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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 27

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Answer: -\sqrt{1-y^{2}}=\sqrt{1-x^{2}}+c \text { or } \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=c

Hint: You must know about the rules of solving differential equation and integration

Given: x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}} d y=0


        y \sqrt{1-x^{2}} d y=-x \sqrt{1-y^{2}} d x

        \int \frac{y}{\sqrt{1-y^{2}}} d y=\int \frac{-x}{\sqrt{1-x^{2}}} d x

        We know \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)=\frac{-x}{\sqrt{1-x^{2}}}

        Integration of \frac{-x}{\sqrt{1-x^{2}}} d x=\sqrt{1-x^{2}}

        Similarly Integration of \frac{y}{\sqrt{1-y^{2}}} d y=-\sqrt{1-y^{2}}     

        \begin{aligned} &\Rightarrow-\sqrt{1-y^{2}}=\sqrt{1-x^{2}}+c \\\\ &\Rightarrow-\sqrt{1-y^{2}}-\sqrt{1-x^{2}}=c \\\\ &\Rightarrow-\left[\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\right]=c \end{aligned}

        \begin{aligned} &\Rightarrow \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=-c \\\\ &\Rightarrow \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=c \end{aligned}

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