Question

(y^2-2xy)dx-(2xy-x^2)dy=0

Answer 1

$({y^2} - 2xy)dx - (2xy - {x^2})dy = 0$

$- \frac{{dy}}{{dx}}( - {x^2} + 2xy) + {y^2} - 2xy = 0$

Let $y(x) = xv(x)$, which gives $\frac{{dy}}{{dx}} = x\frac{{dv}}{{dx}} + v$ :

$- ( - {x^2} + 2{x^2}v)\left( {x\frac{{dv}}{{dx}} + v} \right) + {x^2}{v^2} - 2{x^2}v = 0$

Simplify:

$- {x^2}\left( { - x\frac{{dv}}{{dx}} + {v^2} + v + 2x\frac{{dv}}{{dx}}v} \right) = 0$

Solve for ${\frac{{dv}}{{dx}}}$:

$\frac{{dv}}{{dx}} = \frac{{ - {v^2} - v}}{{x(2v - 1)}}$

Divide both sides by $\frac{{ - {v^2} - v}}{{2v - 1}}:$

$\frac{{\frac{{dv}}{{dx}}(2v - 1)}}{{ - {v^2} - v}} = \frac{1}{x}$

Integrate both sides with respect to x:

$\int {\frac{{\frac{{dv}}{{dx}}(2v - 1)}}{{ - {v^2} - v}}} dx = \int {\frac{1}{x}} dx$

Evaluate the integrals:

$- 3\ln (v + 1) + \ln v = \ln x + C$

Substitute back:

$- 3\ln (\frac{y}{x} + 1) + \ln \frac{y}{x} = \ln x + C$