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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]1 Mark
Question
Find the general solution of $\frac{ d y}{ d x}=\frac{1+y^2}{1+x^2}$
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Answer

$ \frac{ d y}{ d x}=\frac{1+y^2}{1+x^2}$
$\therefore \frac{ d x}{1+x^2}-\frac{ d y}{1+y^2}=0 $
Integrating on both sides, we get
$ \int \frac{ d x}{1+x^2}-\int \frac{ d y}{1+y^2}=0$
$\therefore \tan ^{-1} x -\tan ^{-1} y = c $

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