Yes, it is. If y= sinh(kt) then y'= k cosh(kt) and y"= k^2 sinh(kt). Putting that into the differential equation, y"- 25y= k^2 sinh(kt)- 25 sinh(kt)= (k^2- 25)sinh(kt)= 0 for all t. Since sinh(kt) is not alway7s 0 itself, we must have k^2- 25= 0 so k= 5 or k= -5.

Just DO it! With k= 5, those functions are . Find the second derivative, plug it into the equation and see what happens! With k= -5, you haved . Since sinh is an odd function and cosh is an even function, that is just which is really the same thing since A itself can be positive or negative.but I'm struggling with the other two parts of the question:

For those values of , show that every member of the family of functions (where and are constants) is also a solution of the DE.

[/quote]Come up with a second order DE for which is a solution for the same values of obtained before.[/QUOTE]

So now youy are working with . Take the second derivative, stare at it andthink!