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]

\[ \textbf{Step 1. Solve the Homogeneous Equation} \] We first consider the homogeneous equation associated with the given differential equation: \[ \frac{d^{2} x}{d t^{2}} + 2 \frac{d x}{d t} + 5 x = 0. \] Assume a solution of the form \( x(t) = e^{rt} \). Substituting into the homogeneous equation gives: \[ r^2 e^{rt} + 2r e^{rt} + 5e^{rt} = 0. \] Dividing through by \( e^{rt} \) (which is never zero) yields the characteristic equation: \[ r^2 + 2r + 5 = 0. \] Solve the quadratic using the quadratic formula: \[ r = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(5)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i. \] Thus, the general solution of the homogeneous equation is \[ x_h(t) = e^{-t} \Bigl(C_1 \cos 2t + C_2 \sin 2t \Bigr), \] where \( C_1 \) and \( C_2 \) are arbitrary constants. \[ \textbf{Step 2. Find a Particular Solution} \] We now seek a particular solution \( x_p(t) \) to the non-homogeneous differential equation \[ \frac{d^{2} x}{d t^{2}} + 2 \frac{d x}{d t} + 5 x = 10 \cos 2t. \] Since the forcing term is \( 10 \cos 2t \) and the homogeneous solution is multiplied by \( e^{-t} \), there is no resonance. Therefore, we propose a particular solution of the form: \[ x_p(t) = A \cos 2t + B \sin 2t. \] Differentiate \( x_p(t) \): \[ x_p'(t) = -2A \sin 2t + 2B \cos 2t, \] \[ x_p''(t) = -4A \cos 2t - 4B \sin 2t. \] Substitute \( x_p(t) \), \( x_p'(t) \), and \( x_p''(t) \) into the differential equation: \[ (-4A \cos 2t - 4B \sin 2t) + 2(-2A \sin 2t + 2B \cos 2t) + 5(A \cos 2t + B \sin 2t). \] Simplify the expression by grouping like terms. For the \( \cos 2t \) terms: \[ -4A \cos 2t + 4B \cos 2t + 5A \cos 2t = ( -4A + 4B + 5A ) \cos 2t = ( A + 4B ) \cos 2t. \] For the \( \sin 2t \) terms: \[ -4B \sin 2t - 4A \sin 2t + 5B \sin 2t = ( -4B - 4A + 5B ) \sin 2t = ( B - 4A ) \sin 2t. \] Thus, the left-hand side becomes: \[ ( A + 4B ) \cos 2t + ( B - 4A ) \sin 2t. \] This must equal the right-hand side \( 10 \cos 2t \). Therefore, we equate the coefficients: \[ A + 4B = 10, \] \[ B - 4A = 0. \] Solve the second equation first: \[ B = 4A. \] Substitute \( B = 4A \) into the first equation: \[ A + 4(4A) = A + 16A = 17A = 10 \quad \Longrightarrow \quad A = \frac{10}{17}. \] Then, \[ B = 4A = \frac{40}{17}. \] Thus, 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 of the differential equation is the sum of the homogeneous and particular solutions: \[ x(t) = x_h(t) + x_p(t) = e^{-t} \Bigl(C_1 \cos 2t + C_2 \sin 2t \Bigr) + \frac{10}{17} \cos 2t + \frac{40}{17} \sin 2t. \]