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

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}