Question

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

Answer 1

We have,
\[\sqrt{1 + x^2 + y^2 + x^2 y^2} + xy\frac{dy}{dx} = 0\]
\[ \Rightarrow \sqrt{\left( 1 + x^2 \right)\left( 1 + y^2 \right)} + xy\frac{dy}{dx} = 0\]
\[ \Rightarrow xy\frac{dy}{dx} = - \sqrt{\left( 1 + x^2 \right)\left( 1 + y^2 \right)}\]
\[ \Rightarrow xy\frac{dy}{dx} = - \sqrt{\left( 1 + x^2 \right)}\sqrt{\left( 1 + y^2 \right)}\]
\[ \Rightarrow \frac{y}{\sqrt{\left( 1 + y^2 \right)}}dy = - \frac{\sqrt{\left( 1 + x^2 \right)}}{x}dx\]
Integrating both sides, we get
\[ \Rightarrow \int\frac{y}{\sqrt{\left( 1 + y^2 \right)}}dy = - \int\frac{\sqrt{\left( 1 + x^2 \right)}}{x}dx\]
\[ \Rightarrow \int\frac{y}{\sqrt{\left( 1 + y^2 \right)}}dy = - \int\frac{x\sqrt{\left( 1 + x^2 \right)}}{x^2}dx\]
\[\text{ Putting }1 + y^2 = t\text{ and }1 + x^2 = u^2 \]
\[ \Rightarrow 2y dy = dt \text{ and }2x dx = 2udu\]
\[ \Rightarrow y dy = \frac{dt}{2}\text{ and }xdx = udu\]
\[ \therefore\text{ Integral becomes, }\]
\[\frac{1}{2}\int\frac{dt}{\sqrt{t}} = - \int\frac{u \times u}{u^2 - 1}du\]
\[ \Rightarrow \sqrt{t} = - \int\frac{u^2}{u^2 - 1}du\]
\[ \Rightarrow \sqrt{t} = - \int\frac{u^2}{u^2 - 1}du\]
\[ \Rightarrow \sqrt{t} = - \int1 + \frac{1}{u^2 - 1}du\]
\[ \Rightarrow \sqrt{t} = - \int(1)du - \int\frac{1}{u^2 - 1}du\]
\[ \Rightarrow \sqrt{t} = - u - \frac{1}{2}\log\left| \frac{u - 1}{u + 1} \right| + C\]
\[ \Rightarrow \sqrt{1 + y^2} = - \sqrt{1 + x^2} - \frac{1}{2}\log\left| \frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1} \right| + C\]
\[ \Rightarrow \sqrt{1 + y^2} + \sqrt{1 + x^2} + \frac{1}{2}\log\left| \frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1} \right| = C\]