In each of 12 through 15, use the method of to transform the given equation into the form [p(x)y ] + q(x)y = 0.(1 x2)y xy+ 2y = 0 (Chebyshev equation)

Displacement, velocity and acceleration Ex. a particle moves back and forth along a straight line. Its displacement at time t ( t is in seconds, s(t)) is in feet is s(t)=t^3 12t^2 +36t a. What is the particle’s velocity at time t v(t) = s’(t) =3t^2 24t + 36 ft/sec b. When is the particle at rest Solve v(t)= 3t^2 24t + 36 = 0 Divide by three t^2 8t + 12 = 0 factor (t 2)(t + 6) = 0 t = 2 sec, 6 sec c. what is the initial velocity Plug in 0 for t v(0) = 36 ft/sec d. what is the displacement at t =2 and t =6 sec Plug in 2 and 6 to the original function s(2) = 32 ft s(6) = 0ft Line graph Graph e. When is