Question

Consider the initial value problem \(t^{2} y^{\prime}-y^{2}=0, y(1)=1\). (a) Determine the largest open rectangle in the \(t y\)-plane, containing the point \(\left(t_{0}, y_{0}\right)=(1,1)\), in which the hypotheses of Theorem \(2.2\) are satisfied. (b) A solution of the initial value problem is \(y(t)=t\). This solution exists on \(-\infty

Short answer

#Short Answer# For part (a), the largest open rectangle in the ty-plane containing the point (t_0, y_0) = (1,1) and satisfying the hypotheses of Theorem 2.2 is given by 0 < t, y ∈ R. For part (b), the existence of a solution on the entire domain does not contradict Theorem 2.2. The theorem provides conditions for the existence and uniqueness of a solution in a specific interval containing the initial point, and the existence of a solution on a larger domain is consistent with this but does not contradict the theorem.

Step-by-step solution

Identify the given function and its partial derivative with respect to y

We have the given differential equation: t^2 y' - y^2 = 0 We first need to find the function f(t, y) such that the given equation can be written in the form: y' = f(t, y) By solving for y', we get: y' = \frac{y^2}{t^2} Now, we need to find the partial derivative with respect to y of f(t, y): \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( \frac{y^2}{t^2} \right) \frac{\partial f}{\partial y} = \frac{2y}{t^2}

Determine the continuity of f(t, y) and ∂f/∂y

As we can see, both f(t, y) and ∂f/∂y have denominators in the form of t^2: f(t, y) = \frac{y^2}{t^2} \frac{\partial f}{\partial y} = \frac{2y}{t^2} These functions are continuous everywhere except when their denominator is zero, i.e., t = 0. As long as t ≠ 0, the functions are continuous.

Determine the largest open rectangle satisfying Theorem 2.2

From the previous step, we observed that the given functions are continuous everywhere except t = 0. Hence, the largest open rectangle (in the ty-plane) containing the point (t_0, y_0) = (1,1) and satisfying the hypotheses of Theorem 2.2 should satisfy t ≠ 0, or 0 < t. Taking all these facts into account, the largest open rectangle is given by: 0 < t, y ∈ R ( Note: R denotes the set of all real numbers. )

Analyze if the existence of a solution on the entire domain contradicts Theorem 2.2

We are given a solution to the initial value problem, y(t) = t. This solution exists on the entire domain of t (t ∈ R). However, this fact does not contradict Theorem 2.2 because the theorem provides conditions for the existence and uniqueness of a solution in a certain interval containing the initial point. If the continuity conditions are satisfied, the solution is guaranteed to exist in that specific interval. The existence of a solution on a larger domain is not contradictory, but rather, it provides information beyond the scope of Theorem 2.2. In this case, y(t) = t is a solution that exists on the entire domain of t, which is consistent with our analysis in step 3 (which showed that the given functions are continuous for all t ≠ 0).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are fundamental in expressing how quantities change and are used to model a wide range of real-world problems in physics, engineering, biology, economics, and more. An initial value problem is a specific type of differential equation where the solution is sought given the initial condition of the function at a certain point.

An example of an initial value problem is the equation given in the exercise, which takes the form \(t^{2}y' - y^{2} = 0\), with \(y(1)=1\). Here, the function \(y(t)\) and its derivative \(y'\) with respect to \(t\) are related by the equation, expressing how \(y\) changes over time. The solution \(y(t) = t\) is an explicit representation of how the variable \(y\) changes with respect to \(t\) and satisfies the initial condition.

Existence and Uniqueness Theorem

The existence and uniqueness theorem provides the conditions under which an initial value problem has a unique solution. This theorem ensures that if we start with a particular state of a system, there is exactly one way the system will evolve over time given the differential equation.

In the context of the problem, Theorem 2.2 demands that the function \(f(t, y)\) and its partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y}\), are continuous in some rectangle containing the initial point \((t_0, y_0)\). As we've seen in the exercise, the functions \(f(t, y)\) and \(\frac{\partial f}{\partial y}\) are continuous except when \(t = 0\). The theorem doesn't guarantee the solution exists for all \(t\), but does for an interval around \(t_0\), so having a solution for all \(t\) doesn't contradict the theorem.

Partial Derivatives

Partial derivatives are a type of derivative used when dealing with functions of multiple variables. They measure how a function changes as one variable is varied while keeping other variables constant. In the dialogue of the differential equation, determining partial derivatives is crucial for figuring out whether the conditions of the existence and uniqueness theorem are met.

In the step-by-step solution provided, the calculation \(\frac{\partial f}{\partial y}\) resulted in \(\frac{2y}{t^2}\). This represents how the rate of change of the function \(f(t, y)\) with respect to \(y\) behaves. For our specific initial value problem, it's essential to check that this partial derivative is continuous to satisfy the conditions mentioned in Theorem 2.2.

Continuity

Continuity refers to the property of a function to have no breaks, jumps, or holes in its graph. A function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point. In other words, you can draw the function without lifting your pencil. For the theorem to hold, the function \(f(t, y)\) and its partial derivative must be continuous where the solution is sought.

The step-by-step solution for the exercise establishes that \(f(t, y) = \frac{y^2}{t^2}\) and \(\frac{\partial f}{\partial y} = \frac{2y}{t^2}\) are continuous provided \(t eq 0\). Since the initial condition does not occur at \(t=0\), the requirement for continuity is fulfilled in this problem, affirming that the solution to the initial value problem fits within the constraints of the existence and uniqueness theorem.