Question

Find all the regular singular points of the given differential equation. Determine the indicial equation and the exponents at the singularity for each regular singular point. \(x^{2} y^{\prime \prime}-x(2+x) y^{\prime}+\left(2+x^{2}\right) y=0\)

Short answer

2. Is the singular point a regular or irregular singular point? 3. What are the indicial equation and the exponents at the singularity?

Step-by-step solution

Identify the Singular Points

To find the singular points, first, we need to write down the given differential equation in its standard form by dividing through by the leading coefficient \(x^2\): $$y^{\prime \prime}-\frac{2+x}{x} y^{\prime}+\frac{2+x^{2}}{x^{2}} y=0$$ A singular point of a linear differential equation is a value of x at which the leading coefficient of the highest-order derivative becomes zero. In this case, the singular points are the ones for which the leading coefficient of the highest-order derivative (\(x^2\)) become zero. Since the equation is already in its standard form, we can simply equate the denominator to zero: $$x^2 = 0 \Rightarrow x = 0$$ Thus, the only singular point is \(x=0\).

Determine if the Singular Point is Regular or Irregular

A singular point is considered regular if the coefficients of the lower-order derivatives are analytic functions (i.e., they can be expressed as a convergent power series) about the singular point. In our case, we have the following coefficients: $$P(x) = \frac{2+x}{x} \quad \textrm{and} \quad Q(x) = \frac{2+x^{2}}{x^{2}}$$ By expanding these coefficients using a power series, we get: $$P(x) = \frac{2}{x} + 1 \quad \textrm{and} \quad Q(x) = \frac{2}{x^2} + 1$$ These coefficients can be expressed as convergent power series about the singular point \(x=0\), as they have a finite number of terms which involve negative powers of x. Therefore, the singular point \(x=0\) is a regular singular point.

Find the Indicial Equation and the Exponents at the Singularity

To find the indicial equation and exponents, we use Frobenius' method. We will look for solutions of the form: $$y(x) = x^r \sum_{n=0}^{\infty} a_n x^n$$ Differentiating \(y(x)\) with respect to \(x\) yields: $$y^{\prime}(x) = rx^{r-1} \sum_{n=0}^{\infty} a_n x^n + x^r \sum_{n=0}^{\infty} na_n x^{n-1}$$ And differentiating again with respect to \(x\) gives: $$y^{\prime \prime}(x) = r(r-1)x^{r-2} \sum_{n=0}^{\infty} a_n x^n + 2rx^{r-1} \sum_{n=0}^{\infty} na_n x^{n-1} + x^r \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}$$ Substituting \(y(x)\), \(y^{\prime}(x)\), and \(y^{\prime \prime}(x)\) back into the standard form of the differential equation yields: $$\left[r(r-1)x^{r-2} + 2rx^{r-1} + x^r \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}\right] \\ -\left[\frac{2}{x} + 1 \right] \left[rx^{r-1} + x^r \sum_{n=0}^{\infty} na_n x^{n-1}\right] \\ +\left[\frac{2}{x^2} + 1 \right] \left[x^r \sum_{n=0}^{\infty} a_n x^n\right] = 0$$ Comparing the powers of \(x\), we see that the constant term of this equation is equal to: $$r(r-1) - 2r + 2 = r^2 - 3r + 2 = 0$$ This quadratic equation is the indicial equation. Its solutions are: $$r = \frac{3 \pm \sqrt{3^2 -4 \cdot 1 \cdot 2}}{2} = \frac{3 \pm \sqrt{1}}{2}\Rightarrow r = 1, 2$$ These are the exponents at the regular singular point \(x=0\).

Key concepts

Indicial Equation

The indicial equation plays a crucial role when solving linear differential equations with regular singular points using the Frobenius method. It is a quadratic equation derived from attempting to find a solution in the form of a power series. This method explores a solution of the type:

• \( y(x) = x^r \sum_{n=0}^{\infty} a_n x^n \)

The indicia equation emerges as we substitute this solution back into the differential equation and equate the coefficients of the lowest power of \(x\) to zero.
This gives us a quadratic equation in \(r\), where \(r\) represents the roles of characteristic exponents of the solution. In our example, the indicial equation is \(r^2 - 3r + 2 = 0\), which simplifies to \(r = 1, 2\). These are the values that define the behavior of the solution near the singular point.

Frobenius Method

The Frobenius method is used to find solutions to linear differential equations with regular singular points. This technique is an extension of finding power series solutions. It is valuable because it allows us to consider cases where a regular power series might fail due to singular points.
Here's the general approach for the Frobenius method:

• First, express the differential equation in standard form.
• Assume a solution of the form \( y(x) = x^r \sum_{n=0}^{\infty} a_n x^n \).
• Differentiate as necessary and substitute back into the differential equation.
• Derive the indicial equation to determine possible \(r\) values.
• Use the prime \(r\) values to find the coefficients \(a_n\) of the series.

In our exercise, the Frobenius method helps to solve around the regular singular point \(x=0\), making use of the exponents derived from the indicial equation as part of the series solution.

Linear Differential Equations

Linear differential equations are equations involving derivatives of a function with respect to one variable. These equations appear frequently across physics, engineering, and many other fields and can often be classified by order, degree, and linearity.
A distinguishing feature of linear differential equations is that all the derivatives are linearly dependent on the unknown function and its derivatives, allowing for techniques such as superposition. The standard form of a second-order linear differential equation is:

• \( y'' + P(x)y' + Q(x)y = 0 \)

The focus lies on the solution methods, especially under various conditions such as the presence of singularities, which is where methods like Frobenius come into play.

Singularities in Differential Equations

Singularities in differential equations are points at which the usual rules for solving the equation break down due to behaviors like division by zero. In the given equation, singularities are identified by where the leading coefficient of the highest-order derivative term is zero.
This step involves converting the equation into its standard form, then setting the leading coefficient \(x^2\) to zero, which determines that \(x=0\) is a singular point. Not all singular points are problematic; we distinguish between regular and irregular singular points.

• A regular singular point means the equation's other coefficients are analytic around this point, usable for the Frobenius method.
• An irregular singular point lacks this property, and alternative methods will be needed for solutions.

In our scenario, since \(P(x)\) and \(Q(x)\) allow expansion in terms of power series around \(x=0\), the point is confirmed to be a regular singularity, making it amenable to analytical methods.