Chapter 4: Problem 6
determine intervals in which solutions are sure to exist. $$ \left(x^{2}-4\right) y^{\mathrm{vi}}+x^{2} y^{\prime \prime \prime}+9 y=0 $$
Short Answer
Expert verified
Answer: The intervals in which solutions are sure to exist for the given equation are \((-\infty, -2)\) and \((2, \infty)\).
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. In applications, these functions usually represent physical quantities, and the derivatives represent their rates of change, making differential equations fundamental in describing various phenomena.
In the context of our exercise, we have a sixth-order linear differential equation, which can be particularly challenging due to the high degree of the derivative. This equation tells us how a function, represented by the variable y, is related to its derivatives up to the sixth-order. Solving such an equation involves finding the function y that satisfies the equation for given conditions. The presence of the term \(x^2 - 4\) introduces discontinuities at \(x = \:\pm2\), making the solution process more intricate as we approach these values.
Understanding the structure of these equations and the methods to solve them, like the one adopted in our step-by-step solution, is crucial to mastering the theory and applications of differential equations.
In the context of our exercise, we have a sixth-order linear differential equation, which can be particularly challenging due to the high degree of the derivative. This equation tells us how a function, represented by the variable y, is related to its derivatives up to the sixth-order. Solving such an equation involves finding the function y that satisfies the equation for given conditions. The presence of the term \(x^2 - 4\) introduces discontinuities at \(x = \:\pm2\), making the solution process more intricate as we approach these values.
Understanding the structure of these equations and the methods to solve them, like the one adopted in our step-by-step solution, is crucial to mastering the theory and applications of differential equations.
Boundary Value Problems
Boundary value problems are a type of differential equation where the solution is sought over a range of values with specified conditions at the boundaries of the range. Unlike initial value problems, where the condition of the solution is given at a single point, boundary value problems require the solution to meet conditions at more than one point.
The exercise provided doesn't specify boundary conditions; however, when applying the Existence and Uniqueness Theorem, we implicitly define the boundary by determining that the solutions exist within certain intervals, avoiding the points of discontinuity at \(x = \pm2\). By finding these intervals, we essentially establish 'soft' boundaries where the solutions are valid, which is a critical step in solving boundary value problems. Understanding how to deal with these conditions is essential for students to tackle a variety of problems in engineering and physics.
The exercise provided doesn't specify boundary conditions; however, when applying the Existence and Uniqueness Theorem, we implicitly define the boundary by determining that the solutions exist within certain intervals, avoiding the points of discontinuity at \(x = \pm2\). By finding these intervals, we essentially establish 'soft' boundaries where the solutions are valid, which is a critical step in solving boundary value problems. Understanding how to deal with these conditions is essential for students to tackle a variety of problems in engineering and physics.
Continuity of Functions
Continuity of functions is a key concept in calculus, which ensures that small changes in the input of a function result in small changes in the output. This is a fundamental property that allows us to make sense of functions as models of real-world processes.
In relation to our exercise, the function on the right-hand side of the equation becomes discontinuous when \(x = \pm2\). A function is continuous on an interval if it is continuous at every point within that interval. Our step-by-step solution leverages this concept by determining the intervals where the function is continuous (\(x < -2\) and \(x > 2\)) and thus where the solutions are sure to exist. Grasping the concept of function continuity helps in understanding not only boundary value problems but also the broader reach of differential equations in modeling continuous phenomena.
In relation to our exercise, the function on the right-hand side of the equation becomes discontinuous when \(x = \pm2\). A function is continuous on an interval if it is continuous at every point within that interval. Our step-by-step solution leverages this concept by determining the intervals where the function is continuous (\(x < -2\) and \(x > 2\)) and thus where the solutions are sure to exist. Grasping the concept of function continuity helps in understanding not only boundary value problems but also the broader reach of differential equations in modeling continuous phenomena.
