Question

Consider the initial value problem \(y^{\prime \prime}+0.1 y^{\prime}+y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0\) where \(f(t)\) is the same as in Problem \(13 .\) a. Plot the graph of the solution. Use a large enough value of \(n\) and a long enough \(t\) -interval so that the transient part of the solution has become negligible and the steady state is clearly shown. b. Estimate the amplitude and frequency of the steady-state part of the solution. c. Compare the results of part b with those from Section 3.8 for a sinusoidally forced oscillator.

Short answer

#Short Answer# The general solution of the given initial value problem is: \(y(t) = (\frac{1}{3}) e^{-0.05 t} \cos t + 0.05 e^{-0.05 t} \sin t -\frac{1}{3} \cos(t) - \frac{1}{8}\cos(\sqrt{3}t)\). The estimated amplitude and frequency of the steady-state part of the solution are approximately 0.3136 and 1, respectively. These values should be similar to those obtained for a sinusoidally forced oscillator with similar parameters when exposed to the same forcing function.

Step-by-step solution

Determine the form of the forcing function f(t)

From problem 13, we know that the forcing function f(t) is given by: \(f(t) = \frac{1}{2} \cos (t) + \frac{1}{2} \cos (\sqrt{3} t)\).

Solve the homogeneous equation

The homogeneous equation corresponding to the given initial value problem is: \(y^{\prime \prime} + 0.1 y^{\prime} + y = 0\). To solve this homogeneous equation, we assume \(y_h = e^{rt}\), where 'r' is the root, and then plug it into the equation to get the characteristic equation: \(r^2 + 0.1 r + 1 = 0\). Solve the quadratic to obtain the complex roots conjugate pair \(r_{1,2} = -0.05 \pm i\), and form the general homogeneous solution: \(y_h(t) = c_1 e^{-0.05 t} \cos t + c_2 e^{-0.05 t} \sin t\).

Find a particular solution for the given function f(t)

The forcing function \(f(t)\) consists of two cosine functions, so we assume a particular solution of the form: \(y_p(t) = A\cos(t) + B\sin(t) + C\cos(\sqrt{3}t) + D\sin(\sqrt{3}t)\). Taking the first and second derivatives of the assumed solution and plugging them into the given equation, we can solve for coefficients A, B, C, and D. After solving, the particular solution is \(y_p(t) = -\frac{1}{3} \cos(t) - \frac{1}{8}\cos(\sqrt{3}t)\).

Form the general solution and initial conditions

Now we can form the general solution by adding the homogeneous solution and the particular solution: \(y(t) = y_h(t) + y_p(t) = c_1 e^{-0.05 t} \cos t + c_2 e^{-0.05 t} \sin t -\frac{1}{3} \cos(t) - \frac{1}{8}\cos(\sqrt{3}t)\). Apply the initial conditions: \(y(0) = 0\) and \(y^{\prime}(0) = 0\) to solve for \(c_1\) and \(c_2\). Using the initial conditions, we get that \(c_1=\frac{1}{3}\) and \(c_2=0.05\).

Plot the graph of the solution and estimate the steady-state

Use a numerical method (like Euler's method) to plot the derived solution. Make sure to use enough time points to clearly show the transient and steady-state parts of the response. The steady-state part of the solution is the same as \(y_p(t)\) and it can be written as: \(y_{ss}(t) = -\frac{1}{3} \cos(t) - \frac{1}{8}\cos(\sqrt{3}t)\). From the plot, estimate the amplitude and frequency of the steady-state part of the solution. The amplitude can be estimated as \(\sqrt{(\frac{1}{3})^2 + (\frac{1}{8})^2} = \frac{\sqrt{73}}{24} \approx 0.3136\). The frequency can be estimated as the main component's frequency, which is the same as the forcing function's frequency, which is 1 (from the \(cos(t)\) term).

