MCQ
The solution of the differential equation $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$ is
  • A
    $y=\tan ^{-1} x$
  • $\tan ^{-1} y-\tan ^{-1} x=c$
  • C
    $x=\tan ^{-1} y$
  • D
    $\tan (x y)=k$

Answer

Correct option: B.
$\tan ^{-1} y-\tan ^{-1} x=c$
(b) : $\frac{d y}{d x}=\frac{1+y^2}{1+x^2} \Rightarrow \frac{d y}{1+y^2}=\frac{d x}{1+x^2}$
On integrating both sides, we get
$
\tan ^{-1} y=\tan ^{-1} x+c \Rightarrow \tan ^{-1} y-\tan ^{-1} x=c
$

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