Problem 9 - Advanced Engineering Mathematics

Question

Find an interval centered about \(x=0\) for which the given initial-value problem has a unique solution. $$ (x-2) y^{\prime \prime}+3 y=x, y(0)=0, y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The interval is \((-2, 2)\).

Step 1: Understanding the Problem

We need to find an interval centered at \(x = 0\) where the initial-value problem (IVP) has a unique solution. The given differential equation is \((x-2) y^{\prime\prime} + 3y = x\), and we also have initial conditions \(y(0) = 0\) and \(y'(0) = 1\).

Step 2: Identifying the Singular Point

The coefficient of \(y^{\prime\prime}\) is \(x-2\). The problem will have singular points where this coefficient is zero, which occurs at \(x = 2\). Therefore, \(x = 2\) is a singular point of the differential equation.

Step 3: Finding the Interval of Validity

The differential equation can be rewritten in normal form as \(y^{\prime\prime} + \frac{3}{(x-2)}y = \frac{x}{(x-2)}\). The function \(\frac{3}{x-2}\) becomes undefined at \(x = 2\). Therefore, the interval centered at \(x=0\) can extend until it approaches \(x = 2\) without including it.

Step 4: Defining the Interval

We select an interval symmetric around \(x = 0\) that does not include the singular point \(x = 2\). This means the interval can extend from \(-2\) to slightly less than \(2\), or \((-a, a)\) such that \(a < 2\).

Step 5: Conclusion

The largest interval centered at \(x = 0\) for which the initial-value problem has a unique solution is \((-2, 2)\). Any interval meeting this criterion and not including \(x = 2\) will have a unique solution for the IVP.

Key Concepts

Singular Point

In the study of differential equations, a singular point is a value of the independent variable at which the equation ceases to be well-behaved in some manner, such as being undefined or infinite.
For example, in our given differential equation

• yy^{\prime\prime} + \frac{3}{(x-2)}y = \frac{x}{(x-2)}

the location where the coefficient of the highest derivative, here \ifrac{3}{x-2}\, becomes undefined is called the singular point.
A vital task while solving any differential equation is identifying these points. They help determine where the solutions might not exist or where the solution pattern might change.

In our example, since the coefficient \(x-2\) makes the term undefined at \(x=2\), \(x=2\) is the singular point.
Recognizing this singular point allows us to consider intervals that stay away from it, ensuring the solution remains valid in those regions.

Differential Equation

A differential equation is an equation that involves an unknown function and its derivatives.
These equations are fundamental because they describe various phenomena in physics, engineering, biology, and more.

• They can range from simple forms, like \(y' = ky\), which models exponential growth, to more complex forms that include higher-order derivatives.

The differential equation in our problem

• \((x-2) y'' + 3y = x\)

ncludes a second-order derivative \(y''\), which suggests the function \(y\) involves changes in its rate of change.
Solving a differential equation typically involves finding the unknown function that satisfies it under given conditions, such as initial values or boundary values.

In this case, initial conditions \(y(0) = 0\) and \(y'(0) = 1\) assist in determining a specific solution that fits the particular scenario modeled by the equation.

Interval of Validity

The interval of validity refers to the region along the independent variable, say \(x\), in which the solution to the differential equation is uniquely defined and conforms to the initial or boundary conditions provided.
Identifying this interval is crucial because solutions to differential equations might not exist throughout all \(x\)-values, especially near singular points.
For our initial value problem, the task is to establish where the solution remains valid and unique around \(x=0\), the given starting point.

To achieve this, recognized intervals must avoid singular points.

• Given the singular point at \(x=2\), the interval for a valid and unique solution must stay below \(x=2\).
• This requires selecting \((-a, a)\) such that \(a < 2\), forming a symmetric interval around \(x=0\).

Overall, the interval \((-2, 2)\) fits these conditions, providing a range where the differential equation with its initial conditions is satisfied uniquely without reaching the singular point.