Find the general solution of the following differential equation:
\({2^{ - {\rm{y}}}}{\rm{dx}} = \left( {1 + {{\rm{x}}^2}} \right){\rm{dy}}\)
1
\({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} = {2^{\rm{y}}}\ln 2 + {\rm{\;C}}\)
2
\({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} = \frac{2^{\rm{y}}}{ln 2} + {\rm{\;C}}\)
3
\({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} = {2^{\rm{y}}}\ln {\rm{y}} + {\rm{\;C}}\)
4
\({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{x}} = {{\rm{y}}^2}\ln 2 + {\rm{\;C}}\)