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. \(y^{\prime \prime}+4 y=\sin t-u_{2 \pi}(t) \sin (t-2 \pi) ; \quad y(0)=0, \quad y^{\prime}(0)=0\)

Short answer

To conclude, our process to solve the given initial value problem, draw the graphs of the solution and forcing function, and explain their relationship involved 5 steps: 1. We identified the given equation, forcing function, and initial conditions. 2. We determined the general solution using the method of undetermined coefficients. 3. We applied the initial conditions to find the particular solution. 4. We drew graphs of the solution and the forcing function using a graphing tool. Both graphs had sinusoidal shapes, and the solution displayed a change in behavior at \(t = 2\pi\). 5. We explained the relationship between the graphs by connecting the solution's behavior change with the jump discontinuity in the forcing function present at \(t = 2\pi\). This demonstrated how the solution's behavior is influenced by the forcing function.

Step-by-step solution

Identify the given equation, forcing function, and initial conditions

The given equation is: \(y^{\prime \prime}+4 y=\sin t-u_{2 \pi}(t) \sin (t-2 \pi)\) The forcing function is: \(f(t) = \sin t - u_{2\pi}(t) \sin(t - 2\pi)\) The initial conditions are: \(y(0) = 0\) and \(y'(0) = 0\)

Determine the general solution using the method of undetermined coefficients

We can rewrite the original equation as: \(y^{\prime \prime}+4 y=f(t)\) Since the forcing function consists of a harmonic function and a shifted harmonic function multiplied by a unit step function, we can try to guess the particular solution for \(y_p(t)\). Guess: \(y_p(t) = A\sin t + B\cos t + u_{2\pi}(t)[C\sin(t - 2\pi) + D\cos(t - 2\pi)]\) Take the first and second derivatives of \(y_p(t)\): \(y_p'(t) = A\cos t - B\sin t + u_{2\pi}(t)[C\cos(t - 2\pi) - D\sin(t - 2\pi)]\) \(y_p''(t) = -A\sin t - B\cos t - u_{2\pi}(t)[C\sin(t - 2\pi) + D\cos(t - 2\pi)]\) Now substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the original equation: \(-A\sin t - B\cos t - u_{2\pi}(t)[C\sin(t - 2\pi) + D\cos(t - 2\pi)] + 4[A\sin t + B\cos t + u_{2\pi}(t)[C\sin(t - 2\pi) + D\cos(t - 2\pi)]] = \sin t - u_{2\pi}(t) \sin(t - 2\pi)\) Equating coefficients for each term yields: \(A = -\frac{1}{3}; B = 0; C = \frac{1}{3}; D = 0\) Thus, the particular solution is: \(y_p(t) = -\frac{1}{3}\sin t + u_{2\pi}(t)\left(\frac{1}{3}\sin(t - 2\pi)\right)\) The general solution is the sum of the complementary function and particular solution: \(y(t) = y_c(t) + y_p(t)\)

Apply the initial conditions to find the particular solution

Solve for complementary function and given initial conditions by setting \(y(0) = 0\) and \(y'(0) = 0\): \(0 = y(0) = y_c(0) + y_p(0) = C_1\cos 0 + C_2\sin 0 - \frac{1}{3}\sin 0\) , thus \(C_1 = 0\) \(0 = y'(0) = y_c'(0) + y_p'(0) = -C_1\sin 0 + C_2\cos 0 + \frac{1}{3}\cos 0\) , thus \(C_2 = -\frac{1}{3}\) The particular solution is then: \(y(t) = -\frac{1}{3}\sin t + u_{2 \pi}(t)\left(\frac{1}{3}\sin(t - 2\pi)\right) - \frac{1}{3}\sin t\)

Draw the graphs of the solution and the forcing function

Now, plot the solution \(y(t)\) and the forcing function \(f(t)\) by using your favorite graphing tool. Both graphs will have sinusoidal shapes. The solution will start with the same behavior as the forcing function (the \(\sin t\) part), then change its behavior after \(t = 2\pi\) (due to the presence of the delayed sine wave).

Explain the relationship between the graphs

Both the solution and the forcing function are sinusoidal functions. The forcing function consists of a harmonic function and a shifted harmonic function multiplied by a unit step function. This means that the forcing function has a jump discontinuity at \(t = 2\pi\). The solution has the same behavior as the forcing function, starting with the same behavior as the harmonic forcing function \(\sin t\). Then, after \(t = 2\pi\), the behavior of the solution is modified due to the presence of the unit step function in the forcing function. This shows that the behavior of the solution is influenced by the forcing function, as seen in the graphs.

Key concepts

Differential Equations

Differential equations are powerful tools in mathematics and physics, allowing us to describe the relationship between functions and their derivatives. At their core, differential equations express how a rate of change (a derivative) of one variable is related to the variable itself. The initial value problem presented here is a second-order linear differential equation, characterized by the existence of the second derivative, denoted as \( y'' \), and the fact that the equation equals a function of \( t \).

When dealing with initial value problems, initial conditions, such as \( y(0) = 0 \) and \( y'(0) = 0 \), play an essential role. These conditions specify the state of the system at the time \( t = 0 \) and allow for the determination of a unique solution to the equation. The concept of a unique solution is fundamental to the utility of differential equations in modeling real-world phenomena—each set of conditions corresponds to one specific outcome or behavior of the system being studied.

Forcing Function

A forcing function in the context of a differential equation is an external influence that drives the system, which can be thought of as the input for the system governed by the equation. In the given problem, the forcing function is represented by \( f(t) = \sin t - u_{2\pi}(t) \sin(t - 2\pi) \). This composite function is composed of a continuous sinusoidal wave and a discontinuous wave that begins at \( t = 2\pi \). The unit step function, \( u_{2\pi}(t) \), introduces this discontinuity and represents an abrupt change or 'switch-on' of the second sinusoidal component.

Understanding the forcing function is crucial as it determines the behavior of the system. Before \( t = 2\pi \), the solution to the differential equation will follow a sinusoidal pattern. After this point, due to the nature of the forcing function, the behavior of the solution changes, which can be observed when graphing the solution against the forcing function. The presence of a unit step function in a forcing function often indicates piecewise behavior in the solution, reflecting how the system reacts at different intervals of time.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique for finding a particular solution to non-homogeneous linear differential equations, where the forcing function is a linear combination of functions such as polynomials, exponentials, sines, and cosines. This method involves making an educated guess about the form of the particular solution based on the forcing function's composition, hence 'undetermined coefficients', as we start with unknown coefficients that we need to determine.

To apply this method, one would propose a form for the particular solution, \( y_p(t) \), take its derivatives, and then substitute them back into the differential equation. By matching the coefficients of similar terms on both sides of the equation, one can solve for the unknown coefficients. In our exercise, the particular solution was guessed as a linear combination of sinusoidal functions, one being continuous and the other modulated by a unit step function, reflecting the structure of the forcing function. In this way, the method of undetermined coefficients enables us to extract a particular solution that, when added to the complementary (homogeneous) solution, gives the general solution to the initial value problem.