Exercise 2.7.11

Denote the given set in Theorem 2.34 to be $S$. All the element in the set $S$ is a solution by the proof of the Lemma before Theorem 2.34. Next, we prove that $S$ is linearly independent by induction on the number $k$ of distinct zeroes. For the case $k=1$, it has been proven by the Lemma before Theorem 2.34. Suppose now the set $S$ is linearly independent for the case $k<m$. Assume that

for some coefficient $b_{i,j}$. Observe that

Since any differential operator is linear, we have

Since all terms fo $i=m$ are vanished by the differential operator, we may apply the induction hypothesis and know the coefficients for all terms in the left and side is zero. Observer that the coefficient of the term $t^{n_i-1}{e}^{c_{i}t}$ is ${(c_i-c_m)}^{n_m}{b}_{i,n_i-1}$. This means ${(c_i-c_m)}^{n_m}{b}_{i,n_i-1}=0$ and so $b_{i,n_i-1}=0$ for all $i<m$. Thus we know that the coefficient of the term $t^{n_i-2}{e}^{c_{i}t}$ is ${(c_i-c_m)}^{n_m}{b}_{i,n_i-2}$. Hence $b_{i,n_i-2}=0$ for all $i<m$. Doing this inductively, we get $b_{i,j}=0$ for all $i<m$. Finally, the equality

implies $b_{m,j}=0$ for all $j$ by the Lemma before Theorem 2.34. Thus we complete the proof.