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Chapter 5: Problem 16

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

Short Answer

Expert verified
The singular point for the given differential equation \(x y^{\prime \prime} + y^{\prime} + (\cot x) y = 0\) is \(x=0\). This point is determined to be irregular because as \(x\) approaches \(0\), the coefficient of \(y\), which is \(\cot x\), becomes undefined. Therefore, the coefficient of \(y\) isn't analytic at \(x = 0\), making the singular point irregular.

Step by step solution

01

Identify the Singular Points

We need to identify when the coefficient of the highest derivative becomes zero. In the given equation, the highest derivative is \(y^{\prime \prime}\) and its coefficient is \(x\). So, we should find the values of \(x\) for which the coefficient becomes zero. Set the coefficient equal to zero: \(x=0\) So, there is only one singular point, \(x=0\).
02

Determine the Regular or Irregular Singular Point

Now that we found the singular point \(x=0\), we need to determine if it is regular or irregular. To do this, we will analyze the behavior of the other coefficients in the differential equation (\(y'\) coefficient and \(y\) coefficient) as \(x\) approaches the singular point \(x=0\). The given differential equation is: \(x y^{\prime \prime} + y^{\prime} + (\cot x)y = 0\) As \(x\) approaches \(0\), the coefficient of \(y\) approaches infinity because the cotangent function, \(\cot x\), becomes undefined at \(x=0\). This means that the coefficient of \(y\) isn't analytic at \(x = 0\). Therefore, the singular point is irregular.
03

Result

There is one singular point, \(x=0\), and it is an irregular singular point for the given differential equation.

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Key Concepts

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

Regular and Irregular Singular Points
Singular points are specific values in differential equations where the solutions may behave unusually. They occur when the coefficients of the differential equation's terms go to zero or infinity. There are two main types of singular points: regular and irregular.
  • **Regular Singular Points:** If the coefficients of the lower-order derivatives in a differential equation (after being divided by the leading coefficient) remain well-behaved—meaning they don't approach infinity or become undefined—as you approach the singular point, it's classified as regular.
  • **Irregular Singular Points:** Conversely, if these coefficients do become infinite or undefined, it's termed an irregular singular point.
Identifying whether a singular point is regular or irregular is crucial, as the equation's behavior around these points determines the methods for finding solutions. In this particular problem, we identified that at the singular point (\(x = 0\)) , the behavior of the function \(\cot x\) becomes infinite, indicating that \(x = 0\) is an irregular singular point.
Higher Order Differential Equations
Higher order differential equations involve derivatives of an unknown function that are of second degree or higher. In this exercise, we dealt with a second-order differential equation, which includes the term \(y''\). These types of equations often describe more complex physical systems and phenomena, such as oscillations, heat conduction, and wave propagation. To solve these equations, identifying the nature of singular points is a vital initial step. Singular points can simplify or complicate the solution methods and the type of solutions to expect. Understanding how to handle these complexities helps in predicting and managing behavioral aberrations in the function's solutions.
Analytic Functions
In the study of differential equations, analyzing whether a function is analytic around a point is crucial for determining the regularity of singular points. An analytic function is one that can be represented as a power series at a point and is differentiable in some neighborhood around that point. For a singular point to be regular, the functions made by dividing the coefficients of the derivatives in the differential equation by the leading term need to be analytic. In the provided problem, \(\cot x\) becomes undefined at \(x=0\), indicating it's not analytic, thereby causing the singular point to be irregular. This distinction plays a significant role in choosing the correct method to solve and understand the differential equations effectively.
Differential Equations Analysis
Analyzing differential equations involves assessing their components to determine behavior and potential solutions. To start, identify all singular points by focusing on where the coefficients of derivatives approach zero or infinity. Once singular points are identified, evaluate them to determine whether they are regular or irregular by studying the behavior of the differential equation's coefficients around these points. This analysis allows mathematicians and scientists to tailor their solutions to the peculiarities of each singular point, thereby providing accurate solutions. In practice, the steps involve not just identifying singular points but understanding how these points affect potential solutions' properties and behaviors across the domain. Correct analysis can vastly simplify complex equations and provide insight into their solutions.

