Exercise 2.7.1

Answers

1.: Yes. It comes from Theorem 2.32.
2.: Yes. It comes from Theorem 2.28.
3.: No. The equation $y = 0$ has the auxiliary polynomial $p (t) = 1$ . But $y = 1$ is not a solution.
4.: No. The function $y = e^{t} + e^{- t}$ is a solution to the linear differential equation $y^{″} - y = 0$ .
5.: Yes. The differential operator is linear.
6.: No. The differential equation $y^{″} - 2 y^{'} + y = 0$ has a solution space of dimension two. So ${e^{t}}$ could not be a basis.
7.: Yes. Just pick the differential equation $p (D) (y) = 0$ .

jephianlin

2011-06-27 00:00