For a 2nd order, homogeneous, linear differential equation, if the auxiliary equation has two equal roots, why can't we use $\ y=Ae^{n_1x}+Be^{n_2x} $ as it's solution?
Moreover, what connection does the amount of arbitrary constants has with the differential equation?
So when you try to solve the differential equation (homogeneous second order linear with constant coefficients) you make a guess $f(x) = Ce^{Rx}$ inserting this into the differential equation you get the characteristic equation. Thus $f$ is a solution if and only if $R$ is a root in the characteristic equation.
Now you a case where you get repeated roots. So in effect you have only found one solution, since the other is just the first solution multiplied with a scalar.
Next you utilize the method of "Reduction of order" to get a first order equation that can be used to construct the second solution you are looking for. Please refer to this on how to use reduction of order (this is the reason why the second solution is $xe^{Rx}$ for repeated roots).
For the second part of the equation. The order and the number of arbitrary number of constants are the same - if for instance the Lipschitz condition is met.
EDIT: See this for more information on the Lipschitz condition, the connection with a existence and uniqueness theorem.