If y= A cos kt + B sin kt, where A, B and k are constants, show that y" + k 2y =0

Question

If y= A cos kt + B sin kt, where A, B and k are constants, show that y" + k ²y =0

Answer 1

Given the guess for y(t) = Acos(kt) + Bsin(kt), we can compute the second derivative of y(t) and determine if the differential equation is satisfied...

y(t) = A*cos(kt) + B*sin(kt)

y'(t) = -A*k*sin(kt) + B*k*cos(kt)

y"(t) = -A*k²*cos(kt) - B*k²*sin(kt) = -k² * ( A*cos(kt) + B*sin(kt) )

We can see that y"(t) = -k²*y(t) which leads to...

y" + k²y = 0