Question

In each exercise, (a) Verify that the given functions form a fundamental set of solutions. (b) Solve the initial value problem. \(y^{\prime \prime \prime}+2 y^{\prime \prime}=0 ; \quad y(0)=0, \quad y^{\prime}(0)=3, \quad y^{\prime \prime}(0)=-8\) \(y_{1}(t)=1, \quad y_{2}(t)=t, \quad y_{3}(t)=e^{-2 t}\)

Short answer

Question: Verify if the functions \(y_1(t) = 1\), \(y_2(t) = t\), and \(y_3(t) = e^{-2t}\) form a fundamental set of solutions for a given initial value problem. If so, find the solution to the initial value problem with the conditions: \(y(0) = 0\), \(y'(0) = 3\), and \(y''(0) = -8\). Answer: The functions \(y_1(t) = 1\), \(y_2(t) = t\), and \(y_3(t) = e^{-2t}\) form a fundamental set of solutions. The solution to the initial value problem is \(y(t) = 2 - 2t + 7e^{-2t}\).

Step-by-step solution

Compute the Wronskian of the given functions

To verify if the given functions form a fundamental set of solutions, we will compute their Wronskian (denoted as \(W\)). If the Wronskian is non-zero, then the given functions form a fundamental set of solutions. The Wronskian of three functions \(y_1(t)\), \(y_2(t)\), and \(y_3(t)\) is defined as: $ W(y_1, y_2, y_3) = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{vmatrix}$ Now, we will compute the Wronskian for given functions \(y_1(t) = 1\), $y_2(t) = t\(, and \)y_3(t) = e^{-2t}$.

Compute the derivatives required for the Wronskian

To calculate the Wronskian, we will first compute the first and second derivatives of the given functions. 1. \(y_1(t)=1\): \(y_1'(t) = 0\) \(y_1''(t) = 0\) 2. \(y_2(t)=t\): \(y_2'(t) = 1\) \(y_2''(t) = 0\) 3. \(y_3(t)=e^{-2t}\): \(y_3'(t) = -2e^{-2t}\) \(y_3''(t) = 4e^{-2t}\)

Compute the Wronskian using the derivatives

Using the derivatives calculated in the previous step, we can now compute the Wronskian: $ W(y_1, y_2, y_3) = \begin{vmatrix} 1 & t & e^{-2t} \\ 0 & 1 & -2e^{-2t} \\ 0 & 0 & 4e^{-2t} \end{vmatrix} = (4e^{-2t}) - (-2te^{-2t}) = e^{-2t}(4 + 2t)$ Since the Wronskian is non-zero, the given functions form a fundamental set of solutions.

Form the general solution using the given functions

Since the given functions are a fundamental set of solutions, we can construct the general solution of the given differential equation as a linear combination of the given functions: \(y(t) = C_1y_1(t) + C_2y_2(t) + C_3y_3(t) = C_1 + C_2t + C_3e^{-2t}\)

Apply the initial conditions to find the constants

We are given the initial conditions as \(y(0) = 0\), \(y'(0) = 3\), and \(y''(0) = -8\). To find the values of \(C_1\), \(C_2\), and \(C_3\), we will apply these initial conditions to the general solution and its derivatives: 1. \(y(0) = 0\): \(0 = C_1 + C_2(0) + C_3e^{-2(0)} = C_1 + C_3 \Rightarrow C_1 = -C_3\) 2. \(y'(0) = 3\): \(3 = 0 + C_2 + (-2C_3)e^{-2(0)} = C_2 - 2C_3\) 3. \(y''(0) = -8\): \(-8 = 0 + 0 + 4C_3e^{-2(0)} = 4C_3\) Now, we can solve the system of equations to get the constants \(C_1\), \(C_2\), and \(C_3\). From equation 3: \(C_3 = -2\) From equation 1: \(C_1 = 2\) From equation 2: \(C_2 = 3 + 2(2) = 7\)

Write down the particular solution

With the constants found, we can write down the particular solution to the initial value problem: \(y(t) = 2 -2t + 7e^{-2t}\) So, the solution to the given initial value problem is \(y(t) = 2 -2t + 7e^{-2t}\).

Key concepts

Wronskian

The Wronskian is a powerful tool designed to establish whether a set of functions is a fundamental set of solutions to a differential equation. Functions form a fundamental set if they are linearly independent, meaning that no function in the set can be expressed as a linear combination of the others. The computation of the Wronskian itself involves determinants. For a set of functions, we create a matrix where each row consists of a function and its successive derivatives. Taking the determinant of this matrix gives us the Wronskian.

In the context of our exercise, the functions involved are evaluated, and their derivatives are calculated to fill in the matrix. When we compute the determinant and find that the Wronskian is non-zero at any point, this confirms the fundamental status of the functions. It is important for students to understand that a non-zero Wronskian across an interval of interest is critical for ensuring the solutions are indeed fundamental and unique.

Initial value problem

An initial value problem is a specific type of differential equation problem where we are given not just the differential equation itself but also initial conditions, which are values of the function and its derivatives at the beginning of the interval of interest. The goal is to find a particular solution to the differential equation that satisfies these initial conditions.

In simple terms, you can think of the differential equation as describing a family of possible functions, and the initial conditions select the exact function from this family that corresponds to the specific scenario we're dealing with. This practice is frequent in many applications such as physics and engineering where initial conditions are often known from measurements or boundary conditions.

Differential equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are profoundly significant in math, science, and engineering because they describe a plethora of phenomena, from motion to heat transfer, and from population dynamics to electrical circuits.

The differential equation in our exercise is a third-order equation, denoted by the presence of the third derivative, and it’s a homogeneous linear differential equation because all terms involve the function or its derivatives and equals zero. The solution approach involves finding a general solution that encompasses all possible solutions through the superposition of multiple solutions, and then using initial conditions to determine the specific solution to the problem at hand.

Linear combination

The concept of a linear combination involves combining several functions by multiplying each by a constant and adding the results. Specifically, when we have a set of solutions for a differential equation, the general solution can be expressed as a linear combination of these functions. This is the case because of the principle of superposition, which holds for linear differential equations.

In real-world terms, it's like making a customized recipe from a set of ingredients (the functions) where the constants are akin to the quantity of each ingredient. When we have initial conditions, as in the exercise, they help us 'taste test' and fine-tune the exact 'quantities' of each 'ingredient' required to meet the 'dietary needs'—the initial values given. This tailoring results in the particular solution that fits the initial scenario perfectly.