When working with series solutions near a regular singular point, we apply the method of Frobenius, with at least one solution of the form $$y=x^{\lambda}\sum^{\infty}_{n=0}a_nx^n \tag{1}$$
where $\lambda$ is obtained from the indicial equation.
If $(\lambda_1-\lambda_2) \in \mathbb{N}^+$, $\lambda_1>\lambda_2$ and $\lambda_2$ does not yield a second linearly independent solution $y_2(x)$ from the recurrence formula alone, then $$y_2(x)=\frac{\partial}{\partial \lambda}[(\lambda-\lambda_2)\bar{y}(\lambda,x)]|_{\lambda=\lambda_2} \tag{2}$$
where $\bar{y}(\lambda,x)$ is generated by keeping the recurrence formula in terms of $\lambda$ and using it to find the coefficients $a_n(n \geq1)$ in terms of both $\lambda$ and $a_0$, and substituting into $(1)$.
However I cannot get my head around the use of $(2)$ and the following example should explain what my doubts are:
Consider $$x^2y''+(x^2-2x)y'+2y=0 \tag{3}$$ The recurrence formula for $(3)$ is $$a_n=-\frac{1}{\lambda+n-2}a_{n-1} \tag{4}$$ with $\lambda_1=2$ and $\lambda_2=1$
By substituting $\lambda_1=2$ into $(4)$, we obtain the first solution: $$y_1(x)=a_0x^2\sum^{\infty}_{n=0}\frac{(-1)^n}{n!}x^n=a_0x^2e^{-x} \tag{5}$$
Now turn to $y_2(x)$, the substitution of $\lambda=\lambda_2=1$ into $(4)$ will leave $a_1$ undefined, a solution that is linearly independent on $y_1(x)$ is not obtained and $(2)$ must be used.
We have $$\bar{y}(\lambda,x)=a_0\left(x^{\lambda}-\frac{1}{\lambda-1}x^{\lambda+1}+\frac{1}{\lambda(\lambda-1)}x^{\lambda+2}+...\right) \tag{6}$$ $$(\lambda-\lambda_2)\bar{y}(\lambda,x)=a_0\left((\lambda-1)x^{\lambda}-x^{\lambda+1}+\frac{1}{\lambda}x^{\lambda+2} +...\right) \tag{7}$$ Substitution into $(2)$ leads to $$y_2(x)=a_0\left(x+0-x^2ln(x)-x^3+...\right) \tag{8}$$
But if instead, by evaluating $(2)$ directly, we obtain $$\frac{\partial}{\partial \lambda}[(\lambda-\lambda_2)\bar{y}(\lambda,x)]|_{\lambda=\lambda_2} =\bar{y}(\lambda_2,x) \tag{9}$$ which leads to a wrong answer with undefined terms (because it involves the use of $(4)$ alone).
*So would it be correct to say that the direct expansion with ($\lambda-\lambda_2$) in $(7)$ followed by writing the solution as a series could circumvent the problems in $(9)$ or is $(9)$ itself incorrect?
Can someone please explain where my conceptual errors lie.