Question

The general solution of the equation $(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$ is

• A. $2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k$
• B. $x{e}^{{\tan }^{-1}y}={\tan }^{-1}y+k$
• C. $x{e}^{{\tan }^{-1}y}=e^{{\tan }^{-1}y}+k$
• D. $x=2+k{e}^{-{\tan }^{-1}y}$

Answer 1

Answer: (A)

$2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k$

$(1+y^2)+(x-e^{ta{n}^{-1}y})\frac{dy}{dx}=0\Rightarrow (1+y^2)\frac{dx}{dy}+x=e^{ta{n}^{-1}y}\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{ta{n}^{-1}y}}{1+y^2}$
Taking $I.F=e^{ta{n}^{-1}y}$
We get
$x.I.F=\int y.I.Fdyx{e}^{ta{n}^{-1}y}=\frac{e^{2ta{n}^{-1}y}}{2}+k$