derv:=proc(neqn,t,u,ut)
#
# Procedure derv computes the derivative vector
# of the 1 x 1 ODE problem
#
# Type variables
  local alpha, lambda:
#
# Problem parameters
  alpha:=1.0:
  lambda:=1.0:
#
# Derivative vector
  ut[1]:=lambda*exp(-alpha*t)*u[1]:
#
# End of derv
  end: