Why is it permissible to treat the linear differential operator as a polynomial, while solving a higher order linear differential equation. When I was being taught how to find complimentary solution and particular integral for a second order linear differential equation, I was told that the second or higher derivative can be treated as D to the said higher powers and then I can find the root which shall help me in getting the complimentary solution. Though, I still can't seem to understand why it's mathematically okay to represent second derivative as D^2, and then use algebra.
 A: You can think of multiplication as an "action."  In 3rd grade, we multiply numbers on themselves, and in that sense the numbers "act" on each other.  Later, we multiply matrices on vectors and we start thinking of the matrix "acting" on the vector.  So maybe it would be helpful to think of multiplication as just one example from a larger class of "actions."  Another action would be taking the derivative.   We define $3^n$ as the action multiplying by $3$ $n$-times and we define $D^n$ as acting by differentiation $n$-times.  Multiplication by $3$ is linear action (since $3(x+y) = 3x+3y$ and $3(ax) = a(3x)$.   Likewise $D$ is linear "action", (that is, $D(f+g) = D(f) +D(g)$ and $D(af)=aD(f)$) so it behaves, thanks to our similar notation, just like multiplication by 3.  It's sort of a fundamental principle of algebra that linear combinations of linear actions are again linear actions.  E.g., $3D^3 - 2D+5I$ is a linear combination of the linear actions $D^3$, $D$ and $I$.  So it shouldn't be too surprising that these actions behave algebraically like polynomials. 
