Question

(x² − y²) dx − 2xy dy = 0

Answer 1

We have,
\[\left( x^2 - y^2 \right) dx - 2xy dy = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2 - y^2}{2xy}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x^2 - \left( vx \right)^2}{2x\left( vx \right)}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{x^2 - v^2 x^2}{2v x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 - v^2}{2v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - v^2}{2v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - 3 v^2}{2v}\]
\[ \Rightarrow \frac{2v}{1 - 3 v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{2v}{1 - 3 v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - \frac{1}{3}\int\frac{- 6v}{1 - 3 v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - \frac{1}{3}\log \left| 1 - 3 v^2 \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow \log \left| 1 - 3 v^2 \right| = - 3\log \left| Cx \right|\]
\[ \Rightarrow \log \left| 1 - 3 v^2 \right| = \log \left| \frac{1}{\left( Cx \right)^3} \right|\]
\[ \Rightarrow 1 - 3 v^2 = \frac{1}{\left( Cx \right)^3}\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[1 - 3 \left( \frac{y}{x} \right)^2 = \frac{1}{\left( Cx \right)^3}\]
\[ \Rightarrow \frac{x^2 - 3 y^2}{x^2} = \frac{1}{C^3 x^3}\]
\[ \Rightarrow x\left( x^2 - 3 y^2 \right) = \frac{1}{C^3}\]
\[ \Rightarrow x\left( x^2 - 3 y^2 \right) = K ...........\left(\text{where, }K = \frac{1}{C^3} \right)\]