Originally Posted by

**jordanrs** Stuck on the following problem, unsure of how to go around it

1a.

For what none values of k does the function $\displaystyle y=sinh (kt)$ satisfy the differential equation

$\displaystyle y'' - 25y = 0$

1b.

for those values of k, show that every member of the family of functions $\displaystyle Asinh(kt) + Bcosh(kt)$ (where A and B are constants) is also a solution to the differential equation.

1c.

come up with a second order differntial equation for which $\displaystyle y=sinh (kt)$is a solution for the same values of k obtained in part (a)

im usually ok with differentail equations but unsure when it comes to the worded versions

for 1a i tryed the following but werent sure where else to go

$\displaystyle y=sinh (kt)$

$\displaystyle y'' = k^2sinh(kt)$

$\displaystyle 25y = k^2sinh(kt)$

$\displaystyle 25sinh(kt) = k^2sinh(kt)$

$\displaystyle sinh(kt)(25-k^2) = 0$

so one possible solution is k = +/- 5 but im not sure if that is the right thing to do

Any help is most appreciated