Question

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

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

Answer 1

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

$\begin{array}{c}\left(1+y^2\right)\frac{dx}{dy}=e^{{\tan }^{-1y}}-x\\ \frac{dx}{dy}+\frac{x}{\left(1+y^2\right)}=\frac{e^{{\tan }^{-1y}}}{\left(1+y^2\right)}\end{array}$

Its a linear differential equation with $I.F.=e^{{\tan }^{-1}y}$

So,

We would be $x.e^{{\tan }^{-1}y}$=integral of $\frac{e^{{\tan }^{-1}y}.e^{{\tan }^{-1}y}}{\left(1+y^2\right)}$

Put ${\tan }^{-1y}=t$