Question

The solution of differential 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. $x{e}^{{\tan }^{-1}y}=e^{{\tan }^{-1}y}+k$
• C. $x{e}^{2{\tan }^{-1}y}=e^{{\tan }^{-1}y}+k$
• D. $(x-2)=k{e}^{-{\tan }^{-1}y}$

Answer 1

The correct option is A $2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k$

$(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0\Rightarrow (1+y^2)\frac{dx}{dy}-x=e^{{\tan }^{-1}y}\Rightarrow \frac{dx}{dy}-\frac{x}{1+y^2}=\frac{e^{{\tan }^{-1}y}}{1+y^2}$

Taking $I.F=e^{{\tan }^{-1}y}$

We get

$x.I.F=\int y.I.Fdyx{e}^{{\tan }^{-1}y}=\frac{e^{2{\tan }^{-1}y}}{2}+c$