Question

Solve the differential equation:

$\sqrt{1+x^2}dx+\sqrt{1+y^2}dy=0$

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

Answer 1

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

Given, $\sqrt{1+x^2}dx=-\sqrt{1+y^2}.dy$

It is known that
$\int \sqrt{x^2\pm a^2}dx=\frac{1}{2}x\sqrt{x^2\pm a^2}\pm \frac{a^2}{2}ln|x+\sqrt{x^2\pm a^2}$

Applying this formula gives us

$\int \sqrt{x^2+1}.dx=-\int \sqrt{y^2+1}.dy$
$\Rightarrow \frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}ln|x+\sqrt{x^2+1}|=-[\frac{y}{2}\sqrt{y^2+1}+\frac{1}{2}ln|y+\sqrt{y^2+1}|]+C$
$\Rightarrow x\sqrt{x^2+1}+y\sqrt{y^2+1}+ln|x+\sqrt{x^2+1}|+ln|y+\sqrt{y^2+1}|=C$
$\Rightarrow x\sqrt{x^2+1}+y\sqrt{y^2+1}+ln|(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})|=C$