Question Number 85050 by john santu last updated on 18/Mar/20
$$\left(\mathrm{x}−\mathrm{4y}+\mathrm{3}\right)\mathrm{dx}\:=\:\left(\mathrm{x}−\mathrm{5y}+\mathrm{4}\right)\mathrm{dy} \\ $$
Answered by mr W last updated on 18/Mar/20
$${let} \\ $$$${x}={u}+{a} \\ $$$${y}={v}+{b} \\ $$$${x}−\mathrm{4}{y}+\mathrm{3}={u}−\mathrm{4}{v}+\left({a}−\mathrm{4}{b}+\mathrm{3}\right) \\ $$$${x}−\mathrm{5}{y}+\mathrm{4}={u}−\mathrm{5}{v}+\left({a}−\mathrm{5}{b}+\mathrm{4}\right) \\ $$$${set} \\ $$$${a}−\mathrm{4}{b}+\mathrm{3}=\mathrm{0} \\ $$$${a}−\mathrm{5}{b}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{1},\:{b}=\mathrm{1} \\ $$$$\Rightarrow{x}={u}+\mathrm{1}\:\Rightarrow{dx}={du} \\ $$$$\Rightarrow{y}={v}+\mathrm{1}\:\Rightarrow{dy}={dv} \\ $$$$ \\ $$$$\left({u}−\mathrm{4}{v}\right){du}=\left({u}−\mathrm{5}{v}\right){dv} \\ $$$$\Rightarrow\frac{{dv}}{{du}}=\frac{{u}−\mathrm{4}{v}}{{u}−\mathrm{5}{v}} \\ $$$${let}\:{v}={ut}\:\Rightarrow{t}=\frac{{v}}{{u}} \\ $$$$\frac{{dv}}{{du}}={t}+{u}\frac{{dt}}{{du}} \\ $$$$\Rightarrow{t}+{u}\frac{{dt}}{{du}}=\frac{\mathrm{1}−\mathrm{4}{t}}{\mathrm{1}−\mathrm{5}{t}} \\ $$$$\Rightarrow{u}\frac{{dt}}{{du}}=\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}{\mathrm{1}−\mathrm{5}{t}} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−\mathrm{5}{t}\right){dt}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}=\frac{{du}}{{u}} \\ $$$$\Rightarrow\int\frac{\left(\mathrm{1}−\mathrm{5}{t}\right){dt}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}=\int\frac{{du}}{{u}} \\ $$$$\Rightarrow\int\left(\frac{\mathrm{1}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}−\frac{\mathrm{5}{t}}{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}}\right){dt}=\mathrm{ln}\:\left({cu}\right) \\ $$$$…… \\ $$