Question

In each exercise, (a) Verify that the given functions form a fundamental set of solutions. (b) Solve the initial value problem. \(y^{\prime \prime \prime}-y^{\prime}=0 ; \quad y(0)=4, \quad y^{\prime}(0)=1, \quad y^{\prime \prime}(0)=3\) \(y_{1}(t)=1, \quad y_{2}(t)=e^{t}, \quad y_{3}(t)=e^{-t}\)

Short answer

Question: Verify that the given functions \(y_1(t) = 1\), \(y_2(t) = e^t\), and \(y_3(t) = e^{-t}\) form a fundamental set of solutions and solve the initial value problem with initial conditions \(y(0) = 4\), \(y'(0) = 1\), and \(y''(0) = 3\). Answer: The given functions form a fundamental set of solutions, as verified by computing their Wronskian, which is nonzero. The solution to the initial value problem is \(y(t) = 1 + 2e^t + e^{-t}\).

Step-by-step solution

Verify that the given functions form a fundamental set of solutions

To verify that the given functions form a fundamental set of solutions, we need to calculate their Wronskian: $W(y_1, y_2, y_3) = \begin{vmatrix} 1 & e^t & e^{-t} \\ 0 & e^t & -e^{-t} \\ 0 & e^t & e^{-t} \end{vmatrix}$ Compute the determinant of this matrix: \(W(y_1, y_2, y_3) = 1\left(e^t(-2e^{-t})\right) - e^t\left(e^t(-e^{-t})\right) + e^{-t}\left(0\right) = -2e^t(-2e^{-t}) - e^{2t}e^{-t} = 4 - e^t\) Since the Wronskian is nonzero, the given functions form a fundamental set of solutions.

Form the general solution

A general solution to the given differential equation can be written as a linear combination of the fundamental set of solutions: \(y(t) = C_1y_1(t) + C_2y_2(t) + C_3y_3(t) = C_1 + C_2e^t + C_3e^{-t}\) To solve the initial value problem, we need to find \(C_1\), \(C_2\), and \(C_3\) using the given initial conditions.

Apply the initial conditions

Now, we apply the initial conditions to find the constants \(C_1\), \(C_2\), and \(C_3\). 1. \(y(0) = 4\): \(4 = C_1 + C_2e^0 + C_3e^{-0}\), which simplifies to \(4 = C_1 + C_2 + C_3\). 2. \(y'(0) = 1\): First, find \(y'(t)\): \(y'(t) = 0 + C_2e^t - C_3e^{-t}\). Then apply the initial condition: \(1 = C_2 - C_3\). 3. \(y''(0) = 3\): First, find \(y''(t)\): \(y''(t) = C_2e^t + C_3e^{-t}\). Then apply the initial condition: \(3 = C_2 + C_3\).

Solve for the constants

Now, we have a system of three equations and three unknowns: 1. \(4 = C_1 + C_2 + C_3\) 2. \(1 = C_2 - C_3\) 3. \(3 = C_2 + C_3\) Solve this system, starting with equations (2) and (3): Adding equations (2) and (3): \(4 = 2C_2 \Rightarrow C_2 = 2\). Substituting \(C_2\) into equation (2): \(1 = 2 - C_3 \Rightarrow C_3 = 1\). Substituting \(C_2\) and \(C_3\) into equation (1): \(4 = C_1 + 2 + 1 \Rightarrow C_1 = 1\). Now, we have found \(C_1 = 1\), \(C_2 = 2\), and \(C_3 = 1\).

Write the final solution

Finally, substitute the values of \(C_1\), \(C_2\), and \(C_3\) into the general solution: \(y(t) = 1 + 2e^t + 1e^{-t}\) This is the solution to the initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They serve as a cornerstone in many areas of mathematics and applied science, allowing us to model the behavior of physical systems ranging from motion to heat transfer, and much more.

In our specific case, we are dealing with a third-order linear differential equation, which means it involves the third derivative of the function. When solving an initial value problem, we seek a solution that not only satisfies the differential equation but also meets specific conditions at the starting point or initial values - in this instance, at time zero (t=0).

The process begins by identifying a general solution for the differential equation and then applying the given initial conditions to solve for specific constants and finalize the particular solution.

Wronskian Determinant

Wronskian determinant provides a crucial method for determining whether a set of functions is linearly independent, which is essential for finding a fundamental set of solutions for a differential equation. In other words, the Wronskian helps to ensure that the solutions we obtain can describe the system's behavior fully, without any redundancy.

Computation of the Wronskian requires the construction of a square matrix filled with the functions in question and their derivatives, followed by the calculation of this matrix's determinant. If the Wronskian is nonzero, it conclusively proves that our set of functions is linearly independent. As seen in the exercise, evaluating the Wronskian enabled us to verify that the functions provided indeed form a fundamental set of solutions for the differential equation.

Fundamental Set of Solutions

A fundamental set of solutions to a differential equation is a group of solutions that, when combined, can express any other solution to the differential equation. Think of it as a toolbox from which any needed tool - in this case, a working solution to the differential equation - can be constructed.

The concept of linear independence is key for this set. It guarantees that each solution adds unique information about the differential equation's behavior and none of them can be constructed from a combination of the others. For a third-order differential equation, we need three linearly independent solutions to form a fundamental set, as demonstrated in the problem where we verified the linear independence via the Wronskian and subsequently used those solutions to construct the general solution to the initial value problem.