Concept:

Homogenous equation are of the form \(\frac{{dy}}{{dx}} = \frac{{f\left( {x,y} \right)}}{{\emptyset \left( {x,y} \right)}}\)

Where f(x, y) and ∅(x, y) Homogenous functions of the same degree in x and y.

To solve a homogenous equation

1. Put y = vx, then \(\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\)
2. Separate the variable v and x and integrate.
3. Put \(v = \frac{y}{x}\)

Calculation:

\(3{x^2}dy + \left( {{y^2} - 2xy} \right)dx = 0\)

\(\frac{{dy}}{{dx}} = \frac{{2xy - {y^2}}}{{3{x^2}}}\)

Put y = vx, then \(\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\)

\(v + x\frac{{dv}}{{dx}} = \frac{{2v{x^2} - {v^2}{x^2}}}{{3{x^2}}}\)

\(x\frac{{dv}}{{dx}} = \frac{{ - v - {v^2}}}{3}\)

\(\frac{{dv}}{{{v^2} + v}} = \frac{{ - dx}}{{3x}}\)

\(\left( {\frac{1}{v} - \frac{1}{{v + 1}}} \right)dv = \frac{{ - dx}}{{3x}}\)

By integrating both sides we get

\(\smallint \left( {\frac{1}{v} - \frac{1}{{v + 1}}} \right)dv = \smallint \frac{{ - dx}}{{3x}}\)

\(\ln v - \ln \left( {v + 1} \right) = \frac{{ - 1}}{3}\ln x + \ln c\)

\(\ln \frac{v}{{v + 1}} = \ln \left( {c{x^{\frac{{ - 1}}{3}}}} \right)\)

\(\frac{v}{{v + 1}} = c{x^{\frac{{ - 1}}{3}}}\)

Put v = y/x in the above equation we get

\(\frac{{\frac{y}{x}}}{{\frac{y}{x} + 1}} = c{x^{\frac{{ - 1}}{3}}}\)