Question

determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution. $$ t(t-4) y^{\prime \prime}+3 t y^{\prime}+4 y=2, \quad y(3)=0, \quad y^{\prime}(3)=-1 $$

Short answer

Based on the given initial-value problem, the longest interval in which the problem is certain to have a unique twice differentiable solution is (0, 4).

Step-by-step solution

Identify Coefficients of the Differential Equation

First, let's identify the coefficients of each term in the differential equation: 1. \(t(t-4)\) is the coefficient of the second derivative term, \(y^{\prime \prime}\). 2. \(3t\) is the coefficient of the first derivative term, \(y^{\prime}\). 3. \(4\) is the coefficient of the function term, \(y\).

Determine intervals of continuity for each coefficient

Next, we need to find where each coefficient is continuous: 1. For the \(t(t-4)\) coefficient, we'll examine the zeros (when the coefficient is equal to 0): $$ t(t-4) = 0 $$ Which yields two possible values: \(t=0\) and \(t=4\). Thus, the intervals of continuity are \((-\infty, 0)\), \((0,4)\), and \((4, \infty)\). 2. For the \(3t\) coefficient, we have: $$ 3t = 0 $$ Which has only one zero value of \(t=0\). The intervals of continuity for this coefficient are \((-\infty, 0)\) and \((0, \infty)\). 3. The \(4\) coefficient is constant and thus continuous everywhere.

Find the common interval of continuity containing the initial condition

Now, we have to determine the longest interval where all coefficients are continuous and contains the initial condition, \(y(3)=0\). Since \(3 \in (0, 4)\), it falls in the \((0,4)\) range for the \(t(t-4)\) coefficient, and it also lies within the \((0, \infty)\) range for the \(3t\) coefficient. Both coefficients are continuous in the \((0, 4)\) interval, which is the longest interval containing the initial condition where a unique twice differentiable solution is guaranteed. So the longest interval for the unique solution is: $$ (0, 4) $$

Key concepts

Understanding Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They describe how a quantity changes in relation to others. Here, the given problem involves a second-order differential equation, as it includes the second derivative, denoted by \(y''\). This tells us about how the rate of change itself changes over time.

These equations are fundamental in modeling real-world phenomena in physics, engineering, biology, and more. For example, they can express everything from the motion of planets to changes in population. In our context, each derivative term has a corresponding coefficient in front, such as \(t(t-4)\) for \(y''\). Understanding these coefficients and their behavior is crucial to analyzing and solving differential equations.

The complexity of these equations makes them challenging sometimes. However, they can greatly simplify our understanding of dynamic systems. By solving them, we aim to find a function \(y(t)\) that explains the system or scenario described by the equation. In this case, however, instead of solving, we focus on finding the interval where the solution is unique.

Uniqueness of Solutions in Differential Equations

The uniqueness of solutions in differential equations refers to finding a single function that satisfies both the differential equation and the initial conditions. In the context of an initial value problem, such as the one presented, we need to identify if there is one and only one solution that fits the given conditions \(y(3)=0\) and \(y'(3)=-1\).

A solution's uniqueness is guaranteed when the coefficients of the differential equation are continuous over the interval of interest and include the initial condition point. For this problem, we identify a common interval by examining when the coefficients \(t(t-4)\), \(3t\), and \(4\) are continuous. This ensures that we can confidently predict a single specific behavior of the function \(y(t)\) that fits both the differential equation and these starting points, without ambiguity. This concept is really important in practical applications to ensure consistency and predictability.

Interval of Continuity Explained

An interval of continuity is a range over which a function or its coefficients don't have any breaks, jumps, or points of discontinuity. For differential equations, identifying this interval is crucial because it dictates the range over which a unique solution might exist.

To find such an interval for our given problem, we first determined where the coefficients of each derivative term are continuous. We found that the term \(t(t-4)\) introduces discontinuities at \(t=0\) and \(t=4\). Therefore, the intervals where this term is continuous are \((-\infty, 0)\), \((0,4)\), and \((4, \infty)\).

Next, we check these intervals against the initial condition given in the problem, focusing on the longest interval that includes this condition. Both the \(t(t-4)\) and \(3t\) coefficients were continuous in the interval \((0, 4)\), including the initial condition's point, \(t=3\). Thus, \((0, 4)\) is established as the longest interval of continuity over which a unique, twice differentiable solution is assured for the differential equation, according to the initial conditions.