Question

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

Short answer

Answer: The singular points of the given differential equation are \(x = -2\) and \(x = 1\). Both of these singular points are regular singular points.

Step-by-step solution

Finding Singular Points

First, let's find the singular points (if any) in our given equation by finding the zeros of the coefficient functions. A singular point occurs when the coefficient of the highest order derivative becomes zero. In this case, the coefficient of \(y''\) is \((x+2)^2(x-1)\). So, we can solve for its zeros as follows: \((x+2)^2(x-1) = 0\) Which gives us the singular points \(x = -2\) and \(x = 1\).

Checking If Singular Points Are Regular or Irregular

Next, let's determine whether the singular points are regular or irregular. A singular point is regular if the coefficients of the lower order derivatives, when divided by the highest order coefficient, have a singularity of at most a simple pole at that point. For a differential equation in the general form: \(P(x)y'' + Q(x)y' + R(x)y = 0\) A point \(x = x_0\) is a regular singular point if \(P(x)\) has a zero at \(x_0\), and the functions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) have singularities of at most simple poles at that point. In our case the given differential equation is: \((x+2)^2(x-1)y'' + 3(x-1)y' - 2(x+2)y = 0\) So, \(P(x) = (x+2)^2(x-1)\), \(Q(x) = 3(x-1)\), and \(R(x) = -2(x+2)\) We have already found the singular points \(x = -2\) and \(x = 1\). Now let's check if these points are regular or irregular: 1. For \(x = -2\): \(\frac{Q(x)}{P(x)} = \frac{3(x-1)}{(x+2)^2(x-1)}\), and \(\frac{R(x)}{P(x)} = \frac{-2(x+2)}{(x+2)^2(x-1)}\) At \(x = -2\), both \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) have simple poles. Therefore, \(x = -2\) is a regular singular point. 2. For \(x = 1\): \(\frac{Q(x)}{P(x)} = \frac{3(x-1)}{(x+2)^2(x-1)}\), and \(\frac{R(x)}{P(x)} = \frac{-2(x+2)}{(x+2)^2(x-1)}\) At \(x = 1\), both \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) have simple poles. Therefore, \(x = 1\) is a regular singular point. So, both \(x = -2\) and \(x = 1\) are regular singular points for the given equation.

Key concepts

Regular Singular Points

In differential equations, singular points are the points where the solution to the equation might misbehave or become undefined. These points are crucial in understanding the behavior of solutions around them. Regular singular points are a specific type of singularity where the solution may still be well-behaved under certain conditions.
For a regular singular point to exist at a point \(x = x_0\), the coefficient of the highest derivative, \(P(x)\), must vanish at \(x_0\). Additionally, the expressions \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) must not be more severe than a simple pole at that point. A simple pole means the expression grows like \(\frac{1}{x-x_0}\) as \(x\) approaches \(x_0\).
In the given equation, we perform this analysis for points \(x = -2\) and \(x = 1\), examining \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\) at these points. Since both expressions result in simple poles, these points are regular singular points.

Irregular Singular Points

Irregular singular points arise in differential equations when the behavior of the solution near the singular point is more complicated than that at a regular singular point. These are the singularities where the coefficients divided by the highest coefficient have poles of order higher than one or even essential singularities.
For the differential equation given, we determined whether there are any irregular singular points by examining the nature of the singularity at points \(x = -2\) and \(x = 1\). If either \(\frac{Q(x)}{P(x)}\) or \(\frac{R(x)}{P(x)}\) showed poles higher than first order at these points, they would be classified as irregular singular points.
Since at both \(x = -2\) and \(x = 1\), these functions were merely first-order poles, these points are not considered irregular singular points.

Coefficient Functions in Differential Equations

Coefficient functions in differential equations play a pivotal role in determining the nature of singular points. The equation’s general form \(P(x)y'' + Q(x)y' + R(x)y = 0\) uses three distinct coefficient functions: \(P(x)\), \(Q(x)\), and \(R(x)\).
1. **\(P(x)\)**: This is the coefficient of the highest order derivative. It determines where singular points occur since a singular point arises when \(P(x) = 0\).
2. **\(Q(x)\) and \(R(x)\)**: These coefficients are for the lower order derivatives. Their behavior relative to \(P(x)\) determines if a singular point is regular or irregular by examining the poles of \(\frac{Q(x)}{P(x)}\) and \(\frac{R(x)}{P(x)}\).
In practice, recognizing how these coefficients interact gives insight into the solutions' behavior near singular points. By solving \(P(x) = 0\), we found \(x = -2\) and \(x = 1\) as singular points, which then needed further analysis based on the behavior of \(Q\) and \(R\).