Question: Compute the Frobenius Series about $x=1$ for the following problem:
$(x-1)y"(x)-xy'(x)+y(x)=0$
For those who are not familiar about Frobenius Method, we basically try to compute $y(x)=\sum_{k=0}^{\infty}a_k(x-1)^{k+\alpha}$, such that it is a solution to the above ODE via the idea of recurrence relation.
My work so far
Now, I found the the roots of indicial equation to be $\alpha_1=2$ and $\alpha_2=0$. Subsequently I computed the first Frobenius Series to be $y_1(x)=\sum_{k=0}^{\infty}\frac{2}{(k+2)!}(x-1)^k$, where I took $a_0=1$ for simplicity. (this power series corresponds to $\alpha_1=2$).
Next, I computed power series for $\alpha_2=0$. Now, by considering the recurrence relation, I figured $b_0=b_1$ and $b_2$ is arbitrary. ($b_i$ are coefficients of the second Frobenius Series, $y_2(x)$). Since $b_0$ is again arbitrary, I set it as $1$ again for simplicity. Here I had two choices for $b_2$:
Either set $b_2=0$, which subsequently mean that $b_k=0$ $\forall k\geq 2$, which corresponds to a polynomial, $y_{2_1}(x)=(x-1)+1$.
Or I could set $b_2=\frac{1}{2}b_1=\frac{1}{2}.$ Then by recurrence relation one would obtain the power sees of $y_{2_2}(x)=exp(x-1)$.
My confusion:
ISSUE! Now it seems to me that I have created three linearly independent solutions,$y_1(x), y_{2_1}(x), y_{2_2}(x)$, which I know it cannot be the case, so how could I produce a linear relation between three of them?
I mean, all I could see so far is that $(x-1)^2y_1(x)=2y_{2_2}(x)-2y_{2_1}(x)$. Would this imply they're indeed linearly dependent?
Thank you to everyone who read this far, I really appreciate your patience and kindness. If anyone wants me to post more of my work, I will try my best to do so!
You did not apply the Frobenius form correctly in $y_1$ after computing the coefficients. As you wrote at the start, $$ y_1(x)=\sum_{k=0}^{\infty}a_k(x-1)^{k+\alpha_1} =\sum_{k=0}^\infty\frac{2}{(k+2)!}(x-1)^{k+2} =2\sum_{k=2}^\infty\frac{(x-1)^k}{k!} $$