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Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 38 sub question (iv)

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Answer: \frac{1}{2}\left(\tan ^{-1} x\right)^{2}+\log \left|1+y^{2}\right|=c

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

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

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

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

        Integrating both sides

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

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