Compare the results with those from a sinusoidally forced oscillator

Compare the obtained amplitude and frequency from the initial value problem to those from a sinusoidally forced oscillator from Section 3.8. In general, the response of a sinusoidally forced oscillator depends on its damping ratio, natural frequency, and forcing amplitude and frequency. The amplitude and frequency obtained in this exercise should be similar to those obtained for a damped oscillator with similar parameters when exposed to the same forcing function.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations specifies a function's value and its derivatives at a starting point, typically noted as time zero. This type of problem helps us find a unique solution to a differential equation.
In our exercise, the IVP is given by:

• Differential equation: \(y^{\prime \prime}+0.1 y^{\prime}+y=f(t)\), with \(f(t)\) invoking an external force.
• Initial conditions: \(y(0)=0\) and \(y^{\prime}(0)=0\).

These initial conditions mean that both the function and its derivative start with zero, steering the entire solution path.
This specifics of our IVP allow us to solidly define the unique trajectory our solution will follow as it evolves over time. This sets the stage for analyzing how the system behaves when exposed to external forces, or in this case, the function \(f(t)\).

Homogeneous Solution

In the context of differential equations, the Homogeneous Solution addresses the scenario where there is no external force acting on the system. For this exercise, the problem is represented by:\[y^{\prime \prime} + 0.1 y^{\prime} + y = 0\]We solve this homogeneous equation by first assuming a solution of the form \(y_h = e^{rt}\). We then find the characteristic equation by substituting \(y_h\) into the differential equation:\[r^2 + 0.1r + 1 = 0\]Solving this, we find complex roots, indicating oscillatory behavior dampened over time. These roots are given by \(r_{1,2} = -0.05 \pm i\). Thus, the general homogeneous solution is:\[y_h(t) = c_1 e^{-0.05 t} \cos t + c_2 e^{-0.05 t} \sin t\]This tells us that the system will naturally oscillate, decreasing exponentially due to the \(e^{-0.05t}\) term. Understanding the homogeneous solution helps illustrate how the system behaves internally without external influences.

Particular Solution

The Particular Solution in a differential equation deals with how the system reacts to a specific external force, \(f(t)\). In this problem, the external force is a combination of cosine functions:\[f(t) = \frac{1}{2} \cos (t) + \frac{1}{2} \cos (\sqrt{3} t)\]To find the particular solution, we assume a form \(y_p(t)\) that resembles \(f(t)\):\[y_p(t) = A\cos(t) + B\sin(t) + C\cos(\sqrt{3}t) + D\sin(\sqrt{3}t)\]By substituting \(y_p(t)\) into the differential equation, we solve for constants \(A, B, C,\) and \(D\). After some calculations, the particular solution is:\[y_p(t) = -\frac{1}{3} \cos(t) - \frac{1}{8}\cos(\sqrt{3}t)\]This form shows us how the external forces specifically alter the system's response. The particular solution essentially tailors the ways the system is influenced directly by \(f(t)\).

Steady-State Response

The Steady-State Response represents the behavior of the system as time progresses and transients fade away. It's composed largely of the particular solution because the homogeneous solution's impact diminishes due to exponential decay.
In this exercise, the steady-state part is effectively:\[y_{ss}(t) = -\frac{1}{3} \cos(t) - \frac{1}{8}\cos(\sqrt{3}t)\]This response indicates that the system will continue oscillating according to these frequencies and amplitudes indefinitely once the transient effects vanish. Estimating amplitude and frequency from the steady-state solution involves evaluating the size and periodicity of its terms.

• Amplitude: Calculated by the formula \(\sqrt{(\frac{1}{3})^2 + (\frac{1}{8})^2} = \frac{\sqrt{73}}{24} \approx 0.3136\).
• Frequency: Mostly driven by the primary forcing terms, in this case, the main component \(\cos(t)\) indicates a frequency of 1.

Understanding the steady-state response is crucial as it reflects the regular output of the system under constant excitation in the real world.