e^{e}\)

Given, \(\frac{1}{y}ay=e^{x}dx\) so \(\log y=e^{x}+c\hspace{5 cm}\) Setting \(x=0,y=e\) we get \(c=0\hspace{6cm}\) Setting \(x=1, \log y=e\Rightarrow y=e^{e}\)