Question

The general solution of the differential equation $\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$ is (where $C$ is a constant of integration)

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

Answer 1

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

$\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$
$\sqrt{(1+x^2)(1+y^2)}+xy\frac{dy}{dx}=0$
$\frac{\sqrt{(1+x^2)dx}}{x}=-\frac{y}{\sqrt{1+y^2}}dy$
Intergate the equation
$\int \frac{\sqrt{1+x^2}}{x}dx=-\int \frac{y}{\sqrt{1+y^2}}dy$
$1+x^2=t^2$
$1+y^2=z^2$
$2xdx=2tdt$
$dx=\frac{t}{x}dt$
$2ydy=2zdz$
$\int \frac{t\cdot tdt}{t^2-1}=-\int \frac{zdx}{z}$
$\int \frac{t^2-1+1}{t^2-1}dt=-z+c$
$\int 1dt+\int \frac{1}{t^2-1}dt=-z+c$
$t+\frac{1}{2}\ln \left(\frac{t-1}{t+1}\right)=-z+c$
$\sqrt{1+x^2}+\frac{1}{2}$ In $\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)=-\sqrt{1+y^2}+C$
$\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}\ln \left(\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}-1}\right)+C$