Question

In each of Problems I through 6 determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. $$ (\ln t) y^{\prime}+y=\cot t, \quad y(2)=3 $$

Short answer

Answer: The solution exists in the interval $\frac{\pi}{2} < t < \frac{3\pi}{2}$.

Step-by-step solution

Rewrite the differential equation in terms of \(y'\)

Divide both sides of the equation by \(\ln t\), so that we have: $$ y^{\prime} = -\frac{y}{\ln t} + \frac{\cot t}{\ln t} $$

Identify the functions and their partial derivatives

We have two functions here: 1. \(f(t, y) = -\frac{y}{\ln t}\) 2. \(g(t, y) = \frac{\cot t}{\ln t}\) Now, find the partial derivatives of these functions with respect to y: 1. \(\frac{\partial{f}}{\partial{y}} = -\frac{1}{\ln t}\) 2. \(\frac{\partial{g}}{\partial{y}} = 0\)

Check the continuity of functions and their partial derivatives

We need to verify the continuity of \(f(t, y)\), \(g(t, y)\), \(\frac{\partial{f}}{\partial{y}}\), and \(\frac{\partial{g}}{\partial{y}}\) in a neighborhood around the point (2,3). 1. \(f(t, y)\) is continuous for \(t > 0\) as \(\ln t\) is continuous for positive values of t. 2. \(g(t, y)\) is continuous for \(t \neq k\pi, k \in \mathbb{Z}\) as \(\cot t\) has singularities at multiples of \(\pi\). Also, \(t \neq 0\), to avoid a division by zero. 3. \(\frac{\partial{f}}{\partial{y}}\) is continuous for \(t > 0\) as \(\ln t\) is continuous for positive values of t. 4. \(\frac{\partial{g}}{\partial{y}} = 0\) is continuous for all values of t and y.

Determine an interval for the existence of the solution

The initial condition is given as \(y(2) = 3\). Since all the functions and their partial derivatives with respect to y are continuous for \(t > 0\) and \(t \neq k\pi, k \in \mathbb{Z}\), we can choose an interval for the existence of a solution that satisfies these conditions, for example: $$ \frac{\pi}{2} < t < \frac{3\pi}{2} $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are fundamental in expressing the laws of physics and other disciplines where the rate at which something changes is important to understand. The equation presented in the exercise, \( (\ln t) y^{\prime}+y=\cot t \), is a first-order differential equation because it involves the first derivative of the unknown function, denoted as \( y^{\prime} \).

To solve such an equation, we often need to determine the initial condition, which in this case is \( y(2)=3 \). This information is crucial as it anchors the solution curve at a particular point, and for many differential equations, this initial condition ensures that the solution is unique throughout its domain.

• Rewriting the equation in terms of \( y' \) effectively isolates the derivative on one side, making it easier to analyze and eventually solve.
• Understanding the behavior and characteristics of the differential equation, such as continuity and integrability, is essential to predict the nature of its solutions.

Simplifying the equation and understanding its form helps us identify appropriate methods to find the solution, such as separation of variables, integrating factors, or, in some cases, more advanced techniques like Laplace transforms.

Partial Derivatives

Partial derivatives are an extension of the derivative concept to functions of multiple variables. They describe how a function changes as one variable changes, while other variables are held constant. In the context of differential equations, partial derivatives appear when we examine functions that depend on both the independent variable (usually time or space) and the dependent variable (the function we are trying to solve for).

The exercise involves finding the partial derivatives of two functions, \( f(t, y) = -\frac{y}{\ln t} \) and \( g(t, y) = \frac{\cot t}{\ln t} \), with respect to \( y \). Calculating these partial derivatives, \( \frac{\partial{f}}{\partial{y}} = -\frac{1}{\ln t} \) and \( \frac{\partial{g}}{\partial{y}} = 0 \) respectively, helps us in analyzing the behavior of the differential equation, particularly in terms of the existence and continuity of solutions.

• Checking the continuity of these partial derivatives is a step towards ensuring that the conditions for existence and uniqueness theorems for the differential equation are met.
• The interval determined for the existence of the solution relies on the function and its partial derivatives being continuous within that interval.

It is important to recognize that the continuity of partial derivatives plays a crucial role in the predictability and stability of solutions to initial value problems.

Existence and Uniqueness of Solutions

The existence and uniqueness of solutions to differential equations are topics of significant interest, as they provide us with the assurance that a well-defined, predictable solution exists. An initial value problem is a differential equation paired with an initial condition. The exercise focuses on verifying whether there is a solution to the equation and determining an interval in which this solution is certain to exist without actually solving the equation.

For first-order equations, like in our exercise, the Picard-Lindelöf theorem is a classical result which states that if the function and its partial derivatives are continuous in a neighborhood around the initial condition, there exists a unique solution to the initial value problem in some interval around the initial point.

• The theorem requires the function to satisfy certain conditions known as the Lipschitz condition, which relates to the nature of the growth of the function around a point.
• An interval where the solution can exist is chosen based on where the function and its partial derivatives are continuous. In our case, the interval \( \frac{\pi}{2} < t < \frac{3\pi}{2} \) satisfies the required conditions for the existence of a unique solution near the initial point \( (2, 3) \) since the partial derivatives are continuous and the function does not have singularities within this domain.

This theorem is a cornerstone in understanding the behavior of differential equations and ensures that, under specific conditions, the problems we attempt to solve have meaningful and well-defined answers.