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

$\displaystyle D^2y-Dy-2y=e^{x}$ 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

$\displaystyle y_p=Ae^{x}$