1. ## Operators

Hello I really need help in understanding operators: the best way would be to show me how to solve:
y''-y'-2y=e^(x)
Thank you!

2. Originally Posted by deltafee
Hello I really need help in understanding operators: the best way would be to show me how to solve:

Thank you!
What exactly do you need to know. It can be rewritten as

$D^2y-Dy-2y=e^{x}$ This can be factored as

$(D^2-D-2)y=e^{x} \iff (D-2)(D+1)y=e^{x}$

So the homogeneous system $(D-2)(D+1)y=0$ has solutions

$e^{2x}$ because $(D-2)e^{2x}=0$ and

$e^{-x}$ because $(D+1)e^{-x}=0$

So it has the complimentary solution $y_c=c_1e^{2x}+c_2e^{-x}$

Now we can solve the original

$(D-2)(D+1)y=e^{x}$ by annihilating the right hand side with $(D-1)$ since $(D-1)e^{x}=0$

$(D-1)(D-2)(D+1)y=(D-1)e^{x} \iff
(D-1)(D-2)(D+1)y=0$
since that factor is not repeated the particular solution has the form

$y_p=Ae^{x}$