Question

The solution of the differential equation $\left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0$ is
(where $C$ is constant of integration.)

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

Answer 1

Answer: (A)

$\because \left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0$
$\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{{\tan }^{-1}y}}{1+y^2}$
$\text{ I.F. }=e^{\int \frac{dy}{1+y^2}}=e^{{\tan }^{-1}y}$
$\therefore \text{ Solution is }x{e}^{{\tan }^{-1}y}=\int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy=\frac{e^{2{\tan }^{-1}y}}{2}+\frac{C}{2}$
$\Rightarrow 2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+C,C=\text{ constant of integration. }$