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Question Number 98367 by bobhans last updated on 13/Jun/20

$$\mathrm{show}\mathrm{that}\phi \left(\mathrm{x}\right)={\mathrm{x}}^2-{\mathrm{x}}^{-1}\mathrm{is}\mathrm{an}\mathrm{explicit}$$$$\mathrm{solution}\mathrm{to}\mathrm{linear}\mathrm{equation}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{2\mathrm{y}}{{\mathrm{x}}^2}=0$$

Commented by john santu last updated on 13/Jun/20

$$\mathrm{the}\mathrm{function}\phi \left(\mathrm{x}\right)={\mathrm{x}}^2-{\mathrm{x}}^{-1},{\phi }^{{}^{\prime }}\left(\mathrm{x}\right)=2\mathrm{x}+{\mathrm{x}}^{-2}\mathrm{and}$$$${\phi }^{″}\left(\mathrm{x}\right)=2-{2\mathrm{x}}^{-3}\mathrm{are}\mathrm{defined}\mathrm{for}\mathrm{all}\mathrm{x}\ne 0$$$$\mathrm{substitution}\mathrm{of}\phi \left(\mathrm{x}\right)\mathrm{for}\mathrm{y}\mathrm{in}\mathrm{equation}$$$$\mathrm{gives}(2-{2\mathrm{x}}^{-3})-\frac{2}{{\mathrm{x}}^2}({\mathrm{x}}^2-{\mathrm{x}}^{-1})=$$$$(2-{2\mathrm{x}}^{-3})-(2-{2\mathrm{x}}^{-3})=0.\mathrm{since}\mathrm{this}$$$$\mathrm{is}\mathrm{valid}\mathrm{for}\mathrm{any}\mathrm{x}\ne 0,\mathrm{the}\mathrm{function}$$$$\phi \left(\mathrm{x}\right)={\mathrm{x}}^2-{\mathrm{x}}^{-1}\mathrm{is}\mathrm{an}\mathrm{explicit}\mathrm{solution}$$$$\mathrm{for}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{2\mathrm{y}}{{\mathrm{x}}^2}=0\mathrm{on}(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$$

Commented by bobhans last updated on 13/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by abdomathmax last updated on 13/Jun/20

$$\phi \left(\mathrm{x}\right)={\mathrm{x}}^2-\frac{1}{\mathrm{x}}\Rightarrow {\phi }^{{}^{\prime }}\left(\mathrm{x}\right)=2\mathrm{x}+\frac{1}{{\mathrm{x}}^2}\mathrm{and}$$$${\phi }^{{}^{″}}\left(\mathrm{x}\right)=2-\frac{2\mathrm{x}}{{\mathrm{x}}^4}=2-\frac{2}{{\mathrm{x}}^3}\Rightarrow {\phi }^{{}^{″}}\left(\mathrm{x}\right)-\frac{2}{{\mathrm{x}}^2}\phi \left(\mathrm{x}\right)$$$$=2-\frac{2}{{\mathrm{x}}^3}-\frac{2}{{\mathrm{x}}^2}({\mathrm{x}}^2-\frac{1}{\mathrm{x}})$$$$=2-\frac{2}{{\mathrm{x}}^3}-2+\frac{2}{{\mathrm{x}}^3}=0\mathrm{so}\phi \mathrm{is}\mathrm{solution}\mathrm{of}?\left(\mathrm{de}\right)$$$${\mathrm{y}}^{{}^{″}}-\frac{2\mathrm{y}}{{\mathrm{x}}^2}=0$$$$$$

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