Question

In each exercise, (a) Verify that the given functions form a fundamental set of solutions. (b) Solve the initial value problem. 6\. \(t^{2} y^{\prime \prime \prime}+t y^{\prime \prime}-y^{\prime}=0, \quad t<0 ; \quad y(-1)=1, \quad y^{\prime}(-1)=-1, \quad y^{\prime \prime}(-1)=-1\) \(y_{1}(t)=1, \quad y_{2}(t)=\ln (-t), \quad y_{3}(t)=t^{2}\)

Short answer

Question: Verify that \(y_1(t)=1\), \(y_2(t)=\ln(-t)\), and \(y_3(t)=t^2\) form a fundamental set of solutions for the given differential equation and find the particular solution for the initial value problem with boundary conditions \(y(-1)=1\), \(y'(-1)=-1\), and \(y''(-1)=-1\). Answer: The given functions \(y_1(t)=1\), \(y_2(t)=\ln(-t)\), and \(y_3(t)=t^2\) form a fundamental set of solutions as their Wronskian determinant is non-zero for all \(t<0\). The particular solution for the initial value problem with the given boundary conditions is \(y(t) = 1 - \ln(-t)\).

Step-by-step solution

Calculate derivatives for given functions

First, we need to find the first and second derivatives of the given functions \(y_1(t)\), \(y_2(t)\), and \(y_3(t)\). For \(y_1(t)=1\): - \(y_1'(t)=0\) - \(y_1''(t)=0\) For \(y_2(t)=\ln(-t)\): - \(y_2'(t)=\frac{-1}{t}\) - \(y_2''(t)=\frac{1}{t^2}\) For \(y_3(t)=t^2\): - \(y_3'(t)=2t\) - \(y_3''(t)=2\)

Compute the Wronskian determinant

Now that we have the derivatives, we can compute the Wronskian determinant \(W\) for the given functions. The Wronskian is given by the following determinant: $W = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{vmatrix}$ Plugging in the given functions and their derivatives calculated in Step 1, we get: $W = \begin{vmatrix} 1 & \ln(-t) & t^2 \\ 0 & -\frac{1}{t} & 2t \\ 0 & \frac{1}{t^2} & 2 \end{vmatrix}$

Evaluate the Wronskian determinant

Now we evaluate the Wronskian determinant to determine if it is non-zero: $W = \Bigg|\begin{matrix} 1 & \ln(-t) & t^2 \\ 0 & -\frac{1}{t} & 2t \\ 0 & \frac{1}{t^2} & 2 \end{matrix}\Bigg| = 1 \times \Bigg|\begin{matrix} -\frac{1}{t} & 2t \\ \frac{1}{t^2} & 2 \end{matrix}\Bigg| = 1 \times \left(-\frac{1}{t} \times 2 - 2t \times \frac{1}{t^2}\right) = -2 + \frac{2}{t}$ Since \(-2 + \frac{2}{t}\ne0\) for \(t < 0\), the given functions do form a fundamental set of solutions.

Write down the general solution

Because the given functions form a fundamental set of solutions, we can write down the general solution as a linear combination of these functions: \(y(t) = C_1 y_1(t) + C_2 y_2(t) + C_3 y_3(t) = C_1 + C_2 \ln(-t) + C_3t^2\)

Apply the initial conditions

Now we apply the given initial conditions to find the particular solution: 1. \(y(-1) = 1 \Rightarrow C_1 + C_2 \ln(1) + C_3(-1)^2 = 1 \Rightarrow C_1 - C_3 = 1\) 2. \(y'(-1) = -1 \Rightarrow 0 - \frac{C_2}{-1} + 2(-1)C_3 = -1 \Rightarrow C_2 - 2C_3 = -1\) 3. \(y''(-1) = -1 \Rightarrow 0 + \frac{C_2}{(-1)^2} + 2 = -1 \Rightarrow C_2 = -1\)

Solve for the coefficients

We can now use the results from applying the initial conditions to solve for \(C_1\), \(C_2\), and \(C_3\): 1. \(C_2 = -1\) 2. From \(C_1 - C_3 = 1\), we have \(C_1 = 1 + C_3\). 3. Plugging \(C_2=-1\) into \(C_2 - 2C_3 = -1\), we have \(-1 - 2C_3 = -1 \Rightarrow C_3 = 0\). So \(C_1 = 1\), \(C_2 = -1\), and \(C_3 = 0\).

