Question

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

Short answer

#Answer# The unique solution of the initial value problem for the given differential equation and initial conditions is \(y(t) = -t + 5\).

Step-by-step solution

Verifying Solutions

To verify if the given functions, \(y_1(t) = t+1\) and \(y_2(t) = -t+2\), are solutions to the differential equation, \(y'' = 0\), we will find the second derivative of each function and check if it's equal to zero. For \(y_1(t) = t + 1\), we have: First derivative: \(y_1'(t) = 1\) Second derivative: \(y_1''(t) = 0\) For \(y_2(t) = -t + 2\), we have: First derivative: \(y_2'(t) = -1\) Second derivative: \(y_2''(t) = 0\) Since both second derivatives are equal to 0, both functions are solutions to the given differential equation.

Calculating the Wronskian

Now, we will calculate the Wronskian, \(W(t)\). The Wronskian for two functions \(y_1\) and \(y_2\) is given by the determinant of the following matrix: $W(t) = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \\ \end{vmatrix}$ Using our functions and their first derivatives, we get: $W(t) = \begin{vmatrix} t+1 & -t+2 \\ 1 & -1 \\ \end{vmatrix} = (t+1)(-1) - (-t+2)(1) = -t - 1 + t - 2 = -3$

Checking for a Fundamental Set of Solutions

Two functions form a fundamental set of solutions if their Wronskian is nonzero on the interval of interest. In this case, the Wronskian \(W(t) = -3\), which is a constant and nonzero for all \(t\). Therefore, \(y_1(t)\) and \(y_2(t)\) form a fundamental set of solutions.

Determining the Unique Solution of the Initial Value Problem

Since \(y_1(t)\) and \(y_2(t)\) form a fundamental set of solutions, we can write the general solution of the differential equation as a linear combination of the two functions: \(y(t) = c_1y_1(t) + c_2y_2(t)\) Now, we will use the initial conditions \(y(1) = 4\) and \(y'(1) = -1\) to find the unique solution: 1. \(y(1) = c_1y_1(1) + c_2y_2(1) = c_1(1+1) + c_2(-1+2) = 4\) 2. \(y'(1) = c_1y_1'(1) + c_2y_2'(1) = c_1(1) + c_2(-1) = -1\) Solving these equations, we get \(c_1 = 1\) and \(c_2 = 2\). So, the unique solution of the initial value problem is: \(y(t) = 1(y_1(t)) + 2(y_2(t)) = 1(t+1) + 2(-t+2) = t + 1 - 2t + 4 = -t + 5\)

Key concepts

Wronskian

The Wronskian is a powerful tool used to determine if a set of functions is linearly independent, which is essential when dealing with solutions to differential equations. When working with two functions, say, \(y_1(t)\) and \(y_2(t)\), the Wronskian, denoted as \(W(t)\), is calculated by finding the determinant of a matrix comprising these functions and their first derivatives.

In the context of differential equations, if the Wronskian is nonzero over the interval of interest, it indicates that the functions form what's called a 'fundamental set of solutions.' This implies that the functions in question, \(y_1(t)\) and \(y_2(t)\), are linearly independent and can respectively serve as the building blocks to construct the general solution to the differential equation.

For our specific exercise, the Wronskian \(W(t)\) of the given functions turned out to be a non-zero constant, which satisfies the condition for linear independence, confirming that we indeed have a fundamental set of solutions.

Initial Value Problem

An initial value problem (IVP) in the context of differential equations is a problem where you are given a differential equation along with the value of the function (and possibly its derivatives) at a specific point, referred to as the initial values. Solving an IVP is a two-step process: first, find the general solution to the differential equation, and then use the initial values to determine the specific coefficients that will satisfy those conditions.

For instance, our exercise outlines an IVP where the differential equation \(y''=0\) is accompanied by initial conditions \(y(1)=4\) and \(y'(1)=-1\). These conditions allow us to find the specific values for constants in the general solution, producing a unique solution that passes through the prescribed initial point with the designated slope.

Fundamental Set of Solutions

A fundamental set of solutions is a selection of solutions to a homogeneous differential equation that provides the backbone for creating the general solution. This concept is crucial because for a second-order differential equation, you often need two linearly independent solutions to construct the general solution.

The functions that comprise this set must be independent, which ensures that every possible solution to the differential equation can be expressed as a linear combination of these fundamental solutions. In the exercise we're discussing, the functions \(y_1(t) = t + 1\) and \(y_2(t) = -t + 2\) are confirmed to form a fundamental set, as evidenced by the non-zero Wronskian. Armed with this fundamental set, we can confidently say that any solution to the differential equation can be written in terms of these two functions.

Second Derivative

The second derivative of a function is a measure of how the rate of change of the function's rate of change is varying. It is the derivative of the derivative, and it provides insights into the curvature or concavity of the function's graph.

When dealing with differential equations, the second derivative tells us a lot about the behavior of solutions. For example, if the second derivative of a function is zero — as in our exercise with \(y''=0\) — it signifies that the function is linear since the rate at which its slope changes is constant. This information not only characterizes the solutions themselves but also plays a role in determining whether a set of solutions is indeed fundamental, as it forms part of the criteria to calculate the Wronskian.