Answer:
Option B
Explanation:
$1+y^{2}+(x-e^{\tan^{-1}y})\frac{\text{d}y}{\text{d}x}=0$
$\Rightarrow (1+y^{2}) dx=(e^{\tan^{-1}y}-x)dy$
$\Rightarrow \frac{dy}{dx}=\frac{e^{\tan^{-1}y}-x}{1+y^{2}}$
$\Rightarrow \frac{dy}{dx}+\frac{1}{1+y^{2}}.x=\frac{e^{\tan^{-1}y}}{1+y^{2}}$
Which is the linear different equation of the `
form $ \frac{dx}{dy}+Rx=S$ , where R and S are functions of y or constant (s)
$\therefore$ I.F=$e^{\int_{}^{}\frac{1}{1+y^{2}}.dy }=e^{tan^{-1}y}$
Hence required solution is
x.I.F = $\int_{}^{} \frac{ae^{\tan^{-1}y}}{1+y^{2}}(I.F) dy$
$\Rightarrow x.e^{\tan^{-1}y}=\int\frac{e^{\tan^{-1}y}}{1+y^{2}} (e^{\tan^{-1}y}) dy$
$\Rightarrow x.e^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^{2}} dy$ ......(i)
Put t= tan-1 y
$\therefore$ $\frac{dt}{dy}=\frac{1}{1+y^{2}}\Rightarrow dt=\frac{1}{1+y^{2}} dy $
$\therefore$ $\int\frac{e^{2\tan^{-1}y}}{1+y^{2}} dy=\int e^{2t}.dt=\frac{e^{2t}}{2}+K$
Hence equation(i) becomes
$x e^{\tan^{-1}y}=\frac{1}{2}e^{2t}+K$
$\Rightarrow x e^{\tan^{-1}y}=\frac{1}{2}e^{2tan^{-1}y}+K$
$\Rightarrow 2x e^{\tan^{-1}y}=e^{2tan^{-1}y}+K$
