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Consider the differential equation (x - 1)y" + y' = 0. By using power series so- lution about the ordinary point 0 we obtain: c_{k+1)=[(k+1)/(k+2)] C_(K); k=1,2,3.... c_(K+2)=[(k+1)/(k+2)] C_(K+1); k=0,1,2,3... None of them c_(k+2)=[(k+1)/(k+2)] C_(k+1); k=1,2,3....​

Respuesta :

LammettHash

Given the ODE

(x - 1) y'' + y' = 0

we assume a solution of the form

[tex]y(x) = \displaystyle \sum_{n=0}^\infty c_n x^n[/tex]

with derivatives

[tex]y'(x) = \displaystyle \sum_{n=0}^\infty n c_n x^{n-1} = \sum_{n=0}^\infty (n+1) c_{n+1} x^n[/tex]

[tex]y''(x) = \displaystyle \sum_{n=0}^\infty (n+1) n c_{n+1} x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1) c_{n+2} x^n[/tex]

Substituting these series into ODE gives

[tex]\displaystyle (x - 1) \sum_{n=0}^\infty (n+2)(n+1) c_{n+2} x^n + \sum_{n=0}^\infty (n+1) c_{n+1} x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1) c_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+2)(n+1) c_{n+2} x^n + \sum_{n=0}^\infty (n+1) c_{n+1} x^n = 0[/tex]

[tex]\displaystyle \sum_{n=1}^\infty (n+1)n c_{n+1} x^n - \sum_{n=0}^\infty (n+2)(n+1) c_{n+2} x^n + \sum_{n=0}^\infty (n+1) c_{n+1} x^n = 0[/tex]

[tex]\displaystyle \sum_{n=1}^\infty \bigg((n+1)n c_{n+1} - (n+2)(n+1) c_{n+2} + (n+1) c_{n+1}\bigg) x^n - 2c_2 + c_1 = 0[/tex]

[tex]\displaystyle \sum_{n=1}^\infty \bigg((n+1)^2 c_{n+1} - (n+2)(n+1) c_{n+2}\bigg) x^n - 2c_2 + c_1 = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+1)^2 c_{n+1} - (n+2)(n+1) c_{n+2}\bigg) x^n = 0[/tex]

Then the coefficients [tex]c_n[/tex] are given recursively by

[tex](n+1)^2 c_{n+1} - (n+2)(n+1) c_{n+2} = 0[/tex]

for all n ≥ 0, or equivalently,

[tex]c_{n+2} = \dfrac{n+1}{n+2} c_{n+1}[/tex]

which most closely resembles the second choice.

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