Question

find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing\(t.\) $$ y^{\prime \prime}+y=0, \quad y(\pi / 3)=2, \quad y^{\prime}(\pi / 3)=-4 $$

Short answer

Answer: The specific solution of the given initial value problem is \(y(t) = 4\cos(t) - 2\sqrt{3}\sin(t)\). As t increases, the graph of this solution will be oscillatory, representing a periodic function with a constant amplitude and frequency.

Step-by-step solution

Solve the homogeneous ODE

First, we need to find the general solution of the given ODE. The characteristic equation corresponding to the ODE is given by: $$ r^2 + 1 = 0 $$ Solving for r, we get complex roots: $$ r_1=i, \quad r_2=-i $$ Since we have complex roots, the general solution of the ODE is given by: $$ y(t) = C_1 \cos(t) + C_2 \sin(t) $$

Use initial conditions to find specific solution

We are given that the initial conditions are \(y(\pi / 3) = 2\) and \(y'(\pi / 3) = -4\). We will use these to find the values of \(C_1\) and \(C_2\) in our general solution. First, let's find the derivative of the general solution: $$ y'(t) = -C_1 \sin(t) + C_2 \cos(t) $$ Now, use the initial conditions: $$ y(\pi / 3) = C_1 \cos(\pi / 3) + C_2 \sin(\pi / 3) = 2 $$ $$ y'(\pi / 3) = -C_1 \sin(\pi / 3) + C_2 \cos(\pi / 3) = -4 $$ Using the trigonometric values \(\cos(\pi / 3) = 1/2\) and \(\sin(\pi / 3) = \sqrt{3}/2\), we get: $$ \frac{C_1}{2} + \frac{\sqrt{3}C_2}{2} = 2 $$ $$ -\frac{\sqrt{3}C_1}{2} + \frac{C_2}{2} = -4 $$ Solving for \(C_1\) and \(C_2\), we get \(C_1 = 4\) and \(C_2 = -2\sqrt{3}\). Thus, the specific solution is given by: $$ y(t) = 4\cos(t) - 2\sqrt{3}\sin(t) $$

Sketch the graph and describe its behavior for increasing t

The solution of the given initial value problem is a combination of sine and cosine functions, both of which are periodic in nature. Therefore, the graph of the solution will also be periodic. As t increases, the function will continue to oscillate between its maximum and minimum values, without any change in amplitude or frequency. To summarize, the behavior of the solution y(t) as t increases will be oscillatory, and the graph will represent a periodic function with a constant amplitude and frequency.

Key concepts

Homogeneous ODE

When working with differential equations, a homogeneous ODE (Ordinary Differential Equation) is one where every term contains either the function itself or its derivatives. In other words, it has no external terms or functions added to it. This makes it distinct because the solution depends purely on initial conditions and system dynamics, not external forces or inputs.

The equation from the exercise is homogeneous:

• The given differential equation is: \( y'' + y = 0 \)
• Each term involves the function \( y \) or its derivatives \( y'' \).

Such equations are fundamental in understanding undisturbed systems in nature, such as harmonic oscillators, where the system's inherent properties solely influence the dynamics.

Characteristic Equation

To solve a homogeneous linear differential equation, we often turn to the characteristic equation. This is derived from the original ODE by assuming a solution of the form \( y = e^{rt} \), where \( r \) is a constant to be determined.

For our specific equation \( y'' + y = 0 \), hypothesizing \( y = e^{rt} \) leads to the characteristic equation:

• \( r^2 + 1 = 0 \)

Solving this helps identify the type of roots, which dictates the form of the solution. For this problem, the roots are complex, specifically \( r_1 = i \) and \( r_2 = -i \). This step is crucial as it provides the exponential basis for constructing the general solution to the differential equation.

Complex Roots

Finding complex roots in the characteristic equation, like in our example \( r^2 + 1 = 0 \), is a key indicator of the solution's behavior. When the roots are complex, they manifest as conjugates, here \( i \) and \( -i \). This results in the solutions taking on an oscillatory nature.

With roots \( r_1 = i \) and \( r_2 = -i \), we express the solution through trigonometric functions:

• \( y(t) = C_1 \cos(t) + C_2 \sin(t) \)

These trigonometric forms replace the real \/ exponential forms due to the imaginary components in the roots, characterizing solutions as oscillations. This arises naturally in problems involving circuits and mechanical vibrations, where the inherent system dynamics lead to periodic solutions.

General Solution

The general solution of a homogeneous ODE with complex roots pulls together the understanding of its core mathematical components. In this problem:

• Based on the roots, \( r_1 = i \) and \( r_2 = -i \), the general solution is modeled as \( y(t) = C_1 \cos(t) + C_2 \sin(t) \).

The constants, \( C_1 \) and \( C_2 \), are determined using initial conditions. For our example, finding these constants allowed us to create a specific solution tailored to the initial values given. Applying the initial conditions \( y(\pi/3) = 2 \) and \( y'(\pi/3) = -4 \) determined:

• \( C_1 = 4 \)
• \( C_2 = -2\sqrt{3} \)

By integrating initial conditions, we transition from a broad set of possible solutions to one unique solution, describing the system comprehensively. The general solution framework thus provides a versatile tool for analyzing dynamics in various scientific and engineering applications.