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The homogeneous solution of a differential equation occurs when you solve the equation without any external driving forces or non-homogeneous parts. For our specific differential equation, we set the non-homogeneous part to zero, resulting in: \[ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} + 2 \frac{\mathrm{d} x}{\mathrm{~d} t} + 2x = 0 \] To find solutions for this kind of equation, we assume a solution of the form \( x_h = e^{\lambda t} \). Plugging this into the homogeneous equation yields the characteristic equation: \[ \lambda^2 + 2\lambda + 2 = 0 \] Solving the characteristic equation using the quadratic formula gives us: - \( \lambda = -1 + i \) - \( \lambda = -1 - i \) These complex roots manifest in the general solution for the homogeneous equation: \[ x_h(t) = e^{-t}(C_1 \cos t + C_2 \sin t) \] where \(C_1\) and \(C_2\) are constants determined by initial conditions.

• It represents the natural behavior of the system without external influences.
• The solution typically describes oscillatory decay because of the exponential term \(e^{-t}\).

Finding a particular solution for a differential equation involves solving for the response caused directly by the non-homogeneous part. In our case, the non-homogeneous part is \( \cos t \). We assume a solution of the form: \[ x_p(t) = A \cos t + B \sin t \] This assumption is appropriate because the non-homogeneous term itself is a cosine function. Upon substitution into the differential equation, and matching coefficients, we find: - \( A + 2B = 1 \) - \( -B + 2A = 0 \) Solving these equations gives the particular constants: - \( A = \frac{2}{5} \) - \( B = \frac{1}{5} \) Hence, the particular solution is: \[ x_p(t) = \frac{2}{5} \cos t + \frac{1}{5} \sin t \] This term represents the part of the solution that directly responds to the external forcing function and remains as time progresses.

The steady-state solution of a differential equation refers to the behavior of the system as time approaches infinity. In many practical cases, it's the solution that persists after transient effects die out. For our problem, the particular solution acts as the steady-state solution: \[ x_{ss}(t) = \frac{2}{5} \cos t + \frac{1}{5} \sin t \] To convert this into amplitude-phase form, we use the identity: \[ A \cos t + B \sin t = R \cos(t - \phi) \] Where: - \( R = \sqrt{A^2 + B^2} \) - \( \tan \phi = \frac{B}{A} \) The calculations yield: - Amplitude \( R = \sqrt{\left(\frac{2}{5}\right)^2 + \left(\frac{1}{5}\right)^2} = \frac{1}{\sqrt{5}} \) - Phase shift \( \phi = \tan^{-1}\left(\frac{1}{2}\right) \) So, the steady-state can be represented as: \[ x_{ss}(t) = \frac{1}{\sqrt{5}} \cos(t - \tan^{-1}(\frac{1}{2})) \] This solution shows how the system will oscillate indefinitely under the influence of the external cosine force.

The transient solution is derived from the homogeneous solution, representing the short-term behavior of the system before it settles into the steady state. For our specific example, the transient solution is given by: \[ x_{tr}(t) = e^{-t}((x_0 - \frac{2}{5}) \cos t + (x_1 + x_0 - \frac{3}{5}) \sin t) \] This part of the solution contains terms associated with the decay factor \( e^{-t} \), ensuring it diminishes over time. Key aspects include:

• The decay is due to the exponential term \( e^{-t} \), which decreases to zero as \( t \to \infty \).
• Initially, it might dominate the system's response, but its influence wanes as time progresses.
• The constants \( C_1 \) and \( C_2 \) are adapted according to the initial conditions, making this solution unique to the specific initial state of the system.

The transient solution is crucial for understanding how the system evolves at the onset and how initial conditions affect early behavior.