Question

Show that x=Be^(µt) is a solution of mx ̈+kx=0

Answer 1

firstly, try it as a solution

if: x = B e^(µt)
then: x' = µ B e^(µt)
and: x'' = µ^2 B e^(µt)

so
mx'' + kx = B e^(µt) { m µ^2 + k } = 0

does that look like a solution, or is more needed?

m µ^2 + k = 0 is clearly unhelpful