Final answer:
The differential equation y" – 10y' + 9tan(y) = u(t) is linearized at y = 0 and u = 0 to y" – 10y' + 9y = u(t). The transfer function is obtained by taking the Laplace transform and the steady state value for u(t) = 5t is determined by using the final value theorem.
Step-by-step explanation:
To linearize the given nonlinear differential equation y" – 10y' + 9tan(y) = u(t) at the equilibrium point where y = 0 and u = 0, we can use a Taylor series expansion about the point y = 0. Since the tangent function tan(y) can be approximated as y when y is close to zero, the linearized form of the differential equation around the equilibrium point is y" – 10y' + 9y = u(t). The transfer function model of the linearized system can be found by taking the Laplace transform of each term in the linearized differential equation and solving for Y(s)/U(s), assuming zero initial conditions.
Given the input signal u(t) = 5t, the steady state value of the system, as t approaches infinity, can be evaluated using the final value theorem. The theorem states that the steady state value of the output is the limit as s approaches zero of s times the transfer function Y(s)/U(s) times the Laplace transform of the input U(s). Therefore, without solving the linearized differential equation, we can use this theorem to find the steady state value by analyzing the transfer function and the known input.