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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. \(x y^{\prime \prime}+2 x y^{\prime}+6 e^{x} y=0\)

Short Answer

Expert verified
A: The exponents related to the regular singular point x=0 are r=0 and r=1.

Step by step solution

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01

Identify the singular points

To find the singular points, we examine the coefficients of the given differential equation: \(x y'' + 2x y' + 6e^x y = 0\) The singular points occur when the coefficients become infinite, which is when \(x=0\). So we have one singular point at \(x=0\).
02

Check if the singular point is regular

A singular point is called a regular singular point if the coefficients of the equation \((x-x_0) y'' + p(x) y' + q(x) y = 0\) have power series representations about the singular point \(x_0\) with a positive radius of convergence. So, let's rewrite our given equation as follows: \((x-0) y'' + \frac{2x}{x} y' + \frac{6e^x}{x} y = 0\) Now we have \(p(x) = \frac{2x}{x}\) and \(q(x) = \frac{6e^x}{x}\). We can see that \(p(x)\) and \(q(x)\) have power series representations about the singular point \(x=0\). Therefore, the singular point \(x=0\) is a regular singular point.
03

Find the indicial equation

To find the indicial equation, we will assume that \(y\) can be expressed as a power series about the regular singular point. We will use the Frobenius method, which involves assuming a solution of the form: \(y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\), where \(r\) is the exponent related to the regular singular point. We need to find \(y'(x)\) and \(y''(x)\) to plug into our differential equation: \(y'(x) = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1}\) and \(y''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2}\). Now, plug these into the differential equation and collect the terms with the same power of \(x^{n+r}\). This will give us the indicial equation: \((r)(r-1)a_0 x^{r-2} + 2ra_0 x^{r-1} + 6a_0 x^r = 0\). Now, we'll examine the coefficients: For \(x^{r-2}\), we have \((r)(r-1)a_0 = 0\). For \(x^{r-1}\), we have \(2ra_0 = 0\).
04

Solve the indicial equation for exponents

The indicial equation can be rewritten as two equations: 1. \((r)(r-1)a_0 = 0\) 2. \(2ra_0 = 0\) We can solve these equations to obtain the exponents: 1. If \(a_0 \neq 0\), then \(r(r-1) = 0\), which means \(r = 0\) or \(r = 1\). 2. If \(a_0 = 0\), then the second equation is always satisfied, which means that the first equation gives us the exponents. Hence, the exponents related to the regular singular point \(x=0\) are \(r=0\) and \(r=1\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indicial Equation
When tackling a second-order linear differential equation with variable coefficients, especially near a point where the usual power series method fails to converge, we encounter a situation known as a regular singular point. Near this point, the Frobenius method comes to the rescue, through which the indicial equation becomes a crucial element.

To formulate the indicial equation, we first assume a solution of the form \(y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\), where the parameter \(r\) signifies the exponent at the singularity. Importantly, the indicial equation arises from forcing the lowest power of \(x\) in the substituted series expansion to yield a zero coefficient; this grants us the constraint(s) on the value of \(r\).

In our problem, after identifying \(x=0\) as the regular singular point and using the Frobenius method, we derived the indicial equation by focusing on the lowest powers after substituting the series for \(y\), \(y'\), and \(y''\) into the original differential equation. Upon simplification, the powers of \(x^{r-2}\) and \(x^{r-1}\) provided us with the equations that needed to be satisfied, revealing \(r=0\) and \(r=1\) as the viable exponents at the regular singular point.
Frobenius Method
The Frobenius method is a technique used to solve linear differential equations near a regular singular point. When a power series solution fails to converge due to a singularity, the Frobenius method proves to be a powerful ally—it extends the power series method by considering solutions with fractional power terms.

To apply this methodology, we begin by assuming a solution series in the form of \(y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\), where \(r\) is not necessarily an integer. The values for the coefficient \(a_n\) and the exponent \(r\) are determined through a systematic process of substituting the series back into the differential equation and equating coefficients of corresponding powers of \(x\). In essence, it's akin to fitting a unique, tailor-made garment for the differential equation near the singularity.

In the example we have, after applying the Frobenius method and obtaining the indicial equation, we align the terms of like powers of \(x\) and set them equal to zero. This is how we find the coefficients that will construct the power series solution, reflecting the behavior of the function around the singular point.
Differential Equations
A differential equation is a mathematical equation that involves functions and their derivatives, embodying the relationship between a function and its rates of change. Solving differential equations is akin to decoding the language of change, which describes a vast range of phenomena in the natural and physical sciences.

These equations come in various forms, from simple first-order equations to complex non-linear systems. A second-order linear differential equation, such as the one in our exercise \(x y'' + 2x y' + 6e^x y = 0\), is particularly notable as it commonly arises in mechanical and physical contexts (think harmonic oscillators or electrical circuits).

What makes these equations challenging is that exact solutions are not always readily obtainable, especially when we cannot apply standard methods like separation of variables or integrating factors. This is where advanced techniques like the Frobenius method gain prominence, providing systematic approaches to finding solutions around singular points, thus expanding our toolkit for analyzing and understanding the diverse language of differential equations.

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Most popular questions from this chapter

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \((x-2)^{2} y^{\prime \prime}+5(x-2) y^{\prime}+8 y=0\)

Consider the Euler equation \(x^{2} y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0 .\) Find conditions on \(\alpha\) and \(\beta\) so that (a) All solutions approach zero as \(x \rightarrow 0 .\) (b) All solutions are bounded as \(x \rightarrow 0 .\) (c) All solutions approach zero as \(x \rightarrow \infty\). (d) All solutions are bounded as \(x \rightarrow \infty\). (e) All solutions are bounded both as \(x \rightarrow 0\) and as \(x \rightarrow \infty\).

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}+6 x y^{\prime}-y=0\)

Find the solution of the given initial value problem. Plot the graph of the solution and describe how the solution behaves as \(x \rightarrow 0\). \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0, \quad y(1)=1, \quad y^{\prime}(1)=-1\)

The Bessel equation of order zero is $$ x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0 $$ Show that \(x=0\) is a regular singular point; that the roots of the indicial equation are \(r_{1}=r_{2}=0 ;\) and that one solution for \(x>0\) is $$ J_{0}(x)=1+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n}}{2^{2 n}(n !)^{2}} $$ Show that the series converges for all \(x .\) The function \(J_{0}\) is known as the Bessel function of the first kind of order zero.

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