Question Number 7216 by peter james last updated on 16/Aug/16
Answered by Yozzia last updated on 16/Aug/16
$${ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\frac{\mathrm{3}−{y}}{{x}+\mathrm{3}}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${Let}\:{u}=\frac{{x}+{y}}{{x}+\mathrm{3}}\Rightarrow{y}={u}\left({x}+\mathrm{3}\right)−{x} \\ $$$$\therefore\:{y}'={u}'\left({x}+\mathrm{3}\right)+{u}−\mathrm{1} \\ $$$$\Rightarrow{y}'+\mathrm{1}={u}+{u}'\left({x}+\mathrm{3}\right). \\ $$$$\therefore\:\left({u}+{u}'\left({x}+\mathrm{3}\right)\right){lnu}={u} \\ $$$${u}+{u}'\left({x}+\mathrm{3}\right)=\frac{{u}}{{lnu}} \\ $$$${u}'\left({x}+\mathrm{3}\right)=\frac{{u}}{{lnu}}−{u}=\frac{{u}\left(\mathrm{1}−{lnu}\right)}{{lnu}} \\ $$$${or}\:\frac{{du}}{{dx}}\left({x}+\mathrm{3}\right)=\frac{{u}\left(\mathrm{1}−{lnu}\right)}{{lnu}} \\ $$$$\Rightarrow\int\frac{{lnu}}{{u}\left(\mathrm{1}−{lnu}\right)}{du}=\int\frac{{dx}}{{x}+\mathrm{3}} \\ $$$${Let}\:{k}={lnu}\Rightarrow{dk}=\frac{\mathrm{1}}{{u}}{du} \\ $$$$\therefore\int\frac{{k}\:{dk}}{\mathrm{1}−{k}}={ln}\mid{x}+\mathrm{3}\mid+{D} \\ $$$$\int\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{k}}\right){dk}={ln}\mid{x}+\mathrm{3}\mid+{D} \\ $$$$−{k}−{ln}\mid\mathrm{1}−{k}\mid={ln}\mid{x}+\mathrm{3}\mid+{D} \\ $$$$\Rightarrow{k}+{ln}\mid\mathrm{1}−{k}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${lne}^{{k}} +{ln}\mid\mathrm{1}−{k}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\because\:{ln}\mid\mathrm{1}−{k}\mid={ln}\mid{k}−\mathrm{1}\mid \\ $$$$\Rightarrow{ln}\mid{e}^{{k}} \left({k}−\mathrm{1}\right)\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\Rightarrow{ee}^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)={e}^{{B}−{ln}\mid{x}+\mathrm{3}\mid} \\ $$$$\left({k}−\mathrm{1}\right){e}^{{k}−\mathrm{1}} ={e}^{−\mathrm{1}} ×{A}×\frac{\mathrm{1}}{{x}+\mathrm{3}}=\frac{{F}}{{x}+\mathrm{3}} \\ $$$$\therefore\:{W}\left\{\left({k}−\mathrm{1}\right){e}^{{k}−\mathrm{1}} \right\}={W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$${k}−\mathrm{1}={W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$${k}=\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$$\therefore\:{lnu}=\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right) \\ $$$${u}={exp}\left(\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right)\right) \\ $$$$\therefore\frac{{y}+{x}}{{x}+\mathrm{3}}={exp}\left(\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right)\right) \\ $$$$\Rightarrow{y}=\left({x}+\mathrm{3}\right){exp}\left(\mathrm{1}+{W}\left(\frac{{F}}{{x}+\mathrm{3}}\right)\right)−{x} \\ $$$${W}={Lambert}\:{W}\:{function} \\ $$$${F}={constant} \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$${Alternative}\:{to}\:{the}\:{result}\:{that}\:{uses} \\ $$$${the}\:{W}−{function},\:{an}\:{implicit}\:{equation} \\ $$$${in}\:{x}\:{and}\:{y}\:{can}\:{be}\:{obtained}. \\ $$$${k}+{ln}\mid\mathrm{1}−{k}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\therefore{lnu}+{ln}\mid\mathrm{1}−{lnu}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$$\therefore\:{ln}\frac{{x}+{y}}{{x}+\mathrm{3}}+{ln}\mid\mathrm{1}−{ln}\frac{{x}+{y}}{{x}+\mathrm{3}}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\mathrm{1}+{ln}\left({x}+\mathrm{3}\right)−{ln}\left({x}+{y}\right)\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${For}\:{ln}\left({x}+\mathrm{3}\right)\in\mathbb{R},\:{x}+\mathrm{3}>\mathrm{0}\Rightarrow\mid{x}+\mathrm{3}\mid={x}+\mathrm{3} \\ $$$$\therefore\:{ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\mathrm{1}+{ln}\left({x}+\mathrm{3}\right)−{ln}\left({x}+{y}\right)\mid={B}−{ln}\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$${ln}\left({x}+{y}\right)+{ln}\mid\mathrm{1}+{ln}\left({x}+\mathrm{3}\right)−{ln}\left({x}+{y}\right)\mid+{C}=\mathrm{0} \\ $$$${C}={constant} \\ $$
Commented by peter james last updated on 17/Aug/16
$$\:\boldsymbol{{T}}{hank}\:{you}\:{so}\:{much}\:{sir}\:{for}\:{the}\:{help}. \\ $$