Question Number 196734 by sonukgindia last updated on 30/Aug/23
Answered by qaz last updated on 31/Aug/23
$${y}''{y}+\left({y}'\right)^{\mathrm{2}} =\left({y}'{y}\right)'=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({y}^{\mathrm{2}} \right)'\right)'=\mathrm{3}{x} \\ $$$$\Rightarrow\left({y}^{\mathrm{2}} \right)'=\mathrm{3}{x}^{\mathrm{2}} +{C}_{\mathrm{1}} \:\:\:\:{y}^{\mathrm{2}} ={x}^{\mathrm{3}} +{C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$${C}_{\mathrm{1}} =−\mathrm{6}\:\:\:\:\:\:\:{C}_{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow{y}^{\mathrm{2}} ={x}^{\mathrm{3}} −\mathrm{6}{x}+\mathrm{3} \\ $$