Question Number 122006 by bemath last updated on 13/Nov/20
$$\:\frac{{dy}}{{dx}}\:=\:\frac{{x}−{y}}{{x}+{y}}\:? \\ $$
Answered by bobhans last updated on 13/Nov/20
$${let}\:{y}\:=\:{zx}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$$\Leftrightarrow\:{z}+{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{x}−{zx}}{{x}+{zx}}\:=\:\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}} \\ $$$$\Leftrightarrow\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}}\:−\:{z}\:=\:\frac{\mathrm{1}−\mathrm{2}{z}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}} \\ $$$$\:\Leftrightarrow\:\frac{\left(\mathrm{1}+{z}\right)\:{dz}}{{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}}\:=\:−\frac{{dx}}{{x}} \\ $$$$\Leftrightarrow\:\int\:\frac{{d}\left({z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}}\:=\:−\mathrm{2}\int\:\frac{{dx}}{{x}} \\ $$$$\Leftrightarrow\:\ell{n}\:\mid{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}\mid\:=\:\ell{n}\mid\frac{{C}}{{x}^{\mathrm{2}} }\:\mid \\ $$$$\Leftrightarrow\:\left({z}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{{C}+\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:\left(\frac{{y}+{x}}{{x}}\right)^{\mathrm{2}} =\:\frac{{C}+\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:{y}+{x}\:=\:\pm\:\sqrt{{C}+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:{y}\:=\:−{x}\:\pm\:\sqrt{{C}+\mathrm{2}{x}^{\mathrm{2}} }\:.\:\circleddash \\ $$
Answered by TANMAY PANACEA last updated on 13/Nov/20
$${xdx}−{ydx}={xdy}+{ydy} \\ $$$${xdy}+{ydx}+{ydy}−{xdx}=\mathrm{0}={dC} \\ $$$${d}\left({xy}\right)+{d}\left(\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right)−{d}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)={dC} \\ $$$${xy}+\frac{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}}={C} \\ $$