Question Number 6661 by Tawakalitu. last updated on 09/Jul/16
$${Solve}\:{this}\:{equation}\:{by}\:{reducing}\:{it}\:{from}\:{non}\:{homogeneous} \\ $$$${equation}\:{to}\:{homogeneous}\:{equation} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}\left({x}\:+\:{y}\:−\:\mathrm{1}\right)}{\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{1}} \\ $$$$ \\ $$$${Please}\:{help}\:{with}\:{this}\:{one}\:{too}. \\ $$$${i}\:{was}\:{trying}\:{your}\:{approah}\:{but}\:{not}\:{the}\:{same} \\ $$$$ \\ $$$${thanks}\:{for}\:{your}\:{help}. \\ $$
Commented by prakash jain last updated on 11/Jul/16
$${u}={x}+{a} \\ $$$${v}={y}+{b} \\ $$$$\mathrm{2}{a}+\mathrm{2}{b}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}{a}+{b}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{6}{a}+\mathrm{2}{b}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{4}{a}+\mathrm{3}=\mathrm{0}\Rightarrow{a}=−\mathrm{3}/\mathrm{4} \\ $$$${b}=\mathrm{5}/\mathrm{4} \\ $$$${u}={x}−\mathrm{3}/\mathrm{4} \\ $$$${v}={y}+\mathrm{5}/\mathrm{4} \\ $$$$\frac{{dv}}{{du}}=\frac{\mathrm{2}{u}+\mathrm{2}{v}}{\mathrm{3}{u}+{v}} \\ $$$$\mathrm{Solve}\:\mathrm{by}\:\mathrm{substituting}\:{u}/{v}={t} \\ $$