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Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 16 maths

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

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

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

Solution: y \sqrt{1+x^{2}}+x \sqrt{1+y^{2}} \frac{d y}{d x}=0

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

        \begin{aligned} &1+y^{2}=m^{2} \Rightarrow y=\sqrt{m^{2}-1} \\\\ &2 y d y=2 m d m \Rightarrow d y=\frac{m}{y} d m \\\\ &\text { And Put } 1+x^{2}=n^{2} \Rightarrow x=\sqrt{n^{2}-1} \\\\ &2 x d x=2 n d n \Rightarrow x d x=n d n \Rightarrow d x=\frac{n}{x} d n \end{aligned}

        Integrating both sides

        \begin{aligned} &=>-\int \frac{\sqrt{1+y^{2}}}{y} d y=\int \frac{\sqrt{1+x^{2}}}{x} d x \\\\ &=>-\int \frac{m}{m^{2}-1} d m . m=\int \frac{n}{n^{2}-1} d n \cdot n \end{aligned}

        =>-\int \frac{m^{2}}{m^{2}-1} d m=\int \frac{n^{2}}{n^{2}-1} d n

        =>-\int \frac{m^{2}-1+1}{m^{2}-1} d m=\int \frac{n^{2}-1+1}{n^{2}-1} d n

        \begin{aligned} &=-\int 1 d m-\int \frac{1}{\left(m^{2}-1\right)} d m=\int 1 d n+\int \frac{1}{n^{2}-1} d n \\\\ &=>-m-\frac{1}{2} \log \left(\frac{m-1}{m+1}\right)=n+\frac{1}{2} \log \left(\frac{n-1}{n+1}\right)+c \end{aligned}

        \begin{aligned} &m=\sqrt{1+y^{2}} \& n=\sqrt{1+x^{2}} \\\\ &\Rightarrow-\sqrt{1+y^{2}}-\frac{1}{2} \log \left(\frac{\sqrt{1+y^{2}}-1}{\sqrt{1+y^{2}}+1}\right)=\sqrt{1+x^{2}}+\frac{1}{2} \log \left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+c \end{aligned}

        c=\sqrt{1+x^{2}}+\sqrt{1+y^{2}}+\frac{1}{2} \log \left(\frac{\sqrt{1+y^{2}}-1}{\sqrt{1+y^{2}}+1}\right)+\frac{1}{2} \log \left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)

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