Question

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

Short answer

Short Answer: The function y₁(t) = t³ is a solution to the differential equation, and the functions y₁(t) and y₂(t) form a fundamental set of solutions on (-∞, 0) since the Wronskian W(y₁, y₂) = 4t is not identically zero. The unique solution for the initial value problem with initial conditions y(-1) = 0 and y'(-1) = -2 is y(t) = -t³ + (-t)⁻¹.

Step-by-step solution

1. Verify that the given functions are solutions to the differential equation.

We are given two functions, \(y_1(t) = t^3\) and \(y_2(t) = -t^{-1}\). To verify if they are solutions to the given differential equation, \(t^2 y'' - ty' - 3y = 0\), we need to plug in each function and its derivatives into the equation and check if the equation holds true. For \(y_1(t) = t^3\), we have: - \(y_1'(t) = 3t^2\) - \(y_1''(t) = 6t\) Plugging in these values into the differential equation: \(t^2(6t) - t(3t^2) - 3(t^3) = 6t^3 - 3t^3 - 3t^3 = 0\) The equation holds true for \(y_1(t)\). For \(y_2(t) = -t^{-1}\), we have: - \(y_2'(t) = -(-1)t^{-2} = t^{-2}\) - \(y_2''(t) = 2t^{-3}\) Plugging in these values into the differential equation: \(t^2(2t^{-3}) - t(t^{-2}) - 3(-t^{-1}) = 2t^{-1} - t^{-1} + 3t^{-1} = 4t^{-1} - t^{-1} = 3t^{-1}\) The equation does not hold true for \(y_2(t)\). Therefore, only \(y_1(t)\) is a solution to the differential equation.

2. Calculate the Wronskian.

We can still calculate the Wronskian of the given functions, even though \(y_2(t)\) is not a solution. The Wronskian is defined as \(W(y_1, y_2) = y_1y_2' - y_2y_1'\), where \(y'_1\) and \(y'_2\) are the derivatives of \(y_1\) and \(y_2\) respectively. \(W(y_1, y_2) = (t^3)(t^{-2}) - (-t^{-1})(3t^2) = t + 3t = 4t\)

3. Determine if the functions form a fundamental set of solutions.

A set of functions form a fundamental set of solutions if their Wronskian is nonzero. Since the Wronskian in our case is \(W(y_1, y_2) = 4t\), it is not identically zero for all \(t\), so the given functions form a fundamental set of solutions on \((-\infty, 0)\).

4. Find the unique solution for the initial value problem.

We can now find the unique solution to the initial value problem, given by \(y = c_1y_1(t) + c_2y_2(t)\), where \(c_1\) and \(c_2\) are constants, and the initial conditions are given by \(y(-1)=0\) and \(y'(-1) = -2\). Applying the initial condition \(y(-1) = 0\): \(0 = c_1(-1)^3 + c_2(-1^{-1}) = -c_1 + c_2\) Applying the initial condition \(y'(-1) = -2\): \(-2 = 3c_1(-1)^2 + c_2(-1)^{-2} = 3c_1 + c_2\) Solve this linear system of equations to find the values of \(c_1\) and \(c_2\): \(c_1 = -1\) \(c_2 = 1\) So, the unique solution to the initial value problem is \(y(t) = -t^3 + (-t)^{-1}\).

Key concepts

Wronskian

The Wronskian is a concept named after the Polish mathematician Józef Hoene-Wronski, and it plays a significant role in the theory of linear differential equations. It is utilized to determine whether a set of solutions is linearly independent, which is critical when working with homogenous differential equations.

The formula to compute the Wronskian of two functions, say y₁ and y₂, is given by:

\[W(y_1, y_2) = \begin{vmatrix}y_1 & y_2 \ y_1' & y_2'\end{vmatrix} = y_1y_2'-y_2y_1'\]
If the Wronskian is non-zero for some point in the interval of interest, the functions are considered linearly independent, and thus, form a fundamental set of solutions. Conversely, if the Wronskian is zero everywhere in the interval, the set of solutions is likely to be linearly dependent.

In our exercise, the Wronskian of the given functions y_1(t) = t^3 and y_2(t) = -t^{-1} was calculated and found to be 4t, which is not identically zero for t in the interval (-ifty, 0). Hence, the functions form a fundamental set of solutions in this interval.

Initial Value Problem

The concept of an initial value problem is paramount in the realm of differential equations. It involves finding a specific solution to a differential equation that not only satisfies the equation itself but also meets given conditions at a particular point, typically initial conditions at the starting point of the interval.

An initial value problem is generally presented with a differential equation along with initial conditions such as values of the function and possibly its derivatives at a specific point. In the case of our exercise, the initial conditions were given as y(-1) = 0 and y'(-1) = -2.

These conditions are instrumental in defining the constants of integration that arise when solving differential equations. In the end, we get the unique solution that precisely fits the prescribed initial scenario. This is crucial because, without these conditions, there would be infinitely many solutions to the differential equation.

Solving Differential Equations

Solving differential equations is an extensive subject that encapsulates various methods to find functions which satisfy differential equations. There are different types of differential equations, such as ordinary differential equations (ODEs) and partial differential equations (PDEs), each possessing unique challenges and requiring specific methods for solutions.

In the context of ODEs, which our exercise concerns, the solutions can generally be obtained by separation of variables, integrating factors, or, in the case of higher-order linear differential equations, by the method of undetermined coefficients or variation of parameters. Furthermore, certain ODEs can be solved by transforming them into algebraic equations using techniques like Laplace transforms.

It should be noted that not all differential equations have straightforward analytical solutions. In such cases, numerical methods and computer simulations are employed to approximate their solutions.

Fundamental Set of Solutions

The term 'fundamental set of solutions' in the context of differential equations refers to a set of solutions that are linearly independent and span the solution space for a linear homogenous differential equation. Simply put, any solution to the differential equation can be expressed as a linear combination of the functions in this set.

For second-order linear differential equations, a fundamental set typically consists of two functions. The Wronskian is used to ascertain whether a given pair of solutions form a 'fundamental set'. If the Wronskian is non-zero within the interval considered, then the two functions are independent of each other and constitutes a fundamental set, as observed for functions y_1(t) = t^3 and y_2(t) = -t^{-1} in our given exercise.

In solving initial value problems, once we have a fundamental set of solutions, we can combine them with their corresponding coefficients to construct the unique solution that fits the initial conditions, as was demonstrated with the final solution for our exercise.