Exercise 2.7.13

Answers

1.

The equation could be rewriten as

p (D) (y) = x

, where

p (t)

is the auxiliary polynomial of the equation. Since

D

is surjective by the Lemma 1 after Theorem 2.32, the differential operator

p (D)

is also surjective. Hence we may find some solution

y_{0}

such that

p (D) (y_{0}) = x

2.

Use the same notation in the previous question. We already know that

p (D) (z) = x

. If

w

is also a solution such that

p (D) (w) = x

, then we have

p (D) (w - z) = p (D) (w) - p (D) (z) = x - x = 0 .

So all the solution must be of the form $z + y$ for some $y$ in the solution space $V$ for the homogeneous linear equation.

jephianlin

2011-06-27 00:00