Question

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

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

Answer 1

Answer: (A)

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