Question Number 181615 by Mastermind last updated on 27/Nov/22
$$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{this}\:\mathrm{is}\:\mathrm{Homogenous} \\ $$$$\mathrm{or}\:\mathrm{not} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{y}−\mathrm{2x}} \\ $$$$ \\ $$$$. \\ $$
Answered by hmr last updated on 27/Nov/22
$$\frac{{dy}}{{dx}}\:=\:\frac{\left(\frac{{y}}{{x}}\right)}{\left(\frac{{y}}{{x}}\right)−\mathrm{2}} \\ $$$$ \\ $$$$\bullet\:{z}\:=\:\left(\frac{{y}}{{x}}\right)\:\rightarrow\:\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$$ \\ $$$${z}\:+\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{z}}{{z}\:−\:\mathrm{2}} \\ $$$${x}\:\frac{{dz}}{{dx}}\:=\:\frac{{z}}{{z}\:−\:\mathrm{2}}\:−\:{z}\: \\ $$$${x}\:\frac{{dz}}{{dx}}\:=\:\:\frac{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }{{z}\:−\:\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{{x}}\:+\:\frac{{z}\:−\:\mathrm{2}}{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }\:\frac{{dz}}{{dx}}\:=\:\mathrm{0} \\ $$$$\int−\frac{\mathrm{1}}{{x}}{dx}\:+\:\int\frac{{z}\:−\:\mathrm{2}}{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }\:{dz}\:=\:{c} \\ $$$$ \\ $$$$\bullet\:\frac{{z}\:−\:\mathrm{2}}{\mathrm{3}{z}\:−\:{z}^{\mathrm{2}} }\:=\:\frac{{a}}{\mathrm{3}−{z}}\:+\:\frac{{b}}{{z}} \\ $$$${z}\:−\:\mathrm{2}\:=\:{az}\:+\:{b}\left(\mathrm{3}\:−\:{z}\right) \\ $$$${z}\:−\:\mathrm{2}\:=\:\left({a}\:−\:{b}\right){z}\:+\:\mathrm{3}{b} \\ $$$$\begin{cases}{{a}\:−\:{b}\:=\:\mathrm{1}}\\{\mathrm{3}{b}\:=\:−\mathrm{2}\:\rightarrow\:{b}\:=\:\frac{−\mathrm{2}}{\mathrm{3}}\:\rightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$ \\ $$$$−{ln}\mid{x}\mid\:+\:\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{3}\:−\:{z}\mid\:−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{z}\mid=\:{c} \\ $$$$−{ln}\mid{x}\mid\:+\:\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\mathrm{3}\:−\:\frac{{y}}{{x}}\mid\:−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\frac{{y}}{{x}}\mid=\:{c} \\ $$