Quote:

but I'm struggling with the other two parts of the question:

For those values of $\displaystyle k$, show that every member of the family of functions $\displaystyle Asinhkt+Bcoshkt$ (where$\displaystyle A$ and $\displaystyle B$ are constants) is also a solution of the DE.

Just DO it! With k= 5, those functions are $\displaystyle A sinh(5t)+ B cosh(5t)$. Find the second derivative, plug it into the equation and see what happens! With k= -5, you haved $\displaystyle A sinh(-5t)+ B cosh(-5t)$. Since sinh is an odd function and cosh is an even function, that is just $\displaystyle -A sinh(5t)+ B cosh(5t)$ which is really the same thing since A itself can be positive or negative.