Write down the particular solution

Finally, plug the coefficients \(C_1\), \(C_2\), and \(C_3\) back into the general solution to obtain the particular solution: \(y(t) = 1 - \ln(-t) + 0t^2 = 1 - \ln(-t)\) This is the particular solution to the initial value problem.

Key concepts

Fundamental Set of Solutions

When dealing with ordinary differential equations (ODEs), a fundamental set of solutions is crucial. It's essentially a collection of solutions that can be combined to express the most general solution to a differential equation. In our exercise, the functions \(y_1(t) = 1\), \(y_2(t) = \ln(-t)\), and \(y_3(t) = t^2\) form a fundamental set of solutions. This means that any solution to the differential equation can be written as a linear combination of these functions.

Verifying a fundamental set involves checking if these solutions are linearly independent, which we do using the Wronskian determinant. An important takeaway here is that the solutions are valid for the domain specified; in this case, \(t < 0\). Linear independence implies that no solution in the set can be formed as a linear combination of others. Thus, knowing that you have a fundamental set allows you to construct the general solution to the ODE confidently.

Wronskian Determinant

The Wronskian determinant tests whether a set of functions, like our solution candidates, is linearly independent. For a set to be a fundamental set of solutions, the Wronskian must be non-zero.

In the exercise, we calculated the Wronskian of \(y_1, y_2, y_3\) and found it to be \(-2 + \frac{2}{t}\). Because the Wronskian is non-zero for \(t < 0\), the functions are indeed linearly independent. Calculating this determinant involves derivatives and setting them in a matrix format.

The process involves evaluating a 3x3 determinant where the first row is the functions, the second their first derivatives, and the third their second derivatives. A non-zero result over the interval of interest confirms that the functions form a fundamental set.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives. These are equations like \(t^2 y''' + ty'' - y' = 0\), representing relationships between the functions' various derivatives.

ODEs appear frequently across science and engineering, modeling phenomena like motion or heat. We approach solving ODEs by finding solutions that satisfy both the equation and any initial conditions given. These solutions often involve obtaining a general solution first, and then honing in on a particular solution that fits specific criteria.

Techniques for solving ODEs include separation of variables, integrating factors, and, as in our exercise, using linear combinations of known solutions. Understanding the type of ODE, such as whether it is linear or nonlinear, can guide the approach to finding solutions.

Linear Combination

A linear combination involves adding together multiple functions, each multiplied by a constant coefficient.

In our exercise, the general solution to the ODE is expressed as a linear combination: \(y(t) = C_1 + C_2 \ln(-t) + C_3t^2\). Here, \(C_1, C_2,\) and \(C_3\) are coefficients that adjust each function to satisfy any initial conditions of the problem.

The flexibility of linear combinations allows the formulation of various solutions by adjusting these coefficients. When dealing with a fundamental set of solutions, any solution of the ODE can be written in this way. Once the general solution is determined, specific initial value problems can provide the context to find the particular solution.

Particular Solution

A particular solution is a specific solution to a differential equation that also meets given initial conditions. For the exercise, we derived the particular solution \(y(t) = 1 - \ln(-t)\).

The process begins with the general solution captured through a linear combination of the fundamental set of solutions. Applying initial conditions such as \(y(-1)=1\), \(y'(-1)=-1\), and \(y''(-1)=-1\) lets us solve for unknown coefficients \(C_1, C_2,\) and \(C_3\).

This tailored solution represents a unique path through the overall solution landscape. This ensures the solution not only satisfies the ODE itself but also aligns with real-world constraints or specified scenarios. A particular solution is critical for accurately capturing the behavior of dynamic systems described by the ODE.