Question

The \(t\)-interval of interest is \(-\infty

Short answer

Question: Verify that the functions \(y_1(t) = \cos t\) and \(y_2(t) = \sin t\) are solutions of the given differential equation \(y'' + y = 0\), calculate their Wronskian, and determine the unique solution of the initial value problem with the conditions \(y(\pi / 2)=1\) and \(y'(\pi / 2)=1\). Answer: The functions \(y_1(t) = \cos t\) and \(y_2(t) = \sin t\) are indeed solutions of the given differential equation. Their Wronskian is calculated as \(W(\cos t, \sin t) = 1\), and the unique solution of the initial value problem is \(y(t) = -\cos t + \sin t\).

Step-by-step solution

Verify that \(y_1(t)\) and \(y_2(t)\) are solutions of the differential equation

To verify that the functions \(y_1(t) = \cos t\) and \(y_2(t) = \sin t\) are solutions of the given differential equation, we will plug them into the equation and check if it holds true. The differential equation is: \(y'' + y = 0\). For \(y_1(t) = \cos t\), we calculate the first and second derivatives: \(y_1'(t) = -\sin t\) \(y_1''(t) = -\cos t\) Now, let's plug \(y_1(t)\) and its derivatives into the differential equation: \((-\cos t) + (\cos t) = 0\) which simplifies to \(0=0\), indicating that \(y_1(t)\) is a solution to the differential equation. For \(y_2(t) = \sin t\), we calculate the first and second derivatives: \(y_2'(t) = \cos t\) \(y_2''(t) = -\sin t\) Now, let's plug \(y_2(t)\) and its derivatives into the differential equation: \((-\sin t) + (\sin t) = 0\) which simplifies to \(0=0\), indicating that \(y_2(t)\) is also a solution to the differential equation. Therefore, both \(y_1(t)\) and \(y_2(t)\) are solutions of the given differential equation.

Calculate the Wronskian and determine if the functions form a fundamental set of solutions

To check if the functions form a fundamental set of solutions, we need to calculate the Wronskian. The Wronskian of two functions \(y_1(t)\) and \(y_2(t)\) is given by: \(W(y_1, y_2)(t) = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{vmatrix} = y_1(t)y_2'(t) - y_1'(t)y_2(t)\) We will now compute the Wronskian for the given functions: \(W(\cos t, \sin t) = \cos(t)(\cos t) - (-\sin(t))(\sin t) = \cos^2(t) + \sin^2(t)\) Since \(\cos^2(t) + \sin^2(t) = 1\) for all \(t\), we have \(W(\cos t, \sin t) = 1\) Since the Wronskian is nonzero for all \(t\) in the interval \((-\infty, \infty)\), the functions \(y_1(t) = \cos t\) and \(y_2(t) = \sin t\) form a fundamental set of solutions.

Determine the unique solution of the initial value problem

Since we have verified that the functions \(y_1(t)\) and \(y_2(t)\) form a fundamental set of solutions, we can now express the unique solution of the initial value problem as a linear combination of these functions: \(y(t) = c_1y_1(t) + c_2y_2(t) = c_1\cos t + c_2\sin t\) We are given the initial conditions \(y(\pi / 2)=1\) and \(y'(\pi / 2)=1\). We will use these conditions to determine the values of \(c_1\) and \(c_2\). First, let's find the general expression for the derivative \(y'(t)\): \(y'(t) = -c_1\sin t + c_2\cos t\) Now, let's use the initial conditions: \(y(\pi / 2) = c_1\cos(\pi / 2) + c_2\sin(\pi / 2) = 1 \Rightarrow 0 + c_2 = 1\) So, \(c_2 = 1\). \(y'(\pi / 2) = -c_1\sin(\pi / 2) + c_2\cos(\pi / 2) = 1 \Rightarrow -c_1 + 0 = 1\) So, \(c_1 = -1\). Now we have the unique solution of the initial value problem: \(y(t) = -\cos t + \sin t\)

Key concepts

Wronskian

Understanding the Wronskian is crucial when solving differential equations, particularly when we want to determine if a set of functions can be used to construct every solution to the equation. The Wronskian is a type of determinant which is used to test the linear independence of a set of functions. It is named after the Polish mathematician Józef Hoene-Wroński and is an essential tool in the study of differential equations.

In a practical sense, when the Wronskian of a set of functions is nonzero on an interval, it confirms that these functions are linearly independent on that interval. For the functions related to our exercise, \(y_1(t) = \text{cos} t\) and \(y_2(t) = \text{sin} t\), the Wronskian is calculated as follows:

\[ W(y_1, y_2)(t) = \begin{vmatrix} y_1(t) & y_2(t) \ y_1'(t) & y_2'(t) \/vmatrix} = y_1(t)y_2'(t) - y_1'(t)y_2(t) = \text{cos}^2(t) + \text{sin}^2(t) \]
The identity \(\text{cos}^2(t) + \text{sin}^2(t) = 1\) indicates that the Wronskian is always 1 and never zero, showing that \(y_1(t)\) and \(y_2(t)\) are linearly independent and form a fundamental set of solutions, an important distinction that means any solution to the differential equation can be expressed as a linear combination of these functions.

Fundamental Set of Solutions

When dealing with differential equations, finding a fundamental set of solutions is akin to finding the building blocks from which all other solutions can be constructed. A fundamental set of solutions to a linear homogeneous differential equation consists of solutions that are linearly independent. These solutions span the solution space, which simply refers to the collection of all possible solutions to the equation. This is comparable to the way a set of basis vectors spans a vector space in linear algebra.

For the specific equation \(y''+y=0\), two functions, \(y_1(t)=\text{cos} t\) and \(y_2(t)=\text{sin} t\), make up the fundamental set, as we confirmed from their nonzero Wronskian. Because of this, we can express any other solution \(y(t)\) as a linear combination of these two, where \(y(t) = c_1y_1(t) + c_2y_2(t)\) with coefficients \(c_1\) and \(c_2\) determined using initial conditions or other constraints imposed on the solution.

Initial Value Problem

An initial value problem in the context of differential equations is a problem where we not only look to find solutions to a differential equation but also specify values of the unknown function (and possibly its derivatives) at a particular point. These specified values are called the initial conditions and are vital in determining a unique solution to the differential equation.

As seen in our exercise, given the initial values \(y(\frac{\text{\pi}}{2})=1\) and \(y'(\frac{\text{\pi}}{2})=1\), we apply them to the general solution \(y(t) = c_1\text{cos} t + c_2\text{sin} t\) and its derivative to solve for the coefficients \(c_1\) and \(c_2\). This process furnishes us with the unique solution that passes through a specific point with a given slope, ensuring that our solution curve not only solves the equation but also aligns exactly with the given initial conditions: \(y(t) = -\text{cos} t + \text{sin} t\).