Question

The functions \(y_{1}(t)=\sin (t+\alpha)\) and \(y_{2}(t)=\sin (t-\alpha)\) are solutions of \(y^{\prime \prime}+y=0\) on \(-\infty

Short answer

The functions \(y_1(t) = \sin(t + \alpha)\) and \(y_2(t) = \sin(t - \alpha)\) form a fundamental set of solutions for the differential equation \(y''(t) + y(t) = 0\) if \(\alpha\) is not a multiple of \(\frac{\pi}{2}\).

Step-by-step solution

Determine the derivatives of \(y_1(t)\) and \(y_2(t)\)

We will determine the first and second derivatives of both the functions and then verify if they satisfy the given differential equation. Let's start by finding the first and second derivatives of both the functions \(y_1(t)\) and \(y_2(t)\): The first derivative of \(y_1(t)\) is given by: \(y_1'(t) = \frac{d}{dt}[\sin(t + \alpha)] = \cos(t + \alpha)\). The second derivative of \(y_1(t)\) is given by: \(y_1''(t) = \frac{d^2}{dt^2}[\sin(t + \alpha)] = -\sin(t + \alpha)\). Similarly, The first derivative of \(y_2(t)\) is given by: \(y_2'(t) = \frac{d}{dt}[\sin(t - \alpha)] = \cos(t - \alpha)\). The second derivative of \(y_2(t)\) is given by: \(y_2''(t) = \frac{d^2}{dt^2}[\sin(t - \alpha)] = -\sin(t - \alpha)\).

Verify if the functions satisfy the given equation

We need to check if both functions \(y_1(t)\) and \(y_2(t)\) satisfy the differential equation \(y''(t) + y(t) = 0\). For \(y_1(t)\): \(y_1''(t) + y_1(t) = -\sin(t + \alpha) + \sin(t + \alpha) = 0\). Thus, \(y_1(t)\) satisfies the given equation. For \(y_2(t)\): \(y_2''(t) + y_2(t) = -\sin(t - \alpha) + \sin(t - \alpha) = 0\). Thus, \(y_2(t)\) also satisfies the given equation.

Check for linear independence

In order for \(\{y_1, y_2\}\) to be a fundamental set, \(y_1(t)\) and \(y_2(t)\) must be linearly independent. We can use the Wronskian to check for linear independence: $W(y_1, y_2) = \begin{vmatrix} \sin(t + \alpha) & \sin(t - \alpha) \\ \cos(t + \alpha) & \cos(t - \alpha) \end{vmatrix} = \sin(t + \alpha)\cos(t - \alpha) - \sin(t - \alpha)\cos(t + \alpha)$. Using the product-to-sum identities, we can rewrite the Wronskian as: \(W(y_1, y_2) = \frac{1}{2}[\sin(2\alpha) - \sin(-2\alpha)] = \sin(2\alpha)\). The two functions are linearly independent if the Wronskian is non-zero: \(\sin(2\alpha) \neq 0 \Rightarrow 2\alpha \neq n\pi\), for any integer \(n\). So the condition for linear independence is: \(\alpha \neq \frac{n\pi}{2}\), for any integer \(n\). Therefore, the given set of functions \(\{y_1(t), y_2(t)\}\) forms a fundamental set for the solutions of theHomogeneous second-order linear ODE \(y''(t) + y(t) = 0\) if \(\alpha\) is not a multiple of \(\frac{\pi}{2}\).

Key concepts

Wronskian

In the study of differential equations, the Wronskian plays a crucial role in determining the linear independence of a set of solutions. It's a determinant used to check whether solutions of differential equations are independent or not.
For two functions, the Wronskian is written as:\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix}\]The value of the Wronskian tells us if the functions are linearly independent. If the Wronskian is non-zero at some point, the functions are linearly independent over an interval. In our exercise, we calculated the Wronskian for the functions \(y_1(t) = \sin(t + \alpha)\) and \(y_2(t) = \sin(t - \alpha)\) and found it to be \(\sin(2\alpha)\).
This means our functions are independent unless \(\sin(2\alpha) = 0\), which happens when \(2\alpha\) is an integer multiple of \(\pi\). By setting \(2\alpha eq n\pi\), we ensure the independence necessary for a fundamental set.

Linear Independence

Linear independence is a concept in mathematics that determines whether a set of functions (or vectors) are independent of each other. For a set of solutions to form a fundamental set for a differential equation, they must be linearly independent.
For functions \(y_1(t)\) and \(y_2(t)\), saying they are linearly independent means one function cannot be expressed as a scalar multiple of the other. The easiest way to check this, especially in the context of differential equations, is by using the Wronskian determinant.
If \(W(y_1, y_2) eq 0\) for some \(t\) in the interval, then \(y_1\) and \(y_2\) are linearly independent. In our exercise, this boils down to checking if \(\sin(2\alpha) eq 0\), which simplifies acceptance of \(\alpha\) as values that are not integer multiples of \(\frac{\pi}{2}\).
This concept is essential in ensuring that the solutions we have cover the whole space of possible solutions for a differential equation.

Fundamental Set

A fundamental set of solutions is a collection of linearly independent solutions to a differential equation, typically a homogeneous one. For second-order linear differential equations like the one given in the exercise, this set will consist of two solutions.
The importance of finding a fundamental set lies in the fact that any solution to the given differential equation can be expressed as a combination of the solutions forming this set. Thus, understanding its components helps in solving general solutions.
In the exercise, the functions \(y_1(t) = \sin(t + \alpha)\) and \(y_2(t) = \sin(t - \alpha)\) form a fundamental set given \(\sin(2\alpha) eq 0\). This highlights the need for linear independence between the solutions, allowing any solution of the differential equation \(y'' + y = 0\) to be expressed as a linear combination of \(y_1(t)\) and \(y_2(t)\).

Homogeneous Second-Order Linear ODE

A homogeneous second-order linear ordinary differential equation (ODE) is a differential equation of the form \(ay'' + by' + cy = 0\), where \(a\), \(b\), and \(c\) are constants, and the equation is set to zero, making it homogeneous.
In the exercise, the ODE \(y'' + y = 0\) fits this description with \(a = 1\), \(b = 0\), and \(c = 1\). Solving such equations involves finding solutions that satisfy the equation everywhere over an interval. Typically, the general solution is provided by a linear combination of two linearly independent solutions, forming the fundamental set.
These types of equations often model various physical systems, like simple harmonic oscillators. The method we used in the exercise ensures we get all possible solutions of the system by constructing them from the fundamental set found through the analysis.