Question

Solve:
$\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$

Answer 1

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

$\sqrt{(1+x^2)+y^2(1+x^2)}+xy\frac{dy}{dx}=0$

$\sqrt{(1+x^2)(1+y^2)}+xy\frac{dy}{dx}=0$

$\Rightarrow \int \frac{\sqrt{1+x^2}}{x}dx+\int \frac{ydy}{\sqrt{1+y^2}}=0$

$\Rightarrow \int \frac{1+x^2}{x\sqrt{1+x^2}}dx+\int \frac{ydy}{\sqrt{1+y^2}}=0$

Let $I=\int \frac{dx}{x\sqrt{1+x^2}}+\int \frac{xdx}{\sqrt{1+x^2}}+\int \frac{ydy}{\sqrt{1+y^2}}=0$

$\therefore I=I_1+I_2+I_3$

Consider $I_1=\int \frac{dx}{x\sqrt{1+x^2}}$

Let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

$I_1=\int \frac{\tan \theta {\sec }^2\theta d\theta }{\tan \theta \sqrt{1+{\tan }^2\theta }}$

$=\int \frac{\sec \theta }{\tan \theta }d\theta$

$=\int \frac{\frac{1}{\cos \theta }}{\frac{\sin \theta }{\cos \theta }}d\theta$

$=\int \csc \theta d\theta$

$=-\ln (\csc \theta +\cot \theta )+c_1$

$\therefore I_1=-\ln (\csc \theta +\cot \theta )+c_1$

Consider $I_2=\int \frac{xdx}{\sqrt{1+x^2}}$

Let $t=1+x^2\Rightarrow dt=2xdx$

$\int \frac{xdx}{\sqrt{1+x^2}}$

$=\frac{1}{2}\int \frac{2xdx}{\sqrt{1+x^2}}$

$=\int \frac{dt}{\sqrt{t}}$

$=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c_2$

$=2\sqrt{t}+c_2$

$=2\sqrt{1+x^2}+c_2$ where $t=1+x^2$

Consider $I_3=\int \frac{ydy}{\sqrt{1+y^2}}$

Let $t=1+y^2\Rightarrow dt=2ydy$

$\int \frac{ydy}{\sqrt{1+y^2}}$

$=\frac{1}{2}\int \frac{2ydy}{\sqrt{1+y^2}}$

$=\int \frac{dt}{\sqrt{t}}$

$=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+c_3$

$=2\sqrt{t}+c_3$

$=2\sqrt{1+y^2}+c_3$ where $t=1+y^2$

$I=I_1+I_2+I_3$

$\therefore I=-\ln (\csc \theta +\cot \theta )+c_1+2\sqrt{1+x^2}+c_2+2\sqrt{1+y^2}+c_3$

$=-\ln (\csc \theta +\cot \theta )+2\sqrt{1+x^2}+2\sqrt{1+y^2}+c$ where $c=c_1+c_2+c_3$