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. \(x^{2} y^{\prime \prime}+3(\sin x) y^{\prime}-2 y=0\)

Short answer

Based on the given differential equation, we found that the unique singular point is x = 0. After writing the equation in its standard form and finding the indicial equation, we determined that the exponents at the singularity x = 0 are 0 and 1.

Step-by-step solution

Identify regular singular points

The given differential equation is: \(x^{2} y^{\prime \prime}+3(\sin x) y^{\prime}-2 y=0\) To identify the regular singular points, we first need to write the equation in the standard form. The standard form is given as: \(y^{\prime \prime}+\frac{p(x)}{x}y^{\prime}+\frac{q(x)}{x^{2}}y=0\) Divide the whole equation by \(x^{2}\): \(y^{\prime \prime}+\frac{3(\sin x)}{x}y^{\prime}-\frac{2}{x^{2}}y=0\) Here, \(p(x) = 3(\sin(x))\), and \(q(x) = -2\). Now, we need to determine the points where \(p(x)\) and \(q(x)\) are not analytic. Since \(p(x)\) is \(3(\sin x)\), it is analytic for all x. The function \(q(x)\) is analytic for \(x \neq 0\). Thus, the only possible singular point is \(x = 0\).

Write the equation in its standard form

Since the differential equation is already in its standard form, we can move to the next step. \(y^{\prime \prime}+\frac{3(\sin x)}{x}y^{\prime}-\frac{2}{x^{2}}y=0\)

Find the indicial equation

Now, we will find the indicial equation. For that, we will assume the solution of the given differential equation to be in the form \(y(x) = x^{r}\sum_{n=0}^{\infty}a_n x^n\). Now differentiate \(y(x)\) with respect to \(x\) twice to find \(y'(x)\) and \(y''(x)\): \(y^{\prime}(x) = r x^{r-1}\sum_{n=0}^{\infty}a_n (n+1) x^n\) \(y^{\prime \prime}(x) = r(r-1) x^{r-2}\sum_{n=0}^{\infty}a_n (n+1)(n+2) x^n\) Substitute these values into the standard form of the differential equation: \(x^{2}y^{\prime\prime} + 3(\sin x)xy^{\prime} - 2y(x) = 0\) \(r(r-1) x^{r}\sum_{n=0}^{\infty}a_n (n+1)(n+2) x^n + 3(\sin x) r x^{r}\sum_{n=0}^{\infty}a_n (n+1)x^n -2 x^{r}\sum_{n=0}^{\infty}a_n x^n = 0\) Now, compare the coefficients, we get the indicial equation as: \(r(r-1) = 0\)

Determine the exponents at the singularity

Solving the indicial equation, we get: \(r(r-1)=0 \Rightarrow r = 0\) or \(r = 1\) Therefore, the exponents at the singularity \(x=0\) are 0 and 1.

Key concepts

Indicial Equation

When dealing with differential equations that have singular points, the indicial equation is a key tool to analyze the behavior of solutions near these points. It is derived by substituting a series solution into the differential equation and equating the lowest power of x to zero to ensure that the series is a valid solution.

In the context of the given exercise, we assumed that the solution would be in the form of a power series multiplied by x raised to some power r, which is an unknown exponent to be determined. Differentiating this assumed solution and substituting it back into the differential equation leads us to an equation that allows us to solve for possible values of r.

As an example, the indicial equation derived from the exercise is r(r-1) = 0. Finding the roots of this equation gave us the crucial exponents that indicate how the series solution behaves as it approaches the singularity. This process is essential in determining the nature of the solutions at singular points within differential equations.

Exponents at Singularity

The exponents at singularity, often found by solving the indicial equation, are integral to understanding the behavior of solutions near a singular point. These exponents, denoted generally by r, can tell us whether a solution will be finite, zero, or blow up as we approach the singularity.

In our exercise, we determined that the exponents at the singularity at x=0 are 0 and 1. These values can be interpreted to mean that one solution to the differential equation will be constant as x approaches zero, corresponding to r=0, and another solution will be directly proportional to x, arising from the exponent r=1. This analysis provides critical insight into the behavior of different solutions and forms the foundation for understanding more complex aspects of the series solutions around singular points.

Analytic Functions

Understanding the concept of analytic functions is very significant when solving differential equations, especially around singular points. A function is considered analytic at a point if it can be represented by a power series that converges to the function in some neighborhood around the point.

The differential equation provided exhibits a coefficient 3(sin x), which is an analytic function everywhere because it can be expressed as a convergent power series for all x. On the other hand, -2 is a constant and therefore analytic everywhere except at x=0, since we cannot divide by zero.

It is by understanding the analyticity of functions within a differential equation that we can identify regular singular points and properly apply the Frobenius method to find a series solution. This framework allows us to move through the process systematically, ensuring that our solutions are valid in the context of the differential equation presented.