Question

The given differential equation has a fundamental set of solutions whose Wronskian \(W(t)\) is such that \(W(0)=1\). What is \(W(4) ?\) \(\left(t^{2}+1\right) y^{\prime \prime \prime}-2 t y^{\prime \prime}+y=0\)

Short answer

Answer: Unfortunately, without finding the explicit form of the fundamental set of solutions, it is impossible to find the value of \(W(4)\). The best we can do is to express it symbolically as \(W(4) = k\), where \(k\) is an undetermined proportionality constant.

Step-by-step solution

Finding the fundamental set of solutions

(Write the content here) First, we need to find a fundamental set of solutions for the given differential equation, which is of the form \((t^2+1)y''' - 2ty'' + y = 0\). This is a linear homogeneous differential equation with variable coefficients. Unfortunately, there's no easy way to find a closed form solution for this equation. So, we will assume the existence of a fundamental set of solutions without finding them explicitly. Let \(y_1(t)\), \(y_2(t)\), and \(y_3(t)\) be linearly independent solutions belonging to a fundamental set.

Wronskian definition and computation

Recall that for three linearly independent solutions of a linear homogeneous differential equation, the Wronskian is defined as: \[W(t) = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \\ \end{vmatrix}\] We are also given that \(W(0) = 1\). The goal is to find the value of \(W(4)\).

Wronskian determinant

Since we do not have explicit solutions for \(y_1\), \(y_2\), and \(y_3\), we can only express \(W(4)\) symbolically in terms of these functions and their derivatives for now: \[W(4) = \begin{vmatrix} y_1(4) & y_2(4) & y_3(4) \\ y_1'(4) & y_2'(4) & y_3'(4) \\ y_1''(4) & y_2''(4) & y_3''(4) \\ \end{vmatrix}\]

Recognize the Wronskian relations

From the given information, we have \(W(0) = 1\). It is important to note that the Wronskian is essentially proportional to an unknown constant from \(W(0)\) to \(W(4)\). Using this observation, we can say that \(W(4) = kW(0) = k(1)\), where \(k\) is the proportionality constant. Since we cannot find the explicit form of the Wronskian, we are unable to find the proportionality constant \(k\), and as a result, we cannot find the explicit value of \(W(4)\). In conclusion, without finding the explicit form of \(y_1\), \(y_2\), and \(y_3\), it is impossible to find the value of \(W(4)\). The best we can do is to express it symbolically as \(W(4) = k\), where \(k\) is an undetermined proportionality constant, showing that the Wronskian \(W(4)\) also depends on the particular fundamental set of solutions found.

Key concepts

Fundamental Set of Solutions

A fundamental set of solutions refers to a collection of solutions to a differential equation that are linearly independent from each other. Think of it like a toolkit that has just the right tools—no two do the same job, and together they can solve a variety of problems. For the differential equation \( (t^2+1)y''' - 2ty'' + y = 0 \) we are looking for three independent solutions, let's call them \( y_1(t) \), \( y_2(t) \), and \( y_3(t) \). Each of these represents a unique 'tool' or pathway the system can take.Imagine you have three friends who all take different routes to get to the same place—none of them follow each other, and each of their journeys offers different scenery. In much the same way, a fundamental set of solutions provides us with essential, distinct components that span the solution space of our equation. This set allows us to express any solution to the differential equation as a combination of these independent solutions.

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is a fancy way of saying it's a 'well-behaved' equation with a certain set of rules. It doesn't play dirty; there are no sudden spikes or dips (nonlinear behavior), and it treats zero specially—if you put zero into it, you'll get zero out. The terms 'linear' and 'homogeneous' are the key here: 'Linear' means that variables and their derivatives are to the first power and appear in a straight line when graphed, and 'homogeneous' means that there's no term in the equation that's just a number; everything is tied to the unknown function we're trying to find. In our example \( (t^2+1)y''' - 2ty'' + y = 0 \), each term is a product of the function \( y(t) \) or its derivatives and some other function of \( t \), with no standalone constants or functions of \( t \) by themselves.

Variable Coefficients

In the world of differential equations, variable coefficients are like the changing weather patterns in your local area—they're the parts of the equation that fluctuate. Instead of being constant, these coefficients depend on the independent variable, which is usually time or space. For our equation, it's \( t \), and we see it explicitly in terms like \( t^2+1 \) and \( -2t \). These variable coefficients make our equation more dynamic and often more challenging to solve because they add an extra layer of complexity where the behavior of the solutions can change as \( t \) changes. Just as you might wear a different outfit depending on whether it's sunny or rainy, variable coefficients mean that the nature of the solutions can look quite different at various points in time or space.

Determinant Computation

The determinant is a special number that can be calculated from a matrix. It's a bit like a DNA test for matrices—it tells you about the properties of the system the matrix represents. When we talk about computing the determinant of the Wronskian, it gives us a single value that incorporates all the functions of our solutions and their derivatives.To compute the determinant, think of it as a game of crisscrossing paths—you multiply diagonals and subtract products of the other diagonals (crossed terms). Do this for each combination in the matrix, and sum all those up to get your determinant. In mathematical terms, for a 3x3 matrix as we have with our Wronskian, the formula might look complex but it boils down to this process of multiplication and subtraction of the diagonal products.