Question

(y² − 2xy) dx = (x² − 2xy) dy

Answer 1

We have, 
\[\left( y^2 - 2xy \right) dx = \left( x^2 - 2xy \right) dy\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{v^2 x^2 - 2v x^2}{x^2 - 2v x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{v^2 - 2v}{1 - 2v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{3 v^2 - 3v}{1 - 2v}\]
\[ \Rightarrow \frac{1 - 2v}{3 v^2 - 3v}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1 - 2v}{3 v^2 - 3v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - \int\frac{2v - 1}{3 v^2 - 3v}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - \frac{1}{3}\int\frac{6v - 3}{3 v^2 - 3v}dv = \int\frac{1}{x}dx\]
\[\text{ Putting }3 v^2 - 3v = t\]
\[ \Rightarrow \left( 6v - 3 \right) dv = dt\]
\[ \therefore - \frac{1}{3}\int\frac{1}{t}dt = \int\frac{1}{x}dx\]
\[ \Rightarrow - \frac{1}{3}\log \left| t \right| = \log \left| x \right| + \log C\]
Substituting the value of t, we get
\[ - \frac{1}{3}\log \left| 3 v^2 - 3v \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow - \frac{1}{3}\log \left| v^2 - v \right| - \frac{1}{3}\log3 = \log \left| x \right| + \log C\]
\[ \Rightarrow - \frac{1}{3}\log \left| v^2 - v \right| = \log \left| x \right| + \log C - \frac{1}{3}\log3\]
\[ \Rightarrow - \frac{1}{3}\log \left| v^2 - v \right| = \log \left| x \right| + \log C_1 ...........\left(\text{where, }\log C_1 = \log C - \frac{1}{3}\log3 \right)\]
Substituting the value of v, we get
\[ - \frac{1}{3}\log \left| \left( \frac{y}{x} \right)^2 - \left( \frac{y}{x} \right) \right| = \log \left| x \right| + \log C_1 \]
\[ \Rightarrow - \frac{1}{3}\log \left| \frac{y^2}{x^2} - \frac{y}{x} \right| = \log \left| C_1 x \right|\]
\[ \Rightarrow \log \left| \frac{y^2 - xy}{x^2} \right| = - 3\log \left| C_1 x \right|\]
\[ \Rightarrow \log \left| \frac{y^2 - xy}{x^2} \right| = \log \left| \frac{1}{{C_1}^3 x^3} \right|\]
\[ \Rightarrow \frac{y^2 - xy}{x^2} = \frac{1}{{C_1}^3 x^3}\]
\[ \Rightarrow x y^2 - x^2 y = \frac{1}{{C_1}^3}\]
\[ \Rightarrow x^2 y - x y^2 = - \frac{1}{{C_1}^3}\]
\[ \Rightarrow x^2 y - x y^2 = K ...........\left(\text{where, }\log K = - \frac{1}{{C_1}^3} \right)\]