Question

Find the solution of the given initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related. \(y^{\mathrm{iv}}+5 y^{\prime \prime}+4 y=1-u_{\pi}(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=0\)

Short answer

Short Answer: The solution to the given initial value problem is \(y(t) = -0.5\cos(2t) + \begin{cases} 1 & 0\le t<\pi \\ 0 & t\ge\pi \end{cases}. \) The forcing function (\(1-u_\pi(t)\)) is active when \(0\le t<\pi\) and is turned off when \(t\ge\pi\). The solution vibrates at different frequencies depending on the time range, displaying oscillatory behavior. Overall, the solution is in sync with the forcing function, indicating that the forcing function controls the system's behavior.

Step-by-step solution

Find the complementary solution

To find the complementary solution, we must first find the roots of the characteristic equation. The characteristic equation for this ODE is: \(r^{4} + 5r^{2} + 4 = 0\) Solve for the roots \(r\): \(r^2 = -1, -4\) \(r = \pm i, \pm 2i\) The complementary solution is of the form: \(y_c(t) = C_1 \cos(t) + C_2 \sin(t) + C_3 \cos(2t) + C_4 \sin(2t)\)

Find the particular integral

To find the particular integral, we need to find a function that satisfies the given differential equation with the forcing function. Since the forcing function is a piecewise function, we can split this into two cases: Case 1: \(0 \le t < \pi\) Let \(y_p(t) = k_1\), where \(k_1\) is a constant. Therefore, the forcing function is equal to \(1\). Substituting \(y_p(t)\) into the given ODE, we get: \(k_1 = 1\) Hence, \(y_p(t) = 1\) for \(0 \le t < \pi\). Case 2: \(t \ge \pi\) Let \(y_p(t) = k_2\), where \(k_2\) is a constant. In this case, the forcing function is equal to \(0\). Substituting \(y_p(t)\) into the given ODE, we get: \(k_2 = 0\) Hence, \(y_p(t) = 0\) for \(t \ge \pi\).

Combine the complementary solution and particular integral

Now we can write down the general solution of the given ODE as follows: $y(t) = y_c(t) + y_p(t) = C_1 \cos(t) + C_2 \sin(t) + C_3 \cos(2t) + C_4 \sin(2t) + \begin{cases} 1 & 0\le t<\pi \\ 0 & t\ge\pi \end{cases}$

Apply the initial conditions

Given the initial conditions \(y(0)=0\), \(y'(0)=0\), \(y''(0)=0\), and \(y'''(0)=0\), plug in values and solve for the constants \(C_1\), \(C_2\), \(C_3\), \(C_4\). From \(y(0)=0\): \(C_1 = 0\) From \(y'(0)=0\): \(C_2 = 0\) From \(y''(0)=0\): \(C_3 = -0.5\) From \(y'''(0)=0\): \(C_4 = 0\) Thus, the solution is: $y(t) = -0.5\cos(2t) + \begin{cases} 1 & 0\le t<\pi \\ 0 & t\ge\pi \end{cases} $

Analyze the solution and forcing function

Plot both the solution (\(y(t)\)) and the forcing function (\(1-u_\pi(t)\)) and observe the relationship between them. Notice that the forcing function is active when \(0\le t<\pi\) and turned off when \(t\ge\pi\), meaning the solution to the differential equation vibrates with different frequencies for different segments of time. Overall, the solution is in sync with the forcing function, resulting in oscillations. In summary, the forcing function dictates the behavior of the solution and, in turn, the system shows oscillatory response depending on the forcing function.

Key concepts

Complementary Solution

When dealing with a linear differential equation, one part of the solution is found by ignoring the non-homogeneous term (the forcing function), which leads us to the complementary solution or homogeneous solution. This solution represents the natural behavior of the system without any external influence.

In our exercise, the complementary solution, denoted as \( y_c(t) \), was obtained by solving the characteristic equation associated with the homogeneous part of the differential equation. The characteristic equation is a polynomial whose roots indicate the type of solutions (for example, sinusoidal functions for imaginary roots) that make up \( y_c(t) \).

To find \( y_c(t) \), the roots of the characteristic equation \( r^{4} + 5r^{2} + 4 = 0 \) were determined to be imaginary numbers \( \pm i, \pm 2i \), indicating oscillatory motion. Consequently, the complementary solution consists of a combination of sine and cosine functions with different frequencies corresponding to these roots. The final form of the complementary solution, incorporating constants \( C_1, C_2, C_3, \) and \( C_4 \), is a reflection of the system's potential motion in the absence of external forces.

Particular Integral

While the complementary solution reflects the system's inherent behavior, the particular integral accounts for the influence of the non-homogeneous part of the equation, which in this case is the forcing function. The particular integral, denoted as \( y_p(t) \), is a specific solution that fits the non-homogeneous equation exactly.

To find \( y_p(t) \), the exercise recognized the piecewise nature of the forcing function and tackled it accordingly. For the time interval before \( \pi \), where the forcing function is a constant value of 1, a constant value of \( k_1 = 1 \) for \( y_p(t) \) was a fitting solution. For intervals where the function turned off (\( t \ge \pi \)), another constant, \( k_2 = 0 \), satisfied the non-homogeneous equation.

Thus, the particular integral for different segments of time is determined based on the behavior of the external force, allowing us to construct a full solution tailored to the specific input.

Characteristic Equation

The characteristic equation is central to finding the complementary solution in the process of solving differential equations. It is a polynomial equation derived from the differential equation by assuming a solution of the form \( e^{rt} \), where \( r \) represents the roots of the equation.

In this instance, the characteristic equation \( r^{4} + 5r^{2} + 4 = 0 \) was formulated from the coefficients of the differential equation's terms. Solving this equation provides us with the roots (in our exercise, \( \pm i, \pm 2i \)) that govern the form of the complementary solution.

It's essential to grasp the role of this equation, as it directs us to the type of functions (exponential, sine, cosine) that compose the homogeneous solution. The roots tell us not only about the oscillatory or exponential nature of the system but also give us insights into its stability and response characteristics. The characteristic equation ultimately shapes the complementary solution, which is then fine-tuned using initial conditions to match the specific requirements of the problem at hand.