Question Number 93620 by i jagooll last updated on 14/May/20
$$\left(\mathrm{3x}+\mathrm{3y}−\mathrm{4}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}+\mathrm{y}\: \\ $$
Answered by john santu last updated on 14/May/20
$${r}\:=\:{x}+{y}\:\Rightarrow\:\frac{{dr}}{{dx}}\:=\:\mathrm{1}+\:\frac{{dy}}{{dx}} \\ $$$$\left(\mathrm{3}{r}−\mathrm{4}\right)\left(\frac{{dr}}{{dx}}−\mathrm{1}\right)\:=\:{r}\: \\ $$$$\frac{{dr}}{{dx}}\:=\:\frac{{r}}{\mathrm{3}{r}−\mathrm{4}}\:+\:\frac{\mathrm{3}{r}−\mathrm{4}}{\mathrm{3}{r}−\mathrm{4}}\:=\:\frac{\mathrm{4}{r}−\mathrm{4}}{\mathrm{3}{r}−\mathrm{4}} \\ $$$$\frac{\left(\mathrm{3}{r}−\mathrm{4}\right)}{{r}−\mathrm{1}}\:{dr}\:=\:\mathrm{4}\:{dx}\: \\ $$$$\int\frac{\mathrm{3}\left({r}−\mathrm{1}\right)−\mathrm{1}}{{r}−\mathrm{1}}\:{dr}\:=\:\mathrm{4}{x}\:+\:{c}\: \\ $$$$\mathrm{3}{r}\:−\:\mathrm{ln}\:\left({r}−\mathrm{1}\right)\:=\:\mathrm{4}{x}\:+\:{c}\: \\ $$$$\mathrm{3}{x}+\mathrm{3}{y}−\mathrm{ln}\left({x}+{y}−\mathrm{1}\right)\:=\:\mathrm{4}{x}+{c}\: \\ $$$$\mathrm{3}{y}−\mathrm{ln}\left({x}+{y}−\mathrm{1}\right)−{x}\:=\:{C}\: \\ $$