Question

\[\left[ x\sqrt{x^2 + y^2} - y^2 \right] dx + xy\ dy = 0\]

Answer 1

We have, 
\[\left[ x\sqrt{x^2 + y^2} - y^2 \right]dx + xy\ dy = 0\]
\[\frac{dy}{dx} = \frac{y^2 - x\sqrt{x^2 + y^2}}{xy}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{v^2 x^2 - x\sqrt{x^2 + v^2 x^2}}{v x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v^2 - \sqrt{1 + v^2}}{v}\]
\[ \Rightarrow v + x\frac{dv}{dx} = v - \frac{\sqrt{1 + v^2}}{v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{- \sqrt{1 + v^2}}{v}\]
\[ \Rightarrow \frac{v}{\sqrt{1 + v^2}}dv = - \frac{1}{x}dx\]
\[\text{ Putting }1 + v^2 = t,\text{ we get }\]
\[v\ dv = \frac{dt}{2}\]
\[ \therefore \frac{1}{2\sqrt{t}}dt = - \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int \frac{1}{2\sqrt{t}}dt = - \int\frac{1}{x}dx\]
\[ \Rightarrow \sqrt{t} = - \log \left| x \right| + \log C . . . . . (1)\]
Substituting the value of `t` in (1), we get
\[\sqrt{1 + v^2} = \log \left| \frac{C}{x} \right|\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \sqrt{y^2 + x^2} = x \log \left| \frac{C}{x} \right|\]
\[\text{ Hence, }\sqrt{y^2 + x^2} = x \log \left| \frac{C}{x} \right| \text{ is the required solution.}\]