QUESTION 8 [TOTAL MARKS: 21 MARKS] A machine with a damping system is modelled as a spring-mass-damper system. The displacement \( x(t) \) satisfies the non-homogenous second order differential equation: \[ \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+5 x=10 \cos 2 t \] \( x \) is the displacement (in metres) and \( t \) is the time taken (in seconds). Determine the general function of \( x \). [21 marks]

\[ \textbf{Step 1: Solve the Homogeneous Equation} \] The homogeneous part of the differential equation is \[ \frac{d^2 x}{dt^2} + 2 \frac{dx}{dt} + 5x = 0. \] We assume a solution of the form \[ x = e^{rt}. \] Substituting into the homogeneous equation gives the characteristic equation: \[ r^2 + 2r + 5 = 0. \] Solve the quadratic using the quadratic formula: \[ r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i. \] Thus, the general solution for the homogeneous equation is: \[ x_h(t) = e^{-t}\left(C_1 \cos 2t + C_2 \sin 2t\right), \] where \( C_1 \) and \( C_2 \) are arbitrary constants. \[ \textbf{Step 2: Find a Particular Solution} \] Given the non-homogeneous differential equation: \[ \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+5 x=10 \cos 2t, \] a reasonable guess for a particular solution is: \[ x_p(t)=A\cos 2t+B\sin 2t. \] Compute the derivatives: \[ x_p'(t) = -2A\sin 2t + 2B\cos 2t, \] \[ x_p''(t) = -4A\cos 2t - 4B\sin 2t. \] Substitute \( x_p \), \( x_p' \), and \( x_p'' \) into the differential equation: \[ (-4A\cos 2t - 4B\sin 2t) + 2(-2A\sin 2t + 2B\cos 2t) + 5(A\cos 2t+B\sin 2t)=10\cos 2t. \] Simplify by combining like terms: \[ (-4A\cos 2t - 4B\sin 2t) + (-4A\sin 2t + 4B\cos 2t) + (5A\cos 2t+5B\sin 2t)=10\cos 2t. \] Group the terms for \(\cos 2t\) and \(\sin 2t\): \[ \text{Coefficient of } \cos 2t: \quad -4A + 4B + 5A = (A + 4B), \] \[ \text{Coefficient of } \sin 2t: \quad -4B - 4A+ 5B = (B - 4A). \] Thus, we obtain: \[ (A+4B)\cos 2t + (B-4A)\sin 2t = 10\cos 2t. \] For the equality to hold for all \( t \), equate the coefficients: \[ A+4B=10, \quad B-4A=0. \] Solve the second equation: \[ B = 4A. \] Substitute \( B = 4A \) into the first equation: \[ A + 4(4A) = A + 16A = 17A = 10 \quad \Longrightarrow \quad A = \frac{10}{17}. \] Then, \[ B = 4A = \frac{40}{17}. \] So, the particular solution is: \[ x_p(t)=\frac{10}{17}\cos 2t+\frac{40}{17}\sin 2t. \] \[ \textbf{Step 3: Write the General Solution} \] The general solution \( x(t) \) is the sum of the homogeneous and particular solutions: \[ x(t)=x_h(t)+x_p(t)=e^{-t}\left(C_1 \cos 2t + C_2 \sin 2t\right)+\frac{10}{17}\cos 2t+\frac{40}{17}\sin 2t. \]