Solutions

a)

$$a'(t)=L(a(t)).$$

b) $b(t)=a(t-1)$. Hence,

$$b'(t)=a'(t-1)=L(a(t-1))=L(b(t)).$$

c) $a(t)-b(t)=c$ for some constant $c$. Then,

$$b'(t)=a'(t).$$