Hi there. I have this exercise, which says:(adsbygoogle = window.adsbygoogle || []).push({});

Demonstrate that:

[tex]xy''+(1-x)y'+\lambda y=0[/tex]

has a polynomial solution for some λ values.

Indicate the orthogonality relation between polynomials, the fundamental interval, and the weight function.

So I thought I should solve this using Frobenius method. I have one singular point at x=0, which is regular. I assumed a solution of the form:

[tex]y(x)=\sum_0^{\infty}a_n x^{n+r}[/tex]

And then replacing in the diff. eq. I get:

[tex]\sum_0^{\infty}a_n (n+r)(n+r-1) x^{n+r-1}+\sum_0^{\infty}a_n (n+r)x^{n+r-1}-\sum_0^{\infty}(n+r)a_n x^{n+r}+\lambda \sum_0^{\infty}a_n x^{n+r}=0[/tex]

[tex]\sum_0^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0[/tex]

[tex]a_0r^2x^{r-1}+\sum_1^{\infty}a_n (n+r)^2 x^{n+r-1}-\sum_0^{\infty}a_n (n+r-\lambda) x^{n+r}=0[/tex]

Therefore r=0.

Then replacing r=0, and changing the index for the first summation, with m=n-1, n=m+1:

[tex]\sum_0^{\infty}a_{m+1} (m+1)^2 x^{m}-\sum_0^{\infty}a_n (n-\lambda) x^{n}=0[/tex]

And now calling m=n

[tex]\sum_0^{\infty}x^m \left ( a_{m+1} (m+1)^2 x^{m}-a_m (m-\lambda) \right )=0[/tex]

So I have the recurrence relation:

[tex]a_{m+1}=\frac{a_m(m-\lambda)}{(m+1)^2}[/tex]

Trying some terms:

[tex]a_1=-a_0\lambda[/tex]

[tex]a_2=\frac{a_1(1-\lambda)}{2^2}=-\frac{a_0\lambda(1-\lambda)}{2^2}[/tex]

[tex]a_3=\frac{a_2(2-\lambda)}{3^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)}{2^23^2}[/tex]

[tex]a_4=\frac{a_3(3-\lambda)}{4^2}=-\frac{a_0\lambda(1-\lambda)(2-\lambda)(3-\lambda)}{2^23^24^2}[/tex]

I'm not sure what this gives, I tried this:

[tex]a_n=-\frac{a_0\lambda(n-1-\lambda)!}{(n!)^2}[/tex]

This is wrong, because the factorial in the numerator is only defined for positive values of (n-1-λ), and if n=1 I get (-\lambda)!, which wouldn't work for a_1, unless λ=0, which gives the trivial solution. But I think it works for n>1.

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# Second order diff. eq. Frobenius

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