# Using Frobenius Method

1. Nov 17, 2013

1. The problem statement, all variables and given/known data
I want to find two linearly independent solutions of
$$x^{2}y''-2x^{2}y'+(4x-2)y=0.$$

3. The attempt at a solution
The roots to the indicial polynomial are $r_{1}=2$ and $r_{2}=-1$.
I found one solution which was $x^{2}$ and I am having trouble finding the second solution.
After all of the arithmetic I came up with the equation
$$[r(r-1)-2]c_{0}+\sum\limits_{k=1}^{\infty}\left[(k+r)(k+r-1)-2\right]c_{k}x^{k}+4c_{k-1}x^{k}-2c_{k-1}(k+r-1)x^{k}.$$
I let $r=-1$ so that I get
$$\left[(k-1)(k-2)-2\right]c_{k}+4c_{k-1}-2c_{k-1}(k-2)=0$$
and so
$$c_{k}=\frac{2(k-4)c_{k-1}}{k(k-3)}.$$
The recurrence relation becomes undefined for k=3 and onwards. How do I deal with this? The solution skips $c_{3}$ and seemingly magically obtains a formula for $c_{4}$ and onwards.

2. Nov 17, 2013

### vanhees71

First of all I guess the equation should be
$$x^{2}y''-2x^{2}y'-(4x+2)y=0,$$
because otherwitse $y(x)=x^2$ isn't a solution, and your indical roots are not $2$ and $-1$.

Assuming the above ODE, you get the indical equation to be yours,
$$r(r-1)-2=0 \; \Rightarrow r_1=2, \quad r_2=-1,$$
and the recursion equation reads (after some algebra)
$$C_k k(k+2r-1)-2 C_{k-1}(k+r-3)=0.$$

Now here you have the special case of a Fuchs-class ODE, where the difference of the indical roots is natural and not 0. In this case you always have this "problem" with the coefficient of $c_{r_1-r_2}$ becoming $0$ for the solution for the smaller indical root, $r<2$. For the solution $y_1$ referring to the larger indical root, there is no problem with the recursion. Given this solution, the correct ansatz for the 2nd linearly independent solution is
$$y_2(x)=\gamma y_1(x) \ln x+x^{r_2} \sum_{k=0}^{\infty} C_k x^k.$$
Of course you assume $C_0 \neq 0$. Then for $k<r_1-r_2$ the recursion can be used to get the coefficients for these $k$. For $k=r_1-r_2$ you find that you can set $C_{r_1-r_2}=0$ and then get $\gamma$ uniquely, and for $k>r_1-r_2$ you get another recursion relation which is well-defined for the $C_k$ with $k>r_1-r_2$.

3. Nov 17, 2013

The equation is $x^{2}y''-2x^{2}y'+(4x-2)y=0$. Unless I'm completely crazy, $y(x)=x^{2}$ is a solution.

4. Nov 17, 2013

### vanhees71

Argh. When copying your equation to a piece of paper, I wrote $+2x^2y'$ for the middle term. Of course, you are right, for
$$x^2 y''-2x^2 y'+(4x-2)y=0$$
one solution is $y_1=x^2$.

The rest of my posting is, however correct. For $r=-1$ you make the ansatz
$$y(x)=\gamma x^2 \ln x+z(x), \quad z(x)=\frac{1}{x} \sum_{k=0}^{\infty} d_k x^k.$$
This plugged into the ODE you get after some algebra the equation
$$\sum_{k=1}^{\infty} [d_k k(k-3)+d_{k-1} (8-2k)] x^k=(-3x^3+2x^4)\gamma.$$
Setting $d_0=1$ you get recursively all the $d_k$ by comparing coefficients. $d_3$ is of course arbitrary. You can set it to $0$.

5. Nov 17, 2013

Ah! Thanks! I see what I was doing wrong now. I was plugging in $y(x)=z(x)$ to try to get the second solution instead of the correct $y(x)=cx^{2}\ln x+z(x)$. Thanks for your help!