Classsify the following differential eq.

\(\displaystyle{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}+{\left({2}{y}-{3}{x}\right)}{\left.{d}{y}\right.}={0}\)

Solution:

\(\displaystyle{\left.{d}{y}\right.}{\left({2}{y}-{3}{x}\right)}=-{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}\)

Finding homogeneous

\(\displaystyle{F}{\left(\lambda{x},\lambda{y}\right)}={\frac{{-{\left({2}\lambda{x}-{3}\lambda{y}\right)}}}{{{\left({2}\lambda{y}-{3}\lambda{x}\right)}}}}\)

\(\displaystyle={\frac{{-\lambda{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}\)

\(\displaystyle={\frac{{-{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{x}-{3}{x}\right)}}}}\)

\(\displaystyle=\lambda^{{\circ}}{F}{\left({r},{y}\right)}\)

The given equation is homogenous.

\(\displaystyle{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}+{\left({2}{y}-{3}{x}\right)}{\left.{d}{y}\right.}={0}\)

Solution:

\(\displaystyle{\left.{d}{y}\right.}{\left({2}{y}-{3}{x}\right)}=-{\left({2}{x}-{3}{y}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}\)

Finding homogeneous

\(\displaystyle{F}{\left(\lambda{x},\lambda{y}\right)}={\frac{{-{\left({2}\lambda{x}-{3}\lambda{y}\right)}}}{{{\left({2}\lambda{y}-{3}\lambda{x}\right)}}}}\)

\(\displaystyle={\frac{{-\lambda{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{y}-{3}{x}\right)}}}}\)

\(\displaystyle={\frac{{-{\left({2}{x}-{3}{y}\right)}}}{{{\left({2}{x}-{3}{x}\right)}}}}\)

\(\displaystyle=\lambda^{{\circ}}{F}{\left({r},{y}\right)}\)

The given equation is homogenous.