Question Number 197530 by Erico last updated on 20/Sep/23
$$\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{f}''\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$
Commented by AST last updated on 20/Sep/23
$${f}\left({x}\right)={x}^{\mathrm{2}} \:{works}. \\ $$
Answered by witcher3 last updated on 20/Sep/23
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{a}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{a}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}\left(\mathrm{a}−\mathrm{1}\right)\mathrm{x}^{\mathrm{a}} =\mathrm{0} \\ $$$$\Leftrightarrow\left(−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{a}}{\mathrm{2}}+\mathrm{1}\right)\mathrm{x}^{\mathrm{a}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{2}=\mathrm{0},\mathrm{a}\in\left\{−\mathrm{1},\mathrm{2}\right\} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\mathrm{bx}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{x}},\mathrm{x}^{\mathrm{2}} ,\mathrm{are}\:\mathrm{idependent}\:\mathrm{2}\:\mathrm{dimension}\:\mathrm{space}\Rightarrow \\ $$$$\forall\mathrm{f}\in\mathrm{C}_{\mathrm{2}} \mathbb{R}_{\ast} \mid\mathrm{f}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{f}''=\mathrm{0}\:\exists\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \:\mathrm{such}\:\mathrm{f}=\frac{\mathrm{a}}{\mathrm{x}}+\mathrm{bx}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Answered by sniper237 last updated on 21/Sep/23
$${y}+{xy}'−\left({xy}'+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{y}''\right)=\mathrm{0} \\ $$$$\frac{{d}}{{dx}}\left({xy}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{y}'\right)=\mathrm{0} \\ $$$${xy}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{y}'={C} \\ $$$$\overset{{homogen}\:{solution}} {\Rightarrow}\:\:{y}_{{H}} ={Kx}^{\mathrm{2}} \\ $$$$\overset{{cte}\:{variat\%}} {\Rightarrow}\:−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}{K}'\left({x}\right)={C}\: \\ $$$$\Rightarrow{K}\left({x}\right)=\frac{{A}}{{x}^{\mathrm{3}} }\:+{B} \\ $$$$\overset{{gen}\:{solut\%}} {\Rightarrow}\:{y}=\frac{{A}}{{x}}\:+{Bx}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$