Question Number 159982 by ArielVyny last updated on 23/Nov/21
$$\left(\mathrm{1}+{bf}\left({x}\right)\right){f}''\left({x}\right)=\frac{{p}}{\lambda{a}} \\ $$$${solve}\:{this}\:{equation}:\:{find}\:\:{f}\left({x}\right) \\ $$
Answered by mr W last updated on 23/Nov/21
$${let}\:{y}'={u} \\ $$$${y}''=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$$\left(\mathrm{1}+{by}\right){u}\frac{{du}}{{dy}}=\frac{{p}}{\lambda{a}} \\ $$$${udu}=\frac{{pdy}}{\lambda{a}\left(\mathrm{1}+{by}\right)} \\ $$$$\int{udu}=\frac{{p}}{\lambda{a}}\int\frac{{dy}}{\mathrm{1}+{by}} \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{2}}=\frac{{p}}{\lambda{ab}}\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right) \\ $$$${u}=\pm\sqrt{\frac{\mathrm{2}{p}}{\lambda{ab}}}\:\sqrt{\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right)} \\ $$$$\frac{{dy}}{{dx}}=\pm\sqrt{\frac{\mathrm{2}{p}}{\lambda{ab}}}\:\sqrt{\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right)} \\ $$$$\frac{{dy}}{\:\sqrt{\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right)}}=\pm\sqrt{\frac{\mathrm{2}{p}}{\lambda{ab}}}\:{dx} \\ $$$$\int\frac{{dy}}{\:\sqrt{\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right)}}=\pm\sqrt{\frac{\mathrm{2}{p}}{\lambda{ab}}}\:{x}+{D} \\ $$$$\int\frac{{d}\left(\mathrm{1}+{by}\right)}{\:\sqrt{\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right)}}=\pm\sqrt{\frac{\mathrm{2}{pb}}{\lambda{a}}}\:{x}+{D} \\ $$$$\frac{\sqrt{\pi}\:{erfi}\left(\sqrt{\mathrm{ln}\:{C}\left(\mathrm{1}+{by}\right)}\right)}{{C}}=\pm\sqrt{\frac{\mathrm{2}{pb}}{\lambda{a}}}\:{x}+{D} \\ $$