CUET Preparation Today
| If \(\frac{dy}{dx}=e^{-2y}, y=0\) when \(x=5\), then the value of \(x\) for \(y=3\) is |
\(\frac{e^{6}+9}{2}\) \(e^{6}+9\) \(\log_{e}6\) \(e^{6}+1\) |
| \(\frac{e^{6}+9}{2}\) |
| Given, \(e^{2y}dy=dx\) so, \(\frac{e^{2y}}{y}=x+c\hspace{5cm}\) Setting \(x=5,y=0\) we get \(c=-\frac{9}{2}\hspace{5cm}\) Thus, setting \(y=3\) and \(c=-\frac{9}{2}\) we get, \(x=\frac{e^{6}+9}{2}\) |
