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

Short answer

Answer: (1, ∞)

Step-by-step solution

Analyze the given differential equation

The given initial value problem is: $$ t y^{\prime \prime}+3 y=t, \quad y(1)=1, \quad y^{\prime}(1)=2 $$ This is a second-order linear differential equation, where \(t\) is the independent variable and \(y\) is the dependent variable. The differential equation can be written in the following form: $$ t y^{\prime \prime}+0y^{\prime}+3y = t $$

Identify the coefficients

For this standard form, the coefficients are: $$ p(t) = 0, \quad q(t) = 3/t $$ Note that \(p(t)\) and \(q(t)\) are continuous functions in their respective domains, \((1, \infty)\) and \((0, \infty)\).

Check the conditions for the existence and uniqueness of the solution

According to the theorem of existence and uniqueness for a linear differential equation, the required conditions are that \(p(t)\) and \(q(t)\) should be continuous functions on an interval \((a,b)\) containing the point \(t=1\). We have determined that \(p(t)\) and \(q(t)\) are continuous functions in their respective domains of \((1, \infty)\) and \((0, \infty)\). Since \(t=1\) is within both intervals of \(p(t)\) and \(q(t)\), there exists a unique twice differentiable solution in their intersection, which is \((1, \infty)\).

Determine the longest interval of existence and uniqueness

As we found in the previous step, the intersection of the domains of \(p(t)\) and \(q(t)\) is \((1, \infty)\). Since this interval contains the initial value \(t=1\), this is the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. In conclusion, the longest interval of existence and uniqueness for this initial value problem is \((1, \infty)\).

Key concepts

Second-Order Linear Differential Equation

When we encounter a second-order linear differential equation, we're looking at an equation that can be described by the form \( a(t)y'' + b(t)y' + c(t)y = g(t) \), where \( y'' \) denotes the second derivative of \( y \) with respect to \( t \) (the independent variable), \( b(t) \) and \( c(t) \) are functions of \( t \) that represent the coefficients of \( y' \) (the first derivative) and \( y \) itself, respectively. The term \( g(t) \) stands for a known function, which could be zero or any other function of \( t \).

In our exercise, the given differential equation is \( t y'' + 3y = t \), which fits the necessary form with \( a(t) = t \), \( b(t) = 0 \), and \( c(t) = 3 \). The goal is often to find a function \( y(t) \) whose second and first derivatives satisfy this relationship for every point in a certain interval. Understanding second-order equations is crucial because they frequently appear in physics and engineering, dictating phenomena such as oscillations and vibrations.

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem provides fundamental assurance when solving initial value problems involving differential equations: if certain conditions are met, we're guaranteed that there is exactly one solution that fits the given equation and initial conditions. For second-order linear differential equations, the theorem states that if the coefficient functions and the nonhomogeneous term are continuous on an open interval containing the initial value of \( t \) (let's denote it as \( t_0 \)), there will be a unique solution that extends through this interval.

This theorem is essential for validating that our efforts to solve a differential equation aren't in vain—it tells us that a solution exists and that it's the only one. Applying this to our exercise, since the coefficients \( p(t) = 0 \) and \( q(t) = 3/t \) are continuous on their respective intervals which include \( t=1 \) (our initial condition), the theorem confirms the existence of a unique solution extending through the intersection of these intervals, \( (1, \)\(\infty) \).

Twice Differentiable Solution

A solution to a differential equation being 'twice differentiable' means it's smooth enough to have a second derivative that's continuous. In more practical terms, the solution's graph has no sharp corners or breaks, which is often necessary for models in physics or engineering that require smooth and predictable changes.

In the context of our exercise, we're not asked to find this solution explicitly but to establish confidence in its existence over a particular interval. By ensuring that the coefficients in the differential equation are continuous over an interval that includes the initial condition, we satisfy the requirements that secure a twice differentiable solution, as per the existence and uniqueness theorem. Therefore, for the interval \( (1, \)\(\infty) \), we can affirm the presence of this type of smooth solution.