We want to find all [tex]f(x)[/tex] such that
[tex](D^2-12D-45)[f(x)]=0[/tex]
We have
[tex](D^2-12D-45)[f]=(D-15)(D+3)[f]=0[/tex]
so that either
[tex](D-15)[f]=0\implies f'-15f=0\implies f(x)=e^{15x}[/tex]
or
[tex](D+3)[f]=0\implies f'+3f=0\implies f(x)=e^{-3x}[/tex]
These solutions are linearly independent; we confirm this by computing the Wronskian:
[tex]W=\begin{vmatrix}e^{15x}&e^{-3x}\\15e^{15x}&-3e^{-3x}\end{vmatrix}=-3e^{12x}-15e^{12x}=-18e^{12x}\neq0[/tex]
We can also confirm that these solutions are correct by substituting the solutions for [tex]f(x)[/tex] in the differential equation:
[tex](D^2-12D-45)[e^{15x}]=225e^{15x}-12(15e^{15x})-45e^{15x}=0[/tex]
[tex](D^2-12D-45)[e^{-3x}]=9e^{-3x}-12(-3e^{-3x})-45e^{-3x}=0[/tex]
