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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 17 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 17 maths textbook solution

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Answer: \log \left(y+\sqrt{1+y^{2}}\right)\left(x+\sqrt{1+x^{2}}\right)=C

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

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

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

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

          Integrating both sides

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

        \text { Let } I=\int \frac{d x}{\sqrt{1+x^{2}}}

        Put

        x=\tan \theta

        d x=\sec ^{2} \theta d \theta

        \begin{aligned} &I=\int \frac{1}{\sqrt{1+\tan ^{2} \theta}} \sec ^{2} \theta d \theta \\\\ &=\int \frac{1}{\sqrt{\sec ^{2} \theta}} \sec ^{2} \theta d \theta \\\\ &\left(\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}\right) \end{aligned}

        \begin{aligned} &I=\int \sec \theta d \theta \\\\ &=\log |\sec \theta+\tan \theta| \\\\ &\tan \theta=x \& \sec \theta=\sqrt{1+\tan ^{2} \theta}=\sqrt{1+x^{2}} \\\\ &I=\log \left|x+\sqrt{1+x^{2}}\right|+c \end{aligned}       

        Similarly,\int \frac{1}{\sqrt{1+y^{2}}}=\log \left|y+\sqrt{1+y^{2}}\right|+c

        Hence,

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

 

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