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. $$ t(t-4) y^{\prime \prime}+(t-2) y^{\prime}+y=0, \quad y(2)=1 $$

Short answer

Based on the Existence and Uniqueness Theorem, it has been determined that the solution of the initial value problem exists in the interval (0, 4) or smaller intervals like (1, 3). These intervals include the initial value, t = 2, and contain continuous coefficients for the differential equation.

Step-by-step solution

Analyze the coefficients and their derivatives

Look at the coefficients of our given differential equation: $$ t(t-4)y'' +(t-2)y' + y = 0 $$ The three coefficients are: 1. \(t(t-4)\) for the second derivative, \(y''\) 2. \((t-2)\) for the first derivative, \(y'\) 3. \(1\) for the function \(y\) Notice that all these coefficients are continuous functions everywhere in their respective domains, and their domains are as follows: 1. \(t(t-4)\): continuous for all \(t \neq 0, 4\) (due to multiplication by \(t\) and \(t-4\)) 2. \((t-2)\): continuous for all \(t \neq 2\) (due to multiplication by \(t-2\)) 3. \(1\): continuous for all \(t\)

Determine the valid interval

The goal is to find an interval in which all of the coefficients of the given equation are continuous. The interval must contain the initial value, which in this case is \(t = 2\). From the analysis in step 1, the coefficient \((t - 2)\) for the first derivative has a discontinuity at \(t = 2\). However, this discontinuity is removable. Applying the limit: $$ \lim_{t \to 2} (t-2) = 0 $$ Since the limit exists and is finite, this discontinuity is removable, and we can consider \((t - 2)\) continuous at \(t = 2\). Now we have to find the interval we are looking for. Let's analyze the coefficients: 1. \(t(t-4)\): continuous at \(t = 2\) 2. \((t-2)\): continuous at \(t = 2\) 3. \(1\): continuous at \(t = 2\) Since all coefficients have removable discontinuities, we should choose an interval to avoid the points of discontinuity except for \(t = 2\). One suitable interval is \( (0, 4) \) which satisfies the initial value of \(t = 2\). However, smaller intervals around \(2\) are also acceptable, for example, \((1, 3)\).

Provide the final answer

The interval in which the solution of the initial value problem is certain to exist is \((0,4)\). However, other smaller intervals also exist, like \((1,3)\).

Key concepts

Differential Equations

Differential equations are essential mathematical tools used to describe various physical phenomena by relating functions and their derivatives. In the given problem, we have a second-order linear differential equation:

• Second-order: because it involves a second derivative, denoted as \( y'' \).
• Linear: because there are no products of the function and its derivatives, and the equation can be expressed in terms of a linear combination.

Differential equations can model rates of change over time, such as growth, decay, or motion. Solving them provides insight into how a system evolves. Here, our task isn't to solve the equation directly but to understand where a solution might exist, considering the discontinuities present. This understanding is crucial for setting up the right context for solving the problem if further analysis or numerical methods were applied later.

Continuous Functions

Continuous functions are pivotal when determining the existence of solutions for differential equations. A function is continuous if its graph is unbroken, meaning there are no jumps or gaps.
In our differential equation, the coefficients (\( t(t-4), (t-2), 1 \)) must be continuous around the initial value \( t = 2 \) where the solution is said to exist.

To check continuity:

• \( t(t-4) \) is continuous for all \( t eq 0, 4 \), since these terms multiply to create potential zeroes in the denominator, making the function undefined at 0 and 4.
• \( (t-2) \) is continuous for all \( t eq 2 \). However, this is a special case because its discontinuity can be removed.
• The constant function \( 1 \) is always continuous as it doesn't depend on \( t \).

Reviewing these ensures we choose an interval allowing all parts of our equation to work together smoothly without mathematical hiccups.

Existence Interval

An existence interval is a range of input values (in this case, time \( t \)) where the solution to a differential equation is guaranteed to be available. To identify such an interval, we must find where all coefficients of our differential equation remain continuous.

For our exercise, we determined:

• \( t(t-4) \) is not continuous at 0 and 4, hence cannot form part of the edges of our interval.
• \( (t-2) \) appears discontinuous at 2, but its discontinuity is removable, allowing it to be continuous at that point.

Given the requirement for the initial value \( t = 2 \) to be within the interval, one appropriate interval is \((0, 4)\). This avoids non-removable discontinuities while including the value where the initial conditions are set. Other intervals like \((1, 3)\) could work, focusing on regions close to the initial value without exceeding the discontinuous boundaries.

Removable Discontinuity

A removable discontinuity occurs when there is an interruption in the graph of a function, but this disruption can be 'patched' by redefining the function's value at the point of discontinuity. It is a subtle gap that can be fixed.
In our differential equation, the term \( (t-2) \) is directly associated with a removable discontinuity at \( t = 2 \).

To verify and address the discontinuity:

• Compute the limit: \( \lim_{t \to 2} (t-2) = 0 \).
• This limit being finite and existing means the discontinuity can be resolved, effectively making \( (t-2) \) continuous at \( t = 2 \).

By understanding and fixing such discontinuities, we ensure that our chosen interval for the initial value problem is valid, allowing for the complete and consistent application of mathematical methods.