In each of the following cases find a second-order linear differential equation satisfied by the given and .

- ,
- ,
- .
- .
- .

- We compute the first two derivatives of the given equations,
So, if and are solutions of a linear second order differential equation, then we have

and

Together these imply

Therefore, are solutions of

- We compute the first two derivatives of the given equations,
So, if and are solutions of a linear second order differential equation, then we have

and

Therefore, are solutions of

- Rather than take derivatives of these two functions (which gets messy) we consider the function
Then if we let and we have . Since this gives us . By Theorem 8.7, then, is a solution of for every choice of and . The functions and correspond to the choices and , respectively. Hence, and are solutions of the second order differential equation

- Similar to our strategy in part (c), let
Letting and we obtain . So these are solutions of the equation . Then, since

we have and are solutions to the equation corresponding the choices and , respectively. Therefore and are solutions of

- First, using the definitions of the hyperbolic sine and cosine we have
So if we let

with and we have . Therefore, is the set of solutions to . Since and are the the solutions with and , respectively we have and solutions