Question Number 196843 by sonukgindia last updated on 01/Sep/23
Answered by qaz last updated on 02/Sep/23
$${xyy}''={yy}'+{x}−{x}\left({y}'\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}\left({yy}'\right)'={yy}'+{x} \\ $$$${yy}'={e}^{\int\frac{{dx}}{{x}}} \left({C}_{\mathrm{1}} +\int{e}^{−\int\frac{{dx}}{{x}}} {dx}\right)={C}_{\mathrm{1}} {x}+{xlnx} \\ $$$$\Rightarrow{y}^{\mathrm{2}} ={C}_{\mathrm{1}} {x}^{\mathrm{2}} +{x}^{\mathrm{2}} {lnx}+{C}_{\mathrm{2}} \:\:\:\:,{C}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:,{C}_{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} {lnx}+\frac{\mathrm{7}}{\mathrm{2}} \\ $$