Question

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

Answer 1

We have,
\[\sqrt{1 + x^2} dy + \sqrt{1 + y^2} dx = 0\]
\[\sqrt{1 + x^2} dy = - \sqrt{1 + y^2} dx\]
\[\frac{1}{\sqrt{1 + y^2}}dy = - \frac{1}{\sqrt{1 + x^2}}dx\]
Integrating both sides, we get
\[\int\frac{1}{\sqrt{1 + y^2}}dy = - \int\frac{1}{\sqrt{1 + x^2}}dx\]
\[ \Rightarrow \log \left| y + \sqrt{1 + y^2} \right| = - \log \left| x + \sqrt{1 + x^2} \right| + \log C\]
\[ \Rightarrow \log \left| y + \sqrt{1 + y^2} \right| + \log \left| x + \sqrt{1 + x^2} \right| = \log C\]
\[ \Rightarrow \log \left| \left( y + \sqrt{1 + y^2} \right)\left( x + \sqrt{1 + x^2} \right) \right| = \log C\]
\[ \Rightarrow \left( y + \sqrt{1 + y^2} \right)\left( x + \sqrt{1 + x^2} \right) = C\]
\[\text{ Hence, }\log \left( y + \sqrt{1 + y^2} \right)\left( x + \sqrt{1 + x^2} \right) =\text{ C is the required differential equation .} \]