Question

Solution of the differential equation $(1+y^2)dx=(1-ta{n}^{-1}y-x)dy$ is

• A. $x{e}^{ta{n}^{-1}y}=(1-ta{n}^{-1}y-x)dy$
• B. $x{e}^{ta{n}^{-1}y}=(2-ta{n}^{-1}y)e^{ta{n}^{-1}y}+c$
• C. $x=2-ta{n}^{-1}y+c{e}^{-ta{n}^{-1}y}$
• D. None of these

Answer 1

Answer: (B), (C)

The correct options are
B $x{e}^{ta{n}^{-1}y}=(2-ta{n}^{-1}y)e^{ta{n}^{-1}y}+c$
C $x=2-ta{n}^{-1}y+c{e}^{-ta{n}^{-1}y}$