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Question Number 26837 by sorour87 last updated on 30/Dec/17

$${xy}^{\left(2\right)}=y^{\left(1\right)}\times \ln \frac{y^{\left(1\right)}}{x}$$

Answered by prakash jain last updated on 30/Dec/17

$$y^{\prime }=u\left(x\right)$$$$y^{″}=u^{\prime }\left(x\right)$$$${xu}^{\prime }=u\ln \frac{u}{x}$$$$v=\frac{u}{x}\Rightarrow u=vx$$$$u^{\prime }={xv}^{\prime }+v$$$$x({xv}^{\prime }+v)=xv\ln v$$$${xv}^{\prime }+v=v\ln v$$$$x\frac{dv}{dx}=v\ln v-v\Rightarrow \frac{dv}{v\ln v-v}=\frac{dx}{x}$$$$\mathrm{Equation}\mathrm{be}\mathrm{solved}\mathrm{in}v\mathrm{and}\mathrm{then}u$$$$\mathrm{and}y\mathrm{from}\mathrm{this}\mathrm{point}.$$

Commented by sorour87 last updated on 30/Dec/17

$$pleassolveittotheend.myproblemisinit.$$$$thanks$$

Commented by prakash jain last updated on 30/Dec/17

$$\frac{dv}{v\ln v-v}=\frac{dx}{x}$$$$\Rightarrow \frac{dv}{v(\ln v-1)}=\frac{dx}{x}$$$$\ln v-1=w\Rightarrow \frac{1}{v}dv=dw$$$$\ln (\ln v-1)=\ln x+\ln c=\ln xc$$$$forsimplicitywearewritngcaslnc$$$$\ln v-1=xc$$$$v=e^{xc+1}$$$$u=xv={xe}^{xc+1}$$$$\frac{dy}{dx}={xe}^{xc+1}={exe}^{xc}$$$$dy={exe}^{xc}dx$$$$y=ex\frac{e^{xc}}{c}-\int e\frac{e^{xc}}{c}dx$$$$y=ex\frac{e^{xc}}{c}-e\frac{e^{xc}}{c^2}+c_1$$$$y=\frac{{cxe}^{xc+1}-e^{xc+1}+c_1{c}^2}{c^2}$$$$cand{c}_1\mathrm{are}\mathrm{arbitrary}\mathrm{constant}.$$

Commented by sorour87 last updated on 30/Dec/17

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{v}\mathit{e}\mathit{r}\mathit{y}\mathit{m}\mathit{u}\mathit{c}\mathit{h}$$

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