Case of repeated root of Euler type equation $x^2y''-xy'-\lambda y=0$ Suppose we have an Euler type equation like the following, $$x^2y''-xy'-\lambda y=0$$ and we want to derive the general solution for the case where the indicial equation has a double root. So we know the first solution which is $y_1(x)=x^r$ and we can use the Wronskian technique to derive the second solution $y_2(x)$.
So we have, $W=y_1y_2'-y_1'y_2$ and $W'=y_1y_2''-y_1''y_2$. Also, $y''=\frac{1}{x}y'+\frac{\lambda}{x^2}y$. Therefore, $$W'=y_1\left(\frac{1}{x}y_2'+\frac{\lambda}{x^2}y_2\right)-\left(\frac{1}{x}y_1'+\frac{\lambda}{x^2}y_1\right)y_2$$ which simplifies to $$W'=\frac{1}{x}(y_1y_2'-y_1'y_2)=\frac{1}{x}W.$$ Thus $W=Ax$ where $A=\log(c)$, where $c$ is a constant of integration. Hence, $x^ry_2'-rx^{r-1}y_2=Ax$ which simplifies to $y_2'-\frac{r}{x}y_2=x^{1-r}.$ If we solve this first order ODE we have the integrating factor $\mu=e^{\int -\frac{r}{x}dx}=x^{-r}$ and thus we get $$y_2=x^r\int x^{1-r}x^{-r}dx=x^r\int x^{1-2r}dx.$$ But this integral isn't equal to $x^r\log(x)$ and this confuses me. We know that the general solution  of any Euler type equation with a double root is $y(x)=(c_1+c_2\log(x))x^r$. So, have I committed an error somewhere?
I would appreciate any help or suggestion.

The inidicial polynomial for this equation is $$r(r-1)-r-\lambda$$ which me may rewrite as $$r^2-2r-\lambda.$$
The roots of this polynomial are double (i.e. one root) if and only if $\lambda=-1$. You can easily see that from the quadratic formula.
And for $\lambda = -1$, the double root is $r=1$, and so you are integrating $$ \int x^{1-2} dx = \int \frac 1 x dx = \ln x +C$$ in the last step. That's where the natural $\log$ comes from.