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Most popular questions from this chapter

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that two linearly independent solutions of the Legendre equation for \(|x|<1\) are $$ \begin{aligned} y_{1}(x)=& 1+\sum_{m=1}^{\infty}(-1)^{m} \\ & \times \frac{\alpha(\alpha-2)(\alpha-4) \cdots(\alpha-2 m+2)(\alpha+1)(\alpha+3) \cdots(\alpha+2 m-1)}{(2 m) !} x^{2 m} \\ y_{2}(x)=& x+\sum_{m=1}^{\infty}(-1)^{m} \\ & \times \frac{(\alpha-1)(\alpha-3) \cdots(\alpha-2 m+1)(\alpha+2)(\alpha+4) \cdots(\alpha+2 m)}{(2 m+1) !} x^{2 m+1} \end{aligned} $$

Suppose that \(x^{r}_{1}\) and \(x^{r_{2}}\) are solutions of an Euler equation for \(x>0,\) where \(r_{1} \neq r_{2},\) and \(r_{1}\) is an integer. According to Eq. ( 24) the general solution in any interval not containing the origin is \(y=c_{1}|x|^{r_{1}}+c_{2}|x|^{r_{2}} .\) Show that the general solution can also be written as \(y=k_{1} x^{r}_{1}+k_{2}|x|^{r_{2}} .\) Hint: Show by a proper choice of constants that the expressions are identical for \(x>0,\) and by a different choice of constants that they are identical for \(x<0 .\)

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 \(x \rightarrow 0\). \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0, \quad y(1)=1, \quad y^{\prime}(1)=-1\)

In this section we showed that one solution of Bessel's equation of order zero, $$ L[y]=x^{2} y^{\prime \prime}+x y^{\prime}+x^{2} y=0 $$ is \(J_{0}\), where \(J_{0}(x)\) is given by Fa. ( 7) with \(a_{0}=1\). According to Theorem 5.7 .1 a second solution has the form \((x>0)\) $$ y_{2}(x)=J_{0}(x) \ln x+\sum_{n=1}^{\infty} b_{n} x^{n} $$ (a) Show that $$ L\left[y_{2}\right](x)=\sum_{n=2}^{\infty} n(n-1) b_{n} x^{n}+\sum_{n=1}^{\infty} n b_{n} x^{n}+\sum_{n=1}^{\infty} b_{n} x^{n+2}+2 x J_{0}^{\prime}(x) $$ (b) Substituting the series representation for \(J_{0}(x)\) in Eq. (i), show that $$ b_{1} x+2^{2} b_{2} x^{2}+\sum_{n=3}^{\infty}\left(n^{2} b_{n}+b_{n-2}\right) x^{n}=-2 \sum_{n=1}^{\infty} \frac{(-1)^{n} 2 n x^{2 n}}{2^{2 n}(n !)^{2}} $$ (c) Note that only even powers of \(x\) appear on the right side of Eq. (ii). Show that \(b_{1}=b_{3}=b_{5}=\cdots=0, b_{2}=1 / 2^{2}(1 !)^{2},\) and that $$ (2 n)^{2} b_{2 n}+b_{2 n-2}=-2(-1)^{n}(2 n) / 2^{2 n}(n !)^{2}, \quad n=2,3,4, \ldots $$ Deduce that $$ b_{4}=-\frac{1}{2^{2} 4^{2}}\left(1+\frac{1}{2}\right) \quad \text { and } \quad b_{6}=\frac{1}{2^{2} 4^{2} 6^{2}}\left(1+\frac{1}{2}+\frac{1}{3}\right) $$ The general solution of the recurrence relation is \(b_{2 n}=(-1)^{n+1} H_{n} / 2^{2 n}(n !)^{2}\). Substituting for \(b_{n}\) in the expression for \(y_{2}(x)\) we obtain the solution given in \(\mathrm{Eq} .(10) .\)
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