Question Number 91643 by jagoll last updated on 02/May/20
$$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}}{{x}}\: \\ $$$$ \\ $$
Commented by john santu last updated on 02/May/20
Commented by Prithwish Sen 1 last updated on 02/May/20
$$\mathrm{ydx}−\mathrm{xdy}\:=\:\mathrm{xdy} \\ $$$$−\left(\mathrm{xdy}−\mathrm{ydx}\right)=\mathrm{xdy} \\ $$$$−\frac{\mathrm{xdy}−\mathrm{ydx}}{\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{dy}}{\mathrm{x}} \\ $$$$−\int\mathrm{d}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\int\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$−\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{lnCx}\:\:\:\:\:\:\mathrm{C}=\:\mathrm{constant} \\ $$$$\boldsymbol{\mathrm{Cx}}\:=\:\boldsymbol{\mathrm{e}}^{\frac{−\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}} \\ $$
Commented by jagoll last updated on 02/May/20
$${thank}\:{you}\:{both} \\ $$
Commented by mathmax by abdo last updated on 02/May/20
$$\Rightarrow\frac{{dy}}{{dx}}\:=\frac{{y}}{{x}}−\mathrm{1}\:\:{let}\:\frac{{y}}{{x}}\:={z}\:\Rightarrow{y}\:={xz}\:\Rightarrow\frac{{dy}}{{dx}}\:={z}\:+{x}\:\frac{{dz}}{{dx}}\:\Rightarrow \\ $$$${z}\:+{x}\:{z}^{'} \:={z}−\mathrm{1}\:\Rightarrow{xz}^{'} \:=−\mathrm{1}\:\Rightarrow{z}^{'} =−\frac{\mathrm{1}}{{x}}\:\Rightarrow{z}\:=−{lnx}\:+{k}\:\Rightarrow \\ $$$${y}\:={x}\left(−{lnx}\:+{k}\right)\:={kx}\:−{xlnx} \\ $$$$ \\ $$