I just asked a question about initial value problems but encountered another one immediately.
$ y''(t) - 6y'(t) + 9y(t) = 18 $
$y(0) = 2, $
$y'(0) = 1 $
The homogenous solution is: $c_1e^{2x} + c_2xe^{2x}$
My main problem here is to choose which particular solution to use. Since the "original equation" is = 18, should my particular solution just be the constant K? However, that doesn't suit well with the fact that $y'(0) = 1$.
Hint $$y''(t) - 6y'(t) + 9y(t) = 18$$ $$y''(t) - 6y'(t) + 9(y(t)-2) =0$$ Substitute $ z=y-2$ then: $$z''(t) - 6z'(t) + 9z(t) =0$$ It's homogenous now. Solve for z. And substitute back $z=y-2$
You can conclude that $y=y_h+y_p=y_h+2$