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Question Number 120811 by fajri last updated on 03/Nov/20

Find the singular point in the  differential equation :    (x^3  − x^2  − 9x + 9)(d^2 y/dx^2 ) +2x(dy/dx) + (x − 3)y = 0

$${Find}\:{the}\:{singular}\:{point}\:{in}\:{the} \\ $$$${differential}\:{equation}\:: \\ $$$$ \\ $$$$\left({x}^{\mathrm{3}} \:−\:{x}^{\mathrm{2}} \:−\:\mathrm{9}{x}\:+\:\mathrm{9}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\mathrm{2}{x}\frac{{dy}}{{dx}}\:+\:\left({x}\:−\:\mathrm{3}\right){y}\:=\:\mathrm{0} \\ $$

Answered by 675480065 last updated on 03/Nov/20

(x+1)(x−3)(x+3)(d^2 y/dx^2 )+2x(dy/dx)+(x−3)y=0  S.p  (x+1)(x−3)(x+3)=0  x=−1,3,−3

$$\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}{x}\frac{{dy}}{{dx}}+\left({x}−\mathrm{3}\right){y}=\mathrm{0} \\ $$$$\mathrm{S}.\mathrm{p} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1},\mathrm{3},−\mathrm{3} \\ $$

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