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. $$ y^{\prime \prime}+(\cos t) y^{\prime}+3(\ln |t|) y=0, \quad y(2)=3, \quad y^{\prime}(2)=1 $$

Short answer

Question: Determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution that includes the given initial conditions: $$ y^{\prime \prime}+(\cos t) y^{\prime}+3(\ln |t|) y=0, \quad y(2)=3, \quad y^{\prime}(2)=1 $$ Answer: The longest interval for the unique solution of the given initial value problem is $(0, \infty)$.

Step-by-step solution

Rewrite the equation and initial conditions

The given initial value problem is: $$ y^{\prime \prime}+(\cos t) y^{\prime}+3(\ln |t|) y=0, \quad y(2)=3, \quad y^{\prime}(2)=1 $$

Analyze the coefficients of the differential equation

We need to determine the domains in which the coefficients of the differential equation are continuous: 1. \(\cos t\) is continuous for all real \(t\). 2. \(\ln |t|\) is continuous for \(t \neq 0\). Thus, we only need to focus on determining the domain for the coefficient \(3(\ln |t|)\).

Determine the domain of \(3(\ln |t|)\)

Since the natural logarithm, \(\ln |t|\), is continuous only for \(t \neq 0\), the domain for the coefficient \(3(\ln |t|)\) is also only for \(t \neq 0\). Now, we know that the initial value is given at \(t = 2\), so the interval should include \(t = 2\). Since \(t \neq 0\), we have two cases to handle based on the sign of \(t\): 1. \(t > 0\): In this case, the \(3(\ln |t|)\) function is continuous and defined for \(t > 0\), and the interval will include the point \(t=2\). 2. \(t < 0\): In this case, \(\ln |t|\) is still continuous and defined for \(t < 0\). However, as we are given that \(y(2) = 3\) and \(y'(2) = 1\), the initial conditions apply only for \(t = 2\).

Identify the longest interval

Based on our analysis, the longest interval in which the initial value problem is certain to have a unique twice differentiable solution that includes the given initial conditions would be for \(t > 0\). Therefore, the longest interval is: $$ (0, \infty) $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are prevalent in physics, engineering, economics, and other fields to describe various phenomena. In the given exercise, we encounter a second-order linear differential equation with variable coefficients, which is an equation involving the second derivative of a function, denoted as \( y^{\text{\textquotesingle}\text{\textquotesingle}} \).

The problem applies this concept by presenting an equation of the form \( y^{\text{\textquotesingle}\text{\textquotesingle}}+(\text{cos}\text{ }t) y^{\text{\textquotesingle}}+3(\text{ln} |t|) y=0 \), where \( y \) is the unknown function of the variable \( t \), \( y^{\text{\textquotesingle}\text{\textquotesingle}} \) is the second derivative of \( y \) with respect to \( t \), and \( y^{\text{\textquotesingle}} \) is the first derivative. This particular example features non-constant coefficients, specifically \( (\text{cos}\text{ }t) \) and \( 3(\text{ln} |t|) \), which adds complexity to both the analysis of the solution’s behavior and the process of solving the equation itself.

Existence and Uniqueness of Solutions

Determining the existence and uniqueness of solutions to differential equations is crucial for ensuring that problems are well-posed and thus have meaningful solutions. The given initial value problem asks to verify the requirements under which a unique twice differentiable solution exists, without requiring the actual solution.

The historical cornerstone for such verifications is the Existence and Uniqueness Theorem, stating that if the function and its partial derivatives are continuous over a certain domain, then within that domain, the initial value problem has exactly one solution. Here, the solution’s existence is tied to the continuity of the coefficients \( (\text{cos}\text{ }t) \) and \( 3(\text{ln} |t|) \), as explored in the solution steps. As long as the coefficients are continuous and the initial conditions are known (in this case, at \( t = 2 \)), we have enough guarantees to affirm the existence of a unique solution in a particular interval.

Continuous Coefficients

The coefficients of a differential equation profoundly impact the solution's behavior. For the theorem guaranteeing a unique solution to apply, these coefficients must be continuous in the interval considered. In our problem, the coefficient \( (\text{cos}\text{ }t) \) is continuous for all real \( t \), but \( 3(\text{ln} |t|) \) only maintains continuity for \( t eq 0 \) due to the logarithmic function. To define the longest interval of continuity that incorporates the initial condition \( (y(2)=3, y^{\text{\textquotesingle}}(2)=1) \), we assess the behavior of the coefficients in relation to the point \( t = 2 \).

By recognizing both the limitless continuity of the cosine function and the continuity of \( 3(\text{ln} |t|) \) for all non-zero \( t \), it becomes clear that the longest interval ensuring continuity – and therefore the unique solution for our initial value problem – is \( (0, \text{\textquotesingle}\text{\textquotesingle}) \), which includes \( t = 2 \). This critical assessment guides students in identifying appropriate solution domains without requiring them to solve the actual equation, illustrating the importance of analyzing coefficients in differential equations.