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Question Number 100323 by bobhans last updated on 26/Jun/20

$$\mathbb{S}\mathrm{olve}{\mathrm{x}}^2{\mathrm{y}}^{″}+{2\mathrm{xy}}^{\prime }-2\mathrm{y}=0$$

Commented by bemath last updated on 26/Jun/20

$$\mathrm{since}\mathrm{diff}\mathrm{a}\mathrm{power}\mathrm{pushes}\mathrm{down}\mathrm{the}$$$$\mathrm{exponent}\mathrm{by}\mathrm{one}\mathrm{unit},\mathrm{the}\mathrm{form}\mathrm{of}$$$$\mathrm{this}\mathrm{eq}\mathrm{suggest}\mathrm{that}\mathrm{we}\mathrm{look}\mathrm{for}$$$$\mathrm{possible}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{type}\mathrm{y}={\mathrm{x}}^{\mathrm{n}}.$$$$\mathrm{we}\mathrm{optain}\mathrm{the}\mathrm{quadratic}\mathrm{eq}$$$$\mathrm{n}(\mathrm{n}-1)+2\mathrm{n}-2=0,\mathrm{this}\mathrm{has}\mathrm{roots}$$$$\mathrm{n}=1;-2\mathrm{so}\mathrm{are}\mathrm{solutions}\mathrm{is}$$$$\mathrm{y}={\mathrm{C}}_{1}x+{\mathrm{C}}_2{x}^{-2}★$$

Answered by mathmax by abdo last updated on 27/Jun/20

$${\mathrm{am}}^2+(\mathrm{b}-\mathrm{a})\mathrm{m}+\mathrm{c}=0\mathrm{with}\mathrm{a}=1,\mathrm{b}=2\mathrm{and}\mathrm{c}=-2\Rightarrow$$$${\mathrm{m}}^2+\mathrm{m}-2=0\to \mathrm{\Delta }=1+8=9\Rightarrow {\mathrm{m}}_1=\frac{-1+3}{2}=1\mathrm{and}{\mathrm{m}}_2=\frac{-1-3}{2}=-2\Rightarrow$$$$\mathrm{y}=\mathrm{ax}+\mathrm{b}{\mathrm{x}}^{-2}=\mathrm{ax}+\frac{\mathrm{b}}{{\mathrm{x}}^2}$$

Commented by mathmax by abdo last updated on 27/Jun/20

$$\mathrm{sorry}\mathrm{y}=\alpha \mathrm{x}+\beta {\mathrm{x}}^{-2}=\alpha \mathrm{x}+\frac{\beta }{{\mathrm{x}}^2}$$

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