Answer:
k = -10/9 and k = 10/9
Step-by-step explanation:
given y = cos(kt) and the differential equation 81y'' = -100y
y' = -ksin(kt)
y'' = -k²cos(kt)
Substituting the value of y and y'' in the differential equation we have;
81 (-k²cos(kt))= -100 (cos(kt))
-81k²cos(kt)) = -100cos(kt))
-81k² = -100
k² = 100/81
k = ±[tex]\sqrt{\frac{100}{81} }[/tex]
k = ±10/9
k = -10/9 and k = 10/9
