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Question Number 217046 by OmoloyeMichael last updated on 27/Feb/25
$${form}\:{the}\:{differential}\:{equation}\:{by}\: \\ $$$${eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$${y}^{\mathrm{2}} ={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$
Answered by mr W last updated on 27/Feb/25
$${y}^{\mathrm{2}} ={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$$$\mathrm{2}{yy}'=\mathrm{2}{Ax}+{B} \\ $$$$\mathrm{2}{yy}''+\mathrm{2}\left({y}'\right)^{\mathrm{2}} =\mathrm{2}{A} \\ $$$${yy}''+\left({y}'\right)^{\mathrm{2}} ={A} \\ $$$${yy}'''+{y}'{y}''+\mathrm{2}{y}'{y}''=\mathrm{0} \\ $$$$\Rightarrow{yy}'''+\mathrm{3}{y}'{y}''=\mathrm{0} \\ $$
Commented by OmoloyeMichael last updated on 01/Mar/25
$${thanks}\:{sir} \\ $$