# Operators

• Dec 5th 2010, 06:54 PM
deltafee
Operators
Hello I really need help in understanding operators: the best way would be to show me how to solve:
Quote:

y''-y'-2y=e^(x)
Thank you!
• Dec 5th 2010, 07:18 PM
TheEmptySet
Quote:

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}\$