Question

Solution of $\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$, is:

• A. $\log \left(\frac{x}{1+\sqrt{1+x^2}}\right)+\sqrt{1+x^2}+\sqrt{1+y^2}=c$
• B. $\log \left(\frac{x}{\sqrt{1+x^2}}\right)+\sqrt{1-x^2}+\sqrt{1+y^2}=c$
• C. $\log \left(\frac{x}{\sqrt{1+x^2}}\right)=c$
• D. $\log (\sqrt{1+x^2}-\sqrt{1+y^2})+\log \left(\frac{x}{\sqrt{1+x^2}}\right)=c$

Answer 1

The correct option is A $\log \left(\frac{x}{1+\sqrt{1+x^2}}\right)+\sqrt{1+x^2}+\sqrt{1+y^2}=c$

Given, $\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$
$\Rightarrow \sqrt{1+y^2}\sqrt{1+x^2}=-xy\frac{dy}{dx}$

Integrating both sides, we get

$\Rightarrow \int \frac{\sqrt{1+x^2}}{x}dx=-\int \frac{y}{\sqrt{1+y^2}}dy$
Substitute, $1+y^2=t^2$ $\Rightarrow$ $2ydy=2tdt$
$\Rightarrow$ $\int \frac{\sqrt{1+x^2}}{x}dx=-\int dt=-t+c=-\sqrt{1+y^2}+c$
Now substitute, $1+x^2=k^2$ $\Rightarrow$ $2xdx=2kdk$
$\Rightarrow \int \frac{k^2}{x^2}dk=-\sqrt{1+y^2}+c$
$\Rightarrow \int \frac{k^2}{k^2-1}dk=-\sqrt{1+y^2}+c$
$\Rightarrow \int [\frac{k^2-1}{k^2-1}+\frac{1}{k^2-1}]dk=-\sqrt{1+y^2}+c$
$\Rightarrow k+\frac{1}{2}\log \frac{k-1}{k+1}=-\sqrt{1+y^2}+C$

Since , $\int \frac{1}{x^2-d^2}dx=\frac{1}{2d}\log \frac{x-d}{x+d}+\text{constant}$
$\therefore \sqrt{1+x^2}+\frac{1}{2}\log \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}+\sqrt{1+y^2}=C$

$\Rightarrow \sqrt{1+x^2}+\log \left(\frac{x}{\sqrt{1+x^2}+1}\right)+\sqrt{1+y^2}=C$