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. $$ \left(4-t^{2}\right) y^{\prime}+2 t y=3 t^{2}, \quad y(1)=-3 $$

Short answer

Answer: The solution is certain to exist on the interval (-2, 2).

Step-by-step solution

Rewrite the ODE in the required form

Divide both sides by (4-t^2): $$ y' + \frac{2t}{4-t^2}y = \frac{3t^2}{4-t^2} $$ Now, we can see that F(t, y) = \(\frac{2t}{4-t^2}\) and G(t, y) = \(\frac{3t^2}{4-t^2}\).

Determine continuity of functions F(t, y) and G(t, y)

Both F(t, y) and G(t, y) are continuous everywhere except where the denominator is zero, which is at t = 2 and t = -2. Thus, we have a discontinuity at t = 2 and t = -2.

Define the rectangular region

Since the initial point is (1, -3), we want to find a rectangular region in the plane where both F(t, y) and G(t, y) are continuous. The discontinuities are at t=2 and t=-2. So, we want the rectangle to not include these points.

Determine an interval for the solution

Since we want an interval around the given initial point (1, -3) that does not contain the integer points of discontinuity, we can choose any interval that lies within this space. A simple choice would be to take the open interval (-2, 2), which does not include the discontinuity points and contains the initial point. In conclusion, the solution for the given initial value problem is certain to exist on the interval (-2, 2).

Key concepts

Differential Equations

Differential equations are a type of equation that involve an unknown function and its derivatives. They play a critical role in various fields like physics, engineering, economics, and more, because they model the way things change over time or space. In the given problem, students are introduced to a first-order differential equation, which includes a term with the first derivative of the function y, denoted as y'. The equation \( (4-t^2) y' + 2t y = 3t^2 \) expresses a relationship between a function y(t) and its rate of change y' with respect to the variable t.

The process of solving an initial value problem (IVP) starts with isolating the derivative term y', leading to the rewritten form presented in the solution. In practice, understanding the structure of a differential equation allows students to apply algorithms or techniques such as separation of variables, integrating factors, or exact equations to find the function y(t) that solves the equation under the given initial condition y(1)=-3.

Existence Interval

When working with initial value problems, it is important to determine the existence interval, which is the range of the independent variable, in this case t, over which a solution to the differential equation is guaranteed to exist and be unique. The solution provided smartly outlines how to approach this by examining the continuity of the functions involved. A key point to improve comprehension would be to explain the significanc of continuity for the existence interval.

An existence interval must avoid points of discontinuity to ensure the solution does not become undefined or infinite. For the problem at hand, the functions F(t, y) and G(t, y) in the rewritten equation have discontinuities at t = 2 and t = -2 since these make the denominators zero, leading to division by zero -- an undefined operation. Therefore, the interval chosen for the existence of the solution, (-2, 2), carefully excludes these values. Explaining the rationale behind avoiding discontinuities can enhance a learner's understanding of how to establish existence intervals in future equations they might encounter.

Continuity of Functions

Continuity of functions is a fundamental concept in mathematics, particularly in calculus and analysis. A function is said to be continuous at a point if the function value at that point is equal to the limit of the function as it approaches that point from either direction. Discontinuities, as noted in this problem's Step 2, occur where a function becomes undefined or experiences a sudden jump in value.

For the IVP in the exercise, continuity is significant because the existence and uniqueness theorems for differential equations require the involved functions to be continuous in a neighborhood around the initial value. If there is a discontinuity, like at t = 2 or t = -2 for F(t, y) and G(t, y), then the theorems that guarantee the existence of a unique solution do not apply. Therefore, ensuring that the functions F(t, y) and G(t, y) are continuous in the selected interval (-2, 2) is essential for a valid solution. Illustrating the connection between continuity and the theorems can make students appreciate why such care is taken to find an appropriate existence interval for the solution of the differential equation.