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Question Number 197530 by Erico last updated on 20/Sep/23

$$\mathrm{f}\left(\mathrm{x}\right)-\frac{{\mathrm{x}}^2}{2}{\mathrm{f}}^{″}\left(\mathrm{x}\right)=0$$$$\mathrm{f}\left(\mathrm{x}\right)=?$$

Answered by witcher3 last updated on 20/Sep/23

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{a}}$$$$\Rightarrow {\mathrm{x}}^{\mathrm{a}}-\frac{1}{2}\mathrm{a}(\mathrm{a}-1){\mathrm{x}}^{\mathrm{a}}=0$$$$\iff (-\frac{{\mathrm{a}}^2}{2}+\frac{\mathrm{a}}{2}+1){\mathrm{x}}^{\mathrm{a}}=0$$$$\Rightarrow {\mathrm{a}}^2-\mathrm{a}-2=0,\mathrm{a}\in \{-1,2\}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+{\mathrm{bx}}^2$$$$\frac{1}{\mathrm{x}},{\mathrm{x}}^2,\mathrm{are}\mathrm{idependent}2\mathrm{dimension}\mathrm{space}\Rightarrow$$$$\mathrm{\forall }\mathrm{f}\in {\mathrm{C}}_2{\mathbb{R}}_{\ast }\mid \mathrm{f}-\frac{{\mathrm{x}}^2}{2}{\mathrm{f}}^{″}=0\mathrm{\exists }(\mathrm{a},\mathrm{b})\in {\mathbb{R}}^2\mathrm{such}\mathrm{f}=\frac{\mathrm{a}}{\mathrm{x}}+{\mathrm{bx}}^2$$$$$$$$$$

Answered by sniper237 last updated on 21/Sep/23

$$y+{xy}^{\prime }-({xy}^{\prime }+\frac{x^2}{2}{y}^{″})=0$$$$\frac{d}{dx}(xy-\frac{x^2}{2}{y}^{\prime })=0$$$$xy-\frac{x^2}{2}{y}^{\prime }=C$$$$\stackrel{homogensolution}{\Rightarrow }{y}_H={Kx}^2$$$$\stackrel{ctevariat\mathrm{\%}}{\Rightarrow }-\frac{x^4}{2}{K}^{\prime }\left(x\right)=C$$$$\Rightarrow K\left(x\right)=\frac{A}{x^3}+B$$$$\stackrel{gensolut\mathrm{\%}}{\Rightarrow }y=\frac{A}{x}+{Bx}^2$$$$$$$$$$

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