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Find the solution to the initial value problem \(d^{2} x / d t^{2}+4 d x / d t+13 x=\) \(f(t), x(0)=0, x^{\prime}(0)=0\), where \(f(t)=\) \(\left\\{\begin{array}{l}1,0 \leq t<\pi \\ 1-t, \pi \leq t<2 \pi \\ 0, t \geq 2 \pi\end{array}\right.\)

Short Answer

Expert verified
The solution involves solving separately for each interval: the solutions are combinations of the homogeneous solution and particular solutions specific to each interval.

Step by step solution

01

- Solve the homogeneous equation

First, solve the corresponding homogeneous equation given by \[\frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 13x = 0\]. Find the characteristic equation by substituting \(x = e^{rt}\), leading to \[r^2 + 4r + 13 = 0\]. Solve this quadratic equation to get the roots: \[r = -2 \pm 3i\]. Hence, the general solution to the homogeneous equation is \[x_h(t) = e^{-2t}(C_1 \cos(3t) + C_2 \sin(3t))\].
02

- Solve for the particular solution

To solve for the particular solution, use the method of undetermined coefficients or the method of variation of parameters. Since the non-homogeneous term \(f(t)\) is piecewise, break the solution into intervals based on \(f(t)\). Solve for each interval separately: \(0 \leq t < \pi\), \(\pi \leq t < 2 \pi\), and \(t \geq 2 \pi\).
03

- Solve for 0 ≤ t < π

For \(0 ≤ t < \pi\), \(f(t) = 1\). The particular solution is assumed to be of the form \(x_p(t) = A\), where \(A\) is a constant. Substitute into the differential equation: \[0 + 0 + 13A = 1 \Rightarrow A = \frac{1}{13}\]. So, the particular solution in this interval is \(x_p(t) = \frac{1}{13}\).
04

- Solve for π ≤ t < 2π

For \(π ≤ t < 2π\), \(f(t) = 1 - t\). Assume a particular solution of the form \(x_p(t) = B + Ct\). Substitute into the differential equation: \[0 + 4C + 13(B + Ct) = 1 - t\]. This gives two equations: \[4C = -1 \Rightarrow C = -\frac{1}{4}\] and \[13B = 1 \Rightarrow B = \frac{1}{13}\]. So, the particular solution in this interval is \(x_p(t) = \frac{1}{13} - \frac{t}{4}\).
05

- Solve for t ≥ 2π

For \(t ≥ 2π\), \(f(t) = 0\). The solution is homogeneous, so \(x_p(t) = 0\). Hence, the solution should be of the form \(x_h(t)\) found in Step 1.
06

- Combine solutions with initial conditions

Combine the particular and homogeneous solutions for each interval and apply the initial conditions \(x(0) = 0\) and \(x'(0) = 0\) to find the constants \(C_1\) and \(C_2\). Using the boundary conditions at \(t = \pi\) and \(t = 2Ï€\), match the solutions and their derivatives to ensure continuity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

characteristic equation
The characteristic equation is a fundamental concept in solving linear differential equations with constant coefficients. It arises when we look for solutions of the form \(x = e^{rt}\). This transforms the differential equation into an algebraic equation, which we can solve for \(r\). In our case, from the homogeneous equation \(\frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 13x = 0\), substituting \(x = e^{rt}\) gives us the characteristic equation as follows:

\[r^2 + 4r + 13 = 0\] We solve this quadratic equation using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For \(a = 1\), \(b = 4\), and \(c = 13\), it solves to give:\[r = -2 \pm 3i\]
These roots are complex, indicating that the solutions will involve exponential functions combined with sine and cosine terms.
homogeneous equation
The homogeneous equation is a differential equation set to zero (i.e., no external forcing function). For our problem, it is:

\[\frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 13x = 0\]

The solutions to the characteristic equation \(r = -2 \pm 3i\) give rise to the homogeneous solution:

\[x_h(t) = e^{-2t}(C_1 \cos(3t) + C_2 \sin(3t))\]

Here \(C_1\) and \(C_2\) are constants to be determined by the initial and boundary conditions. This form combines exponential decay, \(e^{-2t}\), and oscillatory components, \(\cos(3t)\) and \( \sin(3t)\). These functions represent the general behavior of the system without any external input.
particular solution
The particular solution deals with finding a specific solution to the non-homogeneous differential equation. In this case, the non-homogeneous term is a piecewise function \(f(t)\). We use different methods to find this solution, like the method of undetermined coefficients or variation of parameters. Since \(f(t)\) is piecewise:
  • For \(0 \leq t < \pi\): The forcing function \(f(t) = 1\). We assume a constant particular solution \(x_p(t) = \frac{1}{13}\).

  • For \(\pi \leq t < 2\pi\): Here \(f(t) = 1 - t\). We assume the particular solution \(x_p(t) = \frac{1}{13} - \frac{t}{4}\).

  • For \(t \geq 2\pi\): \(f(t) = 0\), which makes the particular solution \(x_p(t) = 0\) and only the homogeneous solution contributes to the result.

Combining these with boundary conditions ensures the solution is continuous and differentiable across intervals.
piecewise function
A piecewise function defines different expressions for different intervals of the independent variable. In our problem, the function \(f(t)\) is defined as:

\[ f(t) = \begin{cases}1, & 0 \leq t < \pi \1 - t, & \pi \leq t < 2\pi \0, & t \geq 2\pi\end{cases} \]

This means we need to solve the differential equation for each interval separately:
  • For \(0 \leq t < \pi\): \(f(t) = 1\).

  • For \(\pi \leq t < 2\pi\): \(f(t) = 1 - t\).

  • For \(t \geq 2\pi\): \(f(t) = 0\).

The solutions for these intervals are then combined together, ensuring they fit together smoothly at \(t = \pi\) and \(t = 2\pi\). Piecewise functions make our solutions more flexible and realistic, allowing us to model complex real-world situations.

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