
#1
Jun611, 08:24 AM

P: 458

1. The problem statement, all variables and given/known data
Locate the singular points of [tex]x^3(x1)y''  2(x1)y' + 3xy =0[/tex] and decide which, if any, are regular. 3. The attempt at a solution In standard form the DE is [tex]y''  \frac{2}{x^3} y' + \frac{3}{x^2(x1)} y = 0.[/tex] Are the singular points [tex]x=0,\pm 1\;?[/tex] Regular singular points [tex]x_0[/tex] of [tex]y'' + p(x)y' + q(x)y =0[/tex] satisfy [tex](xx_0)p(x) ,\; (xx_0)^2q(x)[/tex] both finite as [tex]x \to x_0.[/tex] Considering [tex]x \left( \frac{2}{x^3} \right),\; (x1) \left( \frac{2}{x^3} \right) ,\; (x+1) \left( \frac{2}{x^3} \right)[/tex] none of which are finite as [tex]x\to x_0[/tex] Does this mean there are no regular singular points? 



#2
Jun611, 08:34 AM

Math
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P: 38,882





#3
Jun611, 08:40 AM

P: 458




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