Repeated roots in differential equations: The need to find further roots? Given a differential equation with $n$ roots, suppose there exists a 2-fold repeated root $y_1 = Ae^{\lambda_1 x}$
In this case, we are often told that for a $k$-fold repeated root, we need to find an additional $k-1$ roots, by multiplying integer powers of $x$
as such the additional root required for our DE is $y=Axe^{\lambda_1 x}$
such that the $n$ roots are:
$y_1 = Ae^{\lambda_1 x}, \ y_2=Axe^{\lambda_1 x}, \ y_3 = Axe^{\lambda_3 x}, \ \cdots \ y_n=Axe^{\lambda_n x}$$
Intuitively, why do we need to find the additional roots? For example in a quadratic equation $x^2 +2x +1 $, if there are repeated roots. We would just "accept" that there is only one root $(x=1)$. Why can't we do the same for a DE and just "accept" the existence of a repeated root instead of going further to find new ones?
 A: An $n^{th}$ order, linear differential equation has a general solution depending on $n$ arbitrary parameters (constants of integration).
So if you had a 2nd order equation, with "repeated root" you don't just accept this and stop, because you cannot generate all solutions of the differential equation, you just have a subset of them
$$
y_1 = Ae^{\lambda_1 x}
$$
by going further, you find solutions independent of the one you already have.
This is why you consider
$$
y_2 = Bxe^{\lambda_1 x}
$$
Multiplying by $x$ generates the required, second, independent of the first, solution. A combination of all $y_1$ and $y_2$ provide the general solution.
EDIT:
For your trial solution $e^{\lambda x}$ to be an actual solution, you substitute it in to your 2nd order equation and arrive at the condition
$$
(\lambda-\lambda_1)^2e^{\lambda x}=0
$$
because $\lambda_1$ is a repeated root. To generate an additional solution, differentiate wrt $\lambda$. The above condition is still satisfied, so you still have a solution, but this introduces the extra $x$ that magically appeared previously. There are other functions independent of $e^{\lambda x}$ of course, but the process of differentiating generates one that is a solution to the differential equation. The $x$ arises naturally, if you like.
