1. 2.26 / 5.26 Points
DETAILS MY NOTES PREVIOUS ANSWERS ASK YOUR TEACHER
SCalcET9 9.1.016.
(a) For what values of k does the function $y = \cos(kt)$ satisfy the differential equation $25y'' = -81y$? (Enter your answers as a comma-separated list.)
$k = \frac{9}{5}, -\frac{9}{5}$
(b) For those values of k, verify that every member of the family of functions $y = A \sin(kt) + B \cos(kt)$ is also a solution.
We begin by calculating the following.
$y = A \sin(kt) + B \cos(kt) \rightarrow y' = Ak \cos(kt) - Bk \sin(kt) \rightarrow y'' = \boxed{}$
Note that the given differential equation $25y'' = -81y$ is equivalent to $25y'' + 81y = \boxed{0}$
Now, substituting the expressions for y and $y''$ above and simplifying, we have
LHS = $25y'' + 81y = 25(\boxed{x}) + 81(A \sin(kt) + B \cos(kt))$
$= -25(\boxed{x}) - 25Bk^2 \cos(kt) + 81A \sin(kt) + 81B \cos(kt)$
$= (81 - 25k^2)(\boxed{}) + (81 - 25k^2) B \cos(kt)$
$= 0$
since for all value of k found above, $k^2 = \frac{81}{25}$