1. ## Differential equation

For what $nonzero$ values of $k$ does the function $y=sinhkt$ satisfy the DE $y''-25y=0$?

I got $k=5$ and $k=-5$ which I think is correct, but I'm struggling with the other two parts of the question:

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

Come up with a second order DE for which $y=sinkt$ is a solution for the same values of $k$ obtained before.

2. Originally Posted by Konidias
For what $nonzero$ values of $k$ does the function $y=sinhkt$ satisfy the DE $y''-25y=0$?

I got $k=5$ and $k=-5$ which I think is correct
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.

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

For those values of $k$, show that every member of the family of functions $Asinhkt+Bcoshkt$ (where $A$ and $B$ are constants) is also a solution of the DE.
Just DO it! With k= 5, those functions are $A sinh(5t)+ B cosh(5t)$. Find the second derivative, plug it into the equation and see what happens! With k= -5, you haved $A sinh(-5t)+ B cosh(-5t)$. Since sinh is an odd function and cosh is an even function, that is just $-A sinh(5t)+ B cosh(5t)$ which is really the same thing since A itself can be positive or negative.

[/quote]Come up with a second order DE for which $y=sinkt$ is a solution for the same values of $k$ obtained before.[/QUOTE]
So now youy are working with $y= sin(5t)$. Take the second derivative, stare at it and think!