Question Number 227155 by zahraa last updated on 03/Jan/26
Answered by Ghisom_ last updated on 03/Jan/26
![x=2y′+ln y′ e^x =e^(2y′) y′ 2e^x =e^(2y′) 2y′ t=2y′ e^t t=2e^x t=W (2e^x ) y′=((W (2e^x ))/2) y=∫((W (2e^x ))/2)dx= [u=2e^x ] =(1/2)∫((W (u))/u)du= we know (d/dz)[W (z)]=((W (z))/(z(1+W(z)))) ⇒ (d/dz)[W^2 (z)]=2W (z) (d/dz)[W (z)] ⇒ (d/dz)[((W^2 (z))/2)+W (z)]=(1+W (z))(d/dz)[W (z)]= =((W (z))/z) =(((2+W (u))W (u))/4) ⇒ y=(((2+W (2e^x ))W(2e^x ))/4)+C](https://www.tinkutara.com/question/Q227160.png)
$${x}=\mathrm{2}{y}'+\mathrm{ln}\:{y}' \\ $$$$\mathrm{e}^{{x}} =\mathrm{e}^{\mathrm{2}{y}'} {y}' \\ $$$$\mathrm{2e}^{{x}} =\mathrm{e}^{\mathrm{2}{y}'} \mathrm{2}{y}' \\ $$$${t}=\mathrm{2}{y}' \\ $$$$\mathrm{e}^{{t}} {t}=\mathrm{2e}^{{x}} \\ $$$${t}=\mathrm{W}\:\left(\mathrm{2e}^{{x}} \right) \\ $$$${y}'=\frac{\mathrm{W}\:\left(\mathrm{2e}^{{x}} \right)}{\mathrm{2}} \\ $$$${y}=\int\frac{\mathrm{W}\:\left(\mathrm{2e}^{{x}} \right)}{\mathrm{2}}{dx}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{2e}^{{x}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{W}\:\left({u}\right)}{{u}}{du}= \\ $$$$\:\:\:\:\:\mathrm{we}\:\mathrm{know}\:\frac{{d}}{{dz}}\left[\mathrm{W}\:\left({z}\right)\right]=\frac{\mathrm{W}\:\left({z}\right)}{{z}\left(\mathrm{1}+\mathrm{W}\left({z}\right)\right)} \\ $$$$\:\:\:\:\:\Rightarrow\:\frac{{d}}{{dz}}\left[\mathrm{W}^{\mathrm{2}} \:\left({z}\right)\right]=\mathrm{2W}\:\left({z}\right)\:\frac{{d}}{{dz}}\left[\mathrm{W}\:\left({z}\right)\right] \\ $$$$\:\:\:\:\:\Rightarrow\:\frac{{d}}{{dz}}\left[\frac{\mathrm{W}^{\mathrm{2}} \:\left({z}\right)}{\mathrm{2}}+\mathrm{W}\:\left({z}\right)\right]=\left(\mathrm{1}+\mathrm{W}\:\left({z}\right)\right)\frac{{d}}{{dz}}\left[\mathrm{W}\:\left({z}\right)\right]= \\ $$$$\:\:\:\:\:=\frac{\mathrm{W}\:\left({z}\right)}{{z}} \\ $$$$=\frac{\left(\mathrm{2}+\mathrm{W}\:\left({u}\right)\right)\mathrm{W}\:\left({u}\right)}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$${y}=\frac{\left(\mathrm{2}+\mathrm{W}\:\left(\mathrm{2e}^{{x}} \right)\right)\mathrm{W}\left(\mathrm{2e}^{{x}} \right)}{\mathrm{4}}+{C} \\ $$