$\displaystyle (t^2D^2 - 4tD - 6I)[y] = (6 + 16t - 5t^2 - 14t^3 - 2t^4 + t^5)e^{-t}

$

Printable View

- Mar 29th 2010, 08:24 PMballajrFind the general solution to this linear operator problem:
$\displaystyle (t^2D^2 - 4tD - 6I)[y] = (6 + 16t - 5t^2 - 14t^3 - 2t^4 + t^5)e^{-t}

$ - Mar 30th 2010, 04:28 AMshawsend
I mean what do you do if you walk into class and the professor gives you this problem and says nothing about it? What do you mean I? I'm thinking an integral operator maybe so why not just differentiate both sides to eliminate it. Sure, you get a 3rd order DE and I don't know even if I'm right but that's what he's gettin' from me:

$\displaystyle \frac{d}{dt}\left\{\left(t^2D^2-4tD-6I\right)y=f(t)\right\}$

$\displaystyle \left(t^2D^3+2tD^2-4(tD^2+D)-6\right)y=f'(t)$

. . . then I guess try Laplace transforms. . . . right back at ya prof.