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Identify all the singular points of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) in the interval \(-10

Short Answer

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Question: Identify the singular points of the given second-order differential equation in the interval -10 < t < 10: \((1-t^2)y^{\prime\prime} + ty^{\prime} + (\csc(t))y = 0\). Answer: The singular points of the given equation in the interval -10 < t < 10 are: \(t = -3\pi, -2\pi, -\pi, -1, 0, 1, \pi, 2\pi, 3\pi\).

Step by step solution

01

In our given equation, the coefficient of \(y^{\prime \prime}\) is \((1-t^2)\). We need to find the values of \(t\) for which this coefficient becomes zero. So, we need to solve the following equation: $$ 1 - t^2 = 0 $$ #Step 2: Solve the equation for \(t\)#

To solve \(1 - t^2 = 0\), we can factor the equation as follows: $$ (1 - t)(1 + t) = 0 $$ This equation has two solutions: \(t = 1\) and \(t = -1\). Although these are the points where the coefficient of \(y^{\prime\prime}\) becomes zero, they are still regular singular points, which means that we can apply the Frobenius method to find a solution in the form of power series in the vicinity of these points. We will keep these values in mind when we analyze the second coefficient of the equation. #Step 3: Identify singularities in the coefficient \(q(t) = \csc (t)\)#
02

\(\csc(t)\) is the reciprocal of the sine function, so we know that it is undefined when the sine function is equal to zero. The sine function is equal to zero at integer multiples of \(\pi\). We should find the integer multiples of \(\pi\) in the interval \(-10

To find integer multiples of \(\pi\) in the interval \(-10<t<10\), we can write the equation \(t = n\pi\), where \(n\) is an integer. To find the range of \(n\) values, we can simply divide the endpoints of the given interval by \(\pi\), which gives us the approximation: $$ -3.18 < n < 3.18 $$ #Step 5: List the integer values of \(n\)#
03

From the range of \(n\) values obtained in the previous step, we can list the integer values of \(n\): $$ n = -3, -2, -1, 0, 1, 2, 3 $$ #Step 6: Determine the singular points in the interval \(-10

We can now identify all the singular points in the given interval: - Singular points due to the coefficient \((1-t^2)\): \(t = 1\) and \(t = -1\). - Singular points due to the coefficient \(q(t) = \csc (t)\): \(t = -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi\). We have now identified all the singular points of the given equation in the interval \(-10<t<10\): \(t = -3\pi, -2\pi, -\pi, -1, 0, 1, \pi, 2\pi, 3\pi\).

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Key Concepts

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

Frobenius Method
The Frobenius method is a powerful tool used to find solutions to second-order linear differential equations, particularly near singular points. It combines the concept of power series solutions with the complexity posed by singularities. Typically, when we confront a differential equation like \( y'' + p(t)y' + q(t)y = 0 \), the first step is to check whether we can apply standard power series methods by ensuring the functions \( p(t) \) and \( q(t) \) are analytic at the point of interest.

However, if \( p(t) \) or \( q(t) \) have singularities, we must adjust our strategy. This is where Frobenius steps in, allowing us to work around regular singular points by assuming a solution in the form of a power series with a shifted index, \( y(t) = t^r \sum_{n=0}^\infty a_n t^n \) where \( r \) is a number we solve for. It is essential to identify these points correctly since applying the Frobenius method at an ordinary point, or an irregular singular point would lead us astray from a valid solution.
Regular Singular Points
Regular singular points play a significant role in analyzing differential equations. To put it simply, a point \( t_0 \) is a singular point if either the coefficient of \( y' \) or \( y'' \) in our differential equation \( y'' + p(t)y' + q(t)y = 0 \) becomes infinite or discontinuous at \( t_0 \). However, not all singular points are dealt with equally.

A singular point is considered 'regular' if the limits \( (t - t_0)p(t) \) and \( (t - t_0)^2q(t) \) exist and are finite as \( t \) approaches \( t_0 \) from within the interval of interest. Identifying regular singular points is crucial, as it informs us about the potential need to use the Frobenius method that has the capability to find solutions around these specific types of points.
Power Series Solutions
When faced with differential equations, one of the most approachable methods to find solutions is through power series. A power series solution involves representing the unknown function \( y(t) \) as a power series \( \sum_{n=0}^\infty a_n (t - t_0)^n \), where \( a_n \) are constants that need to be determined, and \( t_0 \) is a point around which the series is centered. As we plug this series into the differential equation, we can equate coefficients of like powers of \( t \) to solve for the \( a_n \) values.

The beauty of this method is its simplicity and the fact that it often leads to very accurate approximations of solutions in the vicinity of \( t_0 \). Giving students the ability to handle even complex differential equations by breaking them down into a manageable series makes the method invaluable for solving various physics and engineering problems.
Differential Equations
Differential equations are the backbone of modeling real-world phenomena across all realms of science and engineering. They encapsulate how things change over time or space and are written as equations involving derivatives of one or more variables. The goal when solving a differential equation is to find the unknown function or functions that satisfy the equation.

In our specific exercise, we're dealing with a second-order linear homogeneous differential equation with variable coefficients \( p(t) \) and \( q(t) \). Navigating such equations may seem daunting, but understanding the nature of the points in the interval, utilizing power series solutions, and the Frobenius method for singularities, provides a structured pathway to finding the solutions. Embracing these concepts ensures that students can not just find the solutions, but also deeply understand the process and reasoning behind each step.

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Most popular questions from this chapter

In each exercise, determine the polynomial \(P(t)\) of smallest degree that causes the given differential equation to have the stated properties. \(y^{\prime \prime}+\frac{1}{t P(t)} y^{\prime}+\frac{1}{t^{3}} y=0 \quad \begin{aligned}&\text { There are irregular singular points at } t=0 \text { and } t=\pm 1 \\\&\text { All other points are ordinary points. }\end{aligned}\)

In each exercise, an initial value problem is given. Assume that the initial value problem has a solution of the form \(y(t)=\sum_{n=0}^{\infty} a_{n} t^{n}\), where the series has a positive radius of convergence. Determine the first six coefficients, \(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\). Note that \(y(0)=a_{0}\) and that \(y^{\prime}(0)=a_{1}\). Thus, the initial conditions determine the arbitrary constants. In Exercises 40 and 41 , the exact solution is given in terms of exponential functions. Check your answer by comparing it with the Maclaurin series expansion of the exact solution. $$ y^{\prime \prime}+t y^{\prime}-2 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 $$

In each exercise, the exponent \(n\) in the given differential equation is a nonnegative integer. Determine the possible values of \(n\) (if any) for which (a) \(t=0\) is a regular singular point. (b) \(t=0\) is an irregular singular point. $$ y^{\prime \prime}+\frac{1}{t^{n}} y^{\prime}+\frac{1}{1+t^{2}} y=0 $$

Use series (7a) to determine the first four nonvanishing terms of the Maclaurin series for (a) \(\sinh t=\frac{e^{t}-e^{-t}}{2}\) (b) \(\cosh t=\frac{e^{t}+e^{-t}}{2}\)

In each exercise, (a) Determine all singular points of the given differential equation and classify them as regular or irregular singular points. (b) At each regular singular point, determine the indicial equation and the exponents at the singularity. $$ t^{2}(1-t)^{1 / 3} y^{\prime \prime}+t y^{\prime}-y=0 $$

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