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I am having an issue understanding the Frobenius method. I plan on posting multiple questions in this thread as I go. I searched these forums and saw some other threads pointing to it, but the questions were slightly different. I am able to get through the problem to a certain extent, but after a point, I don't know where to go on.

Example 1:

[tex]x^{2}y''-3xy'+4y = 0[/tex]

This can be put into the format:

[tex]y''-\frac{3x}{x^{2}}y'+\frac{4}{x^{2}}y[/tex]

Which is consistent with the form that can be solved using the Frobenius method.

I assume the solution [tex]y(x)[/tex] and its derivatives to be in this form, multiplying by the power [tex]x^{r}[/tex]:

[tex]y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}[/tex]

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

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

I plug these into the original equation, getting:

[tex]\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r}-3\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r}+4\sum_{n=0}^{\infty}a_{n}x^{n+r} = 0[/tex]

I find that the lowest power in this equation is [tex]x^{r}[/tex]. Plugging [tex]n=0[/tex] into the equation, and pulling out [tex]x^{r}[/tex], I get:

[tex](r-1)ra_{0} - 3ra_{0} + 4a_{0} = 0[/tex]

[tex]a_{0}(r^{2}-r-3r+4) = 0[/tex]

[tex]a_{0}(r^{2}-4r+4) = 0[/tex]

Assuming [tex]a_{0} \neq 0[/tex], making the solution non-trivial, I get the roots for r, which are [tex]r_{1},r_{2} = 2[/tex]

I use one of the roots, r=2, and plug in (here's my attempt to get the TeX to work again)

[tex]\sum_{n=0}^{\infty}(n+1)(n+2)a_{n}x^{n+2}-3\sum_{n=0}^{\infty}(n+2)a_{n}x^{n+2}+4\sum_{n=0}^{\infty}a_{n}x^{n+2}[/tex]

I do not know where to go from this point. I cannot get [tex]a_{n+1}[/tex] or [tex]a_{n+2}[/tex] in terms of [tex]a_{n+1}[/tex] or [tex]a_{0}[/tex]. Every single result results in some constant, multiplied by [tex]a_{n}[/tex] equals 0. So, at this point. I am lost.

NOTE: I don't want the answer. I just want some clues on how I should continue.

Example 1:

[tex]x^{2}y''-3xy'+4y = 0[/tex]

This can be put into the format:

[tex]y''-\frac{3x}{x^{2}}y'+\frac{4}{x^{2}}y[/tex]

Which is consistent with the form that can be solved using the Frobenius method.

I assume the solution [tex]y(x)[/tex] and its derivatives to be in this form, multiplying by the power [tex]x^{r}[/tex]:

[tex]y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}[/tex]

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

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

I plug these into the original equation, getting:

[tex]\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r}-3\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r}+4\sum_{n=0}^{\infty}a_{n}x^{n+r} = 0[/tex]

I find that the lowest power in this equation is [tex]x^{r}[/tex]. Plugging [tex]n=0[/tex] into the equation, and pulling out [tex]x^{r}[/tex], I get:

[tex](r-1)ra_{0} - 3ra_{0} + 4a_{0} = 0[/tex]

[tex]a_{0}(r^{2}-r-3r+4) = 0[/tex]

[tex]a_{0}(r^{2}-4r+4) = 0[/tex]

Assuming [tex]a_{0} \neq 0[/tex], making the solution non-trivial, I get the roots for r, which are [tex]r_{1},r_{2} = 2[/tex]

I use one of the roots, r=2, and plug in (here's my attempt to get the TeX to work again)

[tex]\sum_{n=0}^{\infty}(n+1)(n+2)a_{n}x^{n+2}-3\sum_{n=0}^{\infty}(n+2)a_{n}x^{n+2}+4\sum_{n=0}^{\infty}a_{n}x^{n+2}[/tex]

I do not know where to go from this point. I cannot get [tex]a_{n+1}[/tex] or [tex]a_{n+2}[/tex] in terms of [tex]a_{n+1}[/tex] or [tex]a_{0}[/tex]. Every single result results in some constant, multiplied by [tex]a_{n}[/tex] equals 0. So, at this point. I am lost.

NOTE: I don't want the answer. I just want some clues on how I should continue.

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