Exercise 2.7.14

We use induction on the order $n$ of the equation. Let $p(t)$ be the auxiliary polynomial of the equation. If now $p(t)=t-c$ for some coefficient $c$, then the solution is $C{e}^{\mathit{ct}}$ for some constant $C$ by Theorem 2.34. So if $C{e}^{c{t}_0}=0$ for some $t_0\in ℝ$, then we know that $C=0$ and the solution is the zero function. Suppose the statement is true for $n<k$. Now assume the degree of $p(t)$ is $k$. Let $x$ be a solution and $t_0$ is a real number. For an arbitrary scalar $c$, we factor $p(t)=q(t)(t-c)$ for a polynomial $q(t)$ of degree $k-1$ and set $z=q(D)(x)$. We have $(D-\mathit{cI})(z)=0$ since $x$ is a solution and $z(t_0)=0$ since $x^{(i)}(t_0)=0$ for all $0\le i\le n-1$. Again, $z$ must be of the form $C{e}^{\mathit{ct}}$. And so $C{e}^{c{t}_0}=0$ implies $C=0$. Thus $z$ is the zero function. Now we have $q(D)(x)=z=0$. By induction hypothesis, we get the conclusion that $x$ is identically zero. This completes the proof.