How do I find the second solution using the Frobenius method?

[DeadOriginal]

Homework Statement

I want to find two linearly independent solutions of
$$
x^{2}y''-2x^{2}y'+(4x-2)y=0.
$$

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.

 

[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.

 

[DeadOriginal]

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.

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.

 

[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.

 

[DeadOriginal]

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!