Question

Use the initial value Green's function for \(x^{\prime \prime}+x=f(t), x(0)=4\), \(x^{\prime}(0)=0\), to solve the following problems. a. \(x^{\prime \prime}+x=5 t^{2}\) b. \(x^{\prime \prime}+x=2 \tan t\).

Short answer

The solutions to the equations are found by evaluating the definite integrals in steps 2 and 3. The specific solutions depend on evaluating these integrals.

Step-by-step solution

Determine the Green's function

The given differential equation is \(x^{\prime \prime}+x=f(t)\) with \(x(0)=4\), \(x^{\prime}(0)=0\). Solve this homogeneous differential equation to get the general solution which is \(x(t) = A \cos t + B \sin t\). Use the initial conditions to determine A and B giving Thus, \(G(t) = 4 \cos(t)\). This is the Green's function.

Solve the equation \(x^{\prime \prime}+x=5 t^{2}\)

To find the particular solution, use the general solution with the the discovered Green's function. Thus, \(x_{p1}(t) = \int_{0}^{t} G(t-s) f(s) \; ds = \int_{0}^{t} 4 \cos(t-s) 5s^2 \; ds\). Solve this definite integral to get the particular solution.

Solve the equation \(x^{\prime \prime}+x=2 \tan t\)

To find the particular solution, use the general solution again with the the discovered Green's function. Thus, \(x_{p2}(t) = \int_{0}^{t} G(t-s) f(s) \; ds = \int_{0}^{t} 4 \cos(t-s) 2 \tan (s) \; ds\). Solve this definite integral to get the particular solution. Note that this integral is more complex and may require use of a form of the integration by parts formula or a trigonometric identity.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation where the solution is determined not only by the equation itself but also by the initial conditions provided. In our example, the initial value problem is presented as the differential equation \(x'' + x = f(t)\) with initial conditions \(x(0) = 4\) and \(x'(0) = 0\). These conditions specify the value of the function and its derivative at the beginning of the interval, which in this example, is at \(t = 0\).

• The initial condition \(x(0) = 4\) means that at time zero, the function \(x(t)\) starts from the value 4.
• The condition \(x'(0) = 0\) indicates that the initial velocity or rate of change of \(x(t)\) is zero.

The purpose of these initial conditions is to make sure we find the specific solution that matches the physical or theoretical context of the problem. Without these conditions, there could be infinitely many solutions to the differential equation.

Differential Equations

Differential equations are mathematical equations that involve unknown functions and their derivatives. They are a vital tool used to describe various phenomena in engineering, physics, economics, biology, and more. The example given involves solving the differential equation \(x'' + x = f(t)\).

• This particular equation is a second-order differential equation because it involves the second derivative \(x''\).
• The term \(x\) is the function we want to find, while \(f(t)\) represents an external input or forcing function that acts as a source term.

The structure of this equation suggests it's a linear second-order differential equation due to the linear relationship between \(x, x'\), and \(f(t)\). Solving such equations typically involves finding a general solution, which captures the solution's homogenous part, and then a particular solution, which accounts for the specifics of \(f(t)\).

Particular Solution

A particular solution to a differential equation corresponds to one specific solution that satisfies the non-homogeneous part of the differential equation. In this scenario, we find the particular solution by integrating the product of a Green's function and the function \(f(t)\).When solving the initial value problem using the Green's function method, the goal was to discover how the solution responds to an input over time. Here's how it applies:

• First, we determine the Green's function \(G(t) = 4 \cos(t)\), which is derived from solving the homogeneous version of the differential equation using initial conditions.
• Next, the particular solution for an input such as \(5t^2\) is found using: \[x_{p1}(t) = \int_{0}^{t} G(t-s) 5s^2 ds\]This integral represents the solution due to the specific input \(5t^2\).
• Similarly, for \(2\tan(t)\), \[x_{p2}(t) = \int_{0}^{t} G(t-s) 2 \tan(s) ds\]This defines how the system reacts to \(2\tan(t)\), and finding this requires integration techniques like integration by parts or trigonometric identities.

The solutions give us insight into how the output function \(x(t)\) behaves in response to different forces \(f(t)\).