TITLE: How does the solution of Cauchy-Euler equation get the logarithmic term?
QUESTION [1 upvotes]: The general solution of second-order Cauchy-Euler equation
$$x^2y''(x)+pxy'(x)+qy(x)=0\tag1$$
is given by
$$y(x)=c_1 x^{\alpha_1}+c_2 x^{\alpha_2},\tag2$$
where $$\alpha_{1,2}=\frac{1-p}2\pm\frac{\sqrt{(1-p)^2-4q}}2.\tag3$$
But when $q=\frac14(1-p)^2$, i.e. when $\alpha_1=\alpha_2=\alpha$, the general solution somehow gets a logarithmic term, without which the generality of $(2)$ is lost:
$$y(x)=c_1x^\alpha+c_2x^\alpha\ln x.\tag4$$
I know how to derive this result e.g. by the method of reduction of order, but it doesn't seem to give much intuition on the origin of this logarithm. What is an intuitive explanation of where this logarithm comes from and why the general solution suddenly stops being general for particular combinations of $p$ and $q$?

REPLY [6 votes]: The loss of generality of the solution is a result of an unfortunate choice of arbitrary constants. Let now $c_{1,2}$ in OP's formula $(2)$ depend on $\alpha_{1,2}$:
$$\begin{align}
c_1&=k_1-c_2,\\
c_2&=\frac{k_2}{\alpha_2-\alpha_1},
\end{align}\tag I
$$
where $k_{1,2}$ are new constants.
Then expression $(2)$ in the OP will become
$$y(x)=x^{\alpha_1}\left(k_1+k_2\frac{x^{\alpha_2-\alpha_1}-1}{\alpha_2-\alpha_1}\right).\tag{II}$$
What happens if we take the limit of $\alpha_2\to\alpha_1$ in $(\mathrm{II})$? See,
$$\lim_{\gamma\to0}\frac{x^\gamma-1}\gamma=\left.\frac{\mathrm d}{\mathrm d \gamma}x^\gamma\right|_{\gamma=0}=\ln x.\tag{III}$$
Thus we conclude that
$$\lim_{\alpha_2\to\alpha_1}y(x)=x^{\alpha_1}(k_1 +k_2\ln x),\tag{IV}$$
which is equivalent to $(4)$ in the OP, up to naming of the constants.
This means that the logarithmic term in $(4)$ appears as the limit of power function with the power going to zero, where we prevented the function from "flattening" to constant $x^0$ by dividing it by the power (which approaches $0$).
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