Question

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x(3-x) y^{\prime \prime}+(x+1) y^{\prime}-2 y=0\)

Short answer

The singular points of the given differential equation are \(x=0\) and \(x=3\). Both of them are irregular singular points because the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) are not analytic at these points.

Step-by-step solution

Identify singular points

To find the singular points, we will set the leading coefficient \(P(x)\) equal to zero and solve for \(x\). $$x(3-x) = 0$$ This equation has two solutions, \(x = 0\) and \(x = 3\). So the singular points are \(x=0\) and \(x=3\).

Check for regular singular points

Now that we have identified the singular points, we need to check whether they are regular or irregular by analyzing the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\). Divide \(Q(x)\) and \(R(x)\) by \(P(x)\) to obtain the following functions: $$\frac{Q(x)}{P(x)} = \frac{x+1}{x(3-x)}$$ $$\frac{R(x)}{P(x)} = \frac{-2}{x(3-x)}$$

Analyze the functions at each singular point

We will now analyze the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) at each singular point. 1. At \(x = 0\): $$\frac{Q(0)}{P(0)} = \frac{1}{0} \quad \text{and} \quad \frac{R(0)}{P(0)} = \frac{-2}{0}$$ The functions are not analytic at \(x=0\), so \(x=0\) is an irregular singular point. 2. At \(x = 3\): $$\frac{Q(3)}{P(3)} = \frac{4}{0} \quad \text{and} \quad \frac{R(3)}{P(3)} = \frac{-2}{0}$$ The functions are also not analytic at \(x=3\), so \(x=3\) is an irregular singular point.

Conclusion

The singular points of the given differential equation are \(x=0\) and \(x=3\). Both of them are irregular singular points because the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) are not analytic at these points.

Key concepts

Differential Equations

Differential equations are mathematical expressions that describe relationships between functions and their derivatives. They play a pivotal role in modeling various physical phenomena, from the motion of planets to the flow of currents in electrical circuits. A differential equation will often involve a function, such as y, and its derivatives, labeled as y' (the first derivative) or y'' (the second derivative), and possibly the independent variable, such as x. The equation presented in the exercise, \(x(3-x) y^{prime prime}+(x+1) y^{prime}-2 y=0\), is a second-order linear differential equation, where the order indicates the highest derivative present.

Solving a differential equation generally means finding a function or a set of functions that satisfy the given equation. In higher-order differential equations, such as the one provided, we often look for solutions depending on several initial conditions or boundary values. The process of solving these equations can vary from separation of variables, using integrating factors, or applying transformation techniques like Laplace transforms, to more advanced methods such as singular point analysis, which is utilized in this particular textbook exercise.

Singular Point Analysis

When dealing with differential equations, particularly those with variable coefficients, we encounter the concept of singular points. Singular point analysis involves identifying and categorizing the points at which the usual analytic solution methods do not apply directly due to certain peculiarities in the equation's coefficients. These peculiarities often manifest as the coefficient functions becoming undefined or infinite.

In the provided exercise, the process begins by determining where the leading coefficient, P(x), is equal to zero. These points, as we have found, are x=0 and x=3. The significance of identifying singular points lies in the fact that the nature of the solutions to the differential equation can drastically change around these points, and special methods are required to find a valid solution. In particular, singular point analysis distinguishes between regular and irregular singular points, which affects how solutions near these points are studied and approximated.

Regular Singular Point

A regular singular point in a differential equation is a type of singularity that, despite causing the usual solution methods to be ineffective, still allows the equation to have well-behaved solutions in a specific sense. For a point to be considered a regular singular point, the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) must be analytic, or at most have a pole, at that point, where P(x), Q(x), and R(x) are the coefficients of the second, first, and zeroth derivative terms of y, respectively.

Generally, solutions near a regular singular point can be expressed in a power series that may include logarithmic terms. However, in the exercise we analyzed, neither of the identified singular points, x=0 or x=3, was classified as regular since the relevant functions were not analytic at these points. This would mean that finding a series solution around these points would require different strategies, such as employing the method of Frobenius.

Irregular Singular Point

An irregular singular point is one where the solutions to a differential equation exhibit more unpredictable behavior compared to a regular singular point. For a point to be classified as an irregular singular point, the quotient functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) must be non-analytic at the singular point, which means they could have essential singularities or the behavior of the solutions could be quite complex.

Solutions near an irregular singular point can vary widely and may involve exponential or other non-analytic functions, making them far more challenging to determine. As per the exercise, the singular points at x=0 and x=3 were confirmed to be irregular since the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) are not analytic at these points. This suggests that the solutions may display a behavior that is significantly different from that around ordinary points, or even regular singular points, and special analytical techniques or numerical approximations might be necessary to approach these solutions.