Question

Use the results of Problem 21 to determine whether the point at infinity is an ordinary point, a regular singular point, or an irregular singular point of the given differential equation. \(x^{2} y^{\prime \prime}+x y^{\prime}-4 y=0\)

Short answer

Question: Determine whether the point at infinity is an ordinary point, a regular singular point, or an irregular singular point for the differential equation \(x^{2} y^{\prime \prime}+x y^{\prime}-4 y=0.\) Answer: The point at infinity is a regular singular point.

Step-by-step solution

Identify the Singular Points of the Differential Equation

First, we need to identify the singular points of the given differential equation: \(x^{2} y^{\prime \prime}+x y^{\prime}-4 y=0.\) Recall that singular points occur when the coefficient of the highest derivative vanishes. In this case, singular points occur when the coefficient of \(y^{\prime \prime}\), which is \(x^2\), equals zero, i.e., \(x=0\).

Determine the Regularity of the Singular Point at x=0

At the singular point \(x=0\), we rewrite the given differential equation and expand the terms near the singular point: \(x^{2} y^{\prime \prime}+x y^{\prime}-4 y=(x^2+x-4) y^{\prime \prime}+(x^3+x^2-4x) y^{\prime}+(x^4-4x^3+4x^2) y=0.\) Now let's examine if the singular point \(x=0\) is regular or not. For this, we check whether the limit as \(x \rightarrow 0\) of the following expression is finite: \(\lim \limits_{x \rightarrow 0} \frac{x^2+x-4}{x^3+x^2-4x},\) and the limit of this expression is \(\lim \limits_{x \rightarrow 0}\frac{1+\frac{1}{x}-\frac{4}{x^2}}{1+\frac{1}{x}-\frac{4}{x^3}}.\) The limit is zero, which means that the singular point \(x=0\) is a regular singular point.

Investigate the Point at Infinity

To determine if the point at infinity is an ordinary point, a regular singular point, or an irregular singular point, we first change the variable with the substitution \(x=\frac{1}{z}\). Then, we rewrite the given differential equation with respect to the new variable \(z\): \((\frac{1}{z^2}) \left(\frac{d^2y}{dz^2} \right) + (\frac{1}{z})\left(\frac{dy}{dz}\right) -4y = 0.\) Now, we investigate the regularity of the point at infinity by checking the limit as \(z \rightarrow 0\) of the following expression: \(\lim \limits_{z \rightarrow 0} \frac{\frac{1}{z^2}+\frac{1}{z}-4}{\frac{1}{z^3}+\frac{1}{z^2}-4\frac{1}{z}},\) and we have \(\lim \limits_{z \rightarrow 0}\frac{1+\frac{z}{1-z^2}}{1+\frac{z^2}{1-z^3}}.\) The limit is 1, which means that the point at infinity is a regular singular point.

Conclusion - Determining the nature of the point at infinity

Based on our analysis above, the point at infinity of the given differential equation \(x^{2} y^{\prime \prime}+x y^{\prime}-4 y=0\) is found to be a regular singular point, which means that the points at infinity are not extraordinaries.

Key concepts

Singular Points

In differential equations, a singular point is where the solution to an equation behaves unusually or meets certain restrictions. In simpler terms, it's a specific value of the independent variable where the equation might break down or become challenging to solve.

Singular points are vital because they help us understand the nature and behavior of solutions around these points. Using an example, for the differential equation \(x^{2} y^{\prime\prime} + x y^{\prime} - 4 y = 0\), any value of \(x\) that zeroes out the leading coefficient (\(x^2\)) is a singular point. Here, \(x = 0\) is a singular point.

These points are not just potential problems; they can also guide us to understand how the solutions behave. They affect the form and nature of solutions, often leading to intriguing and complex behaviors.

Regular Singular Point

A regular singular point is a type of singularity where the solution to a differential equation behaves in a controlled manner. Even though these points can complicate solving the equation, their impact is predictable.

For a point to be considered a regular singular point, the limits of specific expressions related to the coefficients of the equation should remain finite. In our example equation, \(x^{2} y^{\prime\prime} + x y^{\prime} - 4 y = 0\), we assessed the nature of the singular point at \(x = 0\). By substituting it into specific expressions and examining their limits, it was determined that the limit was finite and therefore, \(x = 0\) is a regular singular point.

This means that the general nature of solutions near this point is well-behaved, although special methods, like the Frobenius method, often need to be employed to find explicit solutions.

Irregular Singular Point

An irregular singular point is far more unpredictable compared to a regular singular point. At these points, the solutions may exhibit extremely erratic behavior, and the mathematics can become quite complex.

Irregular singular points don't always lead to finite limits in the expressions that determine the behavior of singular points. This unpredictability tends to complicate finding solutions, and traditional methods might not always be applicable without further adaptations.

For our specific exercise focused on the given differential equation, \( x^{2} y^{\prime\prime} + x y^{\prime} - 4 y = 0 \) , our analysis confirmed that the points in consideration, such as the point at infinity, did not end up as irregular. Identifying whether a point is irregular is essential as they often require advanced techniques and theories when tackling the corresponding solutions.