Question Number 126460 by BHOOPENDRA last updated on 20/Dec/20
$${yy}''−\left({y}'\right)^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by BHOOPENDRA last updated on 20/Dec/20
$${thanks}\:{alot}\:{sir} \\ $$
Answered by mr W last updated on 20/Dec/20
$${p}={y}' \\ $$$${y}''=\frac{{dp}}{{dx}}={p}\frac{{dp}}{{dy}} \\ $$$${yp}\frac{{dp}}{{dy}}−{p}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{0}\:\Rightarrow{y}={C} \\ $$$$\Rightarrow{y}\frac{{dp}}{{dy}}={p}\:\Rightarrow\frac{{dp}}{{p}}=\frac{{dy}}{{y}}\Rightarrow\mathrm{ln}\:{p}=\mathrm{ln}\:{Cy}\: \\ $$$$\Rightarrow{p}={Cy}\:\Rightarrow\frac{{dy}}{{dx}}={Cy}\Rightarrow\frac{{dy}}{{y}}={Cdx} \\ $$$$\Rightarrow\mathrm{ln}\:{y}={Cx}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} {e}^{{Cx}} \\ $$
Commented by Coronavirus last updated on 20/Dec/20
$${mehode}\:{interessante} \\ $$