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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 18

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 18

Answers (1)

Answer: \sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} l \operatorname{og}\left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+c=0

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

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

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

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

          Integrating both sides

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

        \text { Let } 1+y^{2}=t \Rightarrow 2 y d y=d t \Rightarrow y d y=\frac{d t}{2}

        1+x^{2}=m^{2} \Rightarrow x=\sqrt{m^{2}-1} \Rightarrow 2 x d x=2 m d m \Rightarrow d x=\frac{m}{x} d m

        \Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t=-\int \frac{m}{m^{2}-1} d m \cdot m

        \Rightarrow \frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\int \frac{m^{2}}{m^{2}-1} d m=0

        \begin{aligned} &\Rightarrow \sqrt{t}+\int \frac{m^{2}-1+1}{m^{2}-1} d m=0 \\\\ &\Rightarrow \sqrt{t}+\int \frac{m^{2}-1}{m^{2}-1} d m+\int \frac{1}{m^{2}-1} d m=0 \\\\ &\Rightarrow \sqrt{t}+m+\frac{1}{2} \log \frac{(m-1)}{m+1}=0 \end{aligned}       \cdots \cdot \int \frac{1}{m^{2}-1} d m=\frac{1}{2} \log \frac{(m-1)}{m+1}

        \begin{aligned} &\text { Where } m=\sqrt{1+x^{2}}, 1+y^{2}=t \\\\ &\sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} l \operatorname{og}\left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+c=0 \end{aligned}       

        

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