Hello I really need help in understanding operators: the best way would be to show me how to solve:
Thank you!y''-y'-2y=e^(x)
What exactly do you need to know. It can be rewritten as
$\displaystyle D^2y-Dy-2y=e^{x}$ This can be factored as
$\displaystyle (D^2-D-2)y=e^{x} \iff (D-2)(D+1)y=e^{x}$
So the homogeneous system $\displaystyle (D-2)(D+1)y=0$ has solutions
$\displaystyle e^{2x}$ because $\displaystyle (D-2)e^{2x}=0$ and
$\displaystyle e^{-x}$ because $\displaystyle (D+1)e^{-x}=0$
So it has the complimentary solution $\displaystyle y_c=c_1e^{2x}+c_2e^{-x}$
Now we can solve the original
$\displaystyle (D-2)(D+1)y=e^{x}$ by annihilating the right hand side with $\displaystyle (D-1)$ since $\displaystyle (D-1)e^{x}=0$
$\displaystyle (D-1)(D-2)(D+1)y=(D-1)e^{x} \iff
(D-1)(D-2)(D+1)y=0$ since that factor is not repeated the particular solution has the form
$\displaystyle y_p=Ae^{x}$