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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 31

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Answer: \tan ^{-1} y=\tan ^{-1} x+c

Hint: Separate the terms of x and y and then integrate them.

Given: \left(y^{2}+1\right) d x-\left(x^{2}+1\right) d y=0

Solution: \left(y^{2}+1\right) d x-\left(x^{2}+1\right) d y=0

        \left(y^{2}+1\right) d x=\left(x^{2}+1\right) d y

        \begin{aligned} &\frac{d y}{\left(y^{2}+1\right)}=\frac{d x}{\left(x^{2}+1\right)} \\\\ &\tan ^{-1} y=\tan ^{-1} x+c \end{aligned}


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