Second order homogeneous differential equations: why do repeated roots modify the solution set?

$$y'' + 2ky' + k^2y = 0 \, .$$ $$(y' + ky)' + k(y'+ky) = 0 $$ $$((y' + ky)e^{kt})' = 0 $$ Is equivalent to: $$(ye^{kt})''=0$$ Inetgrate twice and the solution will be obvious.

Taking inspiration from Aryadeva's answer, I have found a direct method of solving the differential equation $$ y'' + by + cy = 0 $$ which takes into account what happens when you have a repeated root. This equation has the auxiliary equation $$ m^2+bm+c=0 \, . $$ Say $\alpha$ and $\beta$ are the roots of the auxiliary equation. Then, the equation can be rewritten as $$ (m-\alpha)(m-\beta)=0 $$ which implies that $b=-(\alpha + \beta)$ and $c=\alpha\beta$. Then, the original equation can be rewritten as $$ y'' -(\alpha+\beta)y' + \alpha\beta y =0 \, . $$ It can then be directly shown that \begin{align} y&=Ae^{\alpha x}+Be^{\beta x} \quad\text{if $\alpha\neq\beta$} \\[4pt] y&=(A+Bx)e^{\alpha x} \quad\text{if $\alpha = \beta$} \, . \end{align} Here is how: \begin{align} y'' -(\alpha+\beta)y' + \alpha\beta y &= 0 \\[6pt] y'' - \alpha y' - \beta y' + \alpha\beta y &= 0 \\[6pt] y'' - \alpha y' - \beta (y' - \alpha y) &= 0 \\[6pt] (y'' - \alpha y')e^{-\beta x} - \beta (y' - \alpha y)e^{-\beta x} &= 0 \\[6pt] \left[(y' - \alpha y)e^{-\beta x}\right]' &= 0 \\[6pt] (y' - \alpha y)e^{-\beta x} &= B \\[6pt] y' - \alpha y &= Be^{\beta x} \\[6pt] e^{-\alpha x}y' - \alpha e^{-\alpha x}y &= Be^{(\beta - \alpha)x} \\[6pt] \left[e^{-\alpha x} y \right]' &= Be^{(\beta - \alpha)x} \end{align} Suppose that $\alpha \neq \beta$. Then, the following manipulations are justified: \begin{align} e^{-\alpha x}y &= \int Be^{(\beta - \alpha)x} \, dx = \frac{Be^{(\beta - \alpha)x}}{\beta - \alpha} + A\\[6pt] y &= \frac{B}{\beta - \alpha}e^{\beta x} + Ae^{\alpha x} \, . \end{align} But then, since $\frac{B}{\beta - \alpha}$ is just some arbitrary constant, we may relabel it as $B$. We are left with $$ \boxed{ \;\\[4pt] \quad y=Ae^{\alpha x}+Be^{\beta x} \quad \\ } $$ as expected. If $\alpha = \beta$, then the above reasoning is invalid since we end up with division by zero. Instead, $$ e^{-\alpha x}y = \int Be^{(\beta - \alpha)x} \, dx = Bx + A $$ and therefore $$ \boxed{ \;\\[4pt] \quad y = (A+Bx)e^{\alpha x} \, . \quad \\ } $$