Question Number 185526 by Matica last updated on 23/Jan/23
$$\:{solve}\:{y}''\:=\:\left(\mathrm{2}{y}+\mathrm{3}\right)\left({y}'\right)^{\mathrm{2}} \\ $$$$ \\ $$
Answered by mr W last updated on 23/Jan/23
$${y}''={y}'\frac{{dy}'}{{dy}} \\ $$$${y}'\frac{{dy}'}{{dy}}=\left(\mathrm{2}{y}+\mathrm{3}\right)\left({y}'\right)^{\mathrm{2}} \\ $$$$\frac{{dy}'}{{y}'}=\left(\mathrm{2}{y}+\mathrm{3}\right){dy} \\ $$$$\mathrm{ln}\:{y}'={y}^{\mathrm{2}} +\mathrm{3}{y}+{C}_{\mathrm{1}} \\ $$$${y}'={Ce}^{{y}^{\mathrm{2}} +\mathrm{3}{y}} \\ $$$$\frac{{dy}}{{dx}}={Ce}^{{y}^{\mathrm{2}} +\mathrm{3}{y}} \\ $$$$\frac{{dy}}{{e}^{{y}^{\mathrm{2}} +\mathrm{3}{y}} }={Cdx} \\ $$$$\int\frac{{dy}}{{e}^{{y}^{\mathrm{2}} +\mathrm{3}{y}} }={Cx}+{D} \\ $$$$\Rightarrow\frac{{e}^{\frac{\mathrm{9}}{\mathrm{4}}} \sqrt{\pi}}{\mathrm{2}}{erf}\left({y}+\frac{\mathrm{3}}{\mathrm{2}}\right)={Cx}+{D} \\ $$