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Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 33 maths textbook solution

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Answer: \tan ^{-1} y=x+\frac{x^{3}}{3}+c

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

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

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

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

          Integrating both sides

        \begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int 1 d x+\int x^{2} d x \\\\ &\tan ^{-1} y=x+\frac{x^{3}}{3}+c \end{aligned}

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