Question

In each exercise, assume that \(y(t)=C_{1} \sin \omega t+C_{2} \cos \omega t\) is the general solution of \(y^{\prime \prime}+\omega^{2} y=0\). Find the unique solution of the given initial value problem. $$y^{\prime \prime}+16 y=0, \quad y(\pi / 4)=1, \quad y^{\prime}(\pi / 4)=-4$$

Short answer

Answer: The unique solution to the initial value problem is \(y(t) = - \cos(4t)\).

Step-by-step solution

Write down the given general solution

The general solution for the given differential equation is: $$y(t) = C_1 \sin(4t) + C_2 \cos(4t)$$

Find the first derivative of the general solution

To find the first derivative of \(y(t)\), we take the derivative with respect to \(t\): $$y'(t) = 4C_1 \cos(4t) - 4C_2 \sin(4t)$$

Apply the initial conditions

Now, we will apply the initial conditions given in the problem. The first initial condition is \(y(\frac{\pi}{4})=1\): $$1 = C_1 \sin(\pi) + C_2 \cos(\pi)$$ The second initial condition is \(y'(\frac{\pi}{4})=-4\): $$-4 = 4C_1 \cos(\pi) - 4C_2 \sin(\pi)$$ Now, we have a system of equations to solve for \(C_1\) and \(C_2\).

Solve the system of equations

From the system of equations, we have: $$1 = C_2\cos(\pi) = -C_2$$ and $$-4 = -4C_1\sin(\pi)=0$$ Solving the second equation for \(C_1\), we get \(C_1 = 0\). Plugging in the value of \(C_1\) in the first equation, we get: $$1 = -C_2$$ Therefore, \(C_2 = -1\).

Write the unique solution for the IVP

Now that we have the values of \(C_1\) and \(C_2\), we can write down the unique solution to the initial value problem: $$y(t) = 0 \sin(4t) - 1 \cos(4t)$$ So, the unique solution for the given initial value problem is: $$y(t) = - \cos(4t)$$

Key concepts

Trigonometric Functions

Understanding trigonometric functions is key to solving the given initial value problem. Trigonometric functions, like sine and cosine, form the building blocks of oscillating functions. These functions are fundamental in various fields including physics and engineering.
Trigonometric functions have specific properties:

• Sine and Cosine: Both functions oscillate between -1 and 1. Their graphs are simple waves that repeat every \(2\pi\) radians.
• Frequency: The inside factor (like \(\omega\) in \(\sin \omega t\) and \(\cos \omega t\)) affects how fast the wave oscillates. It determines the wave's frequency.

In our exercise, we use \(4t\) in \([sin(4t), cos(4t)]\), which indicates oscillation due to the differential equation. Recognizing these function characteristics helps understand how initial conditions impact solutions.

Second Order Differential Equation

A second-order differential equation involves the second derivative of a function. It’s a powerful tool for modeling motion, electrical circuits, and more. The general form of a second order differential equation is \( y'' + p(t)y' + q(t)y = g(t) \). However, in the context of our problem, it simplifies to homogeneous form \( y'' + \omega^2 y = 0\).This specific type describes systems like oscillators. The double prime (") denotes the second derivative, offering insight into concavity and acceleration in modeling plots.
Noteworthy factors include:

• Homogeneous: It means no external forces drive the solution, only inherent properties dictate oscillation.
• Characteristic Equation: Solving \( y'' + \omega^2 y = 0 \) leads to sine and cosine solutions that reflect simple harmonic motion.

Using these principles, the solution can be translated into functions \( y(t) = C_1 \sin(4t) + C_2 \cos(4t) \). Each term ties back to our second-order dynamics.

General Solution

The general solution for a differential equation combines particular and homogeneous solutions. It represents all possible solutions before applying initial conditions. In our exercise, the general solution is given by \( y(t) = C_1 \sin(4t) + C_2 \cos(4t) \). To understand it fully, consider:

• Linear Combination: It’s a mix of sine and cosine solutions weighted by constants \( C_1 \) and \( C_2 \).
• Parameters Identification: \( C_1 \) and \( C_2 \) adjust based on initial or boundary conditions, shaping specific system responses.

The general solution remains broad until narrowed by conditions like initial or boundary values. Here, the initial value problem specifics tailor the general solution to meet conditions set at \( t=\pi/4 \). This leads us to a unique solution that fits the original differential equation and its constraints.

First Derivative

The first derivative of a function provides the rate of change or the slope of the function at any given point. For our problem, calculating the first derivative of the general solution \( y(t) = C_1 \sin(4t) + C_2 \cos(4t) \) results in \( y'(t) = 4C_1 \cos(4t) - 4C_2 \sin(4t) \).Key aspects of first derivatives include:

• Direction and Rate: Reveals how quickly \ y(t) \ changes at each \( t \).
• Relationship Adjustment: Essential in relating functions to conditions like initial values to determine exact constants.

This derivative aids in solving initial conditions because it directly adjusts based on those conditions. For instance, knowing that \( y'(\pi/4) = -4 \) plays a crucial role in deciding constants like \( C_1 \) and \( C_2 \). Hence, the derivative is pivotal for final solution adjustments, ensuring the overall model reflects reality.