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Chapter 5: Problem 6

Determine a lower bound for the radius of convergence of series solutions about each given point \(x_{0}\) for the given differential equation. $$ \left(x^{2}-2 x-3\right) y^{\prime \prime}+x y^{\prime}+4 y=0 ; \quad x_{0}=4, \quad x_{0}=-4, \quad x_{0}=0 $$

Short Answer

Expert verified
The lower bounds for the radii of convergence for the given points are as follows: - For x_0 = 4, the lower bound for the radius of convergence is 1. - For x_0 = -4, the lower bound for the radius of convergence is 3. - For x_0 = 0, the lower bound for the radius of convergence is 1.

Step by step solution

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01

Rewrite the Differential Equation in Standard Form

Divide the differential equation by the coefficient of \(y''\), \((x^2-2x-3)\). $$ y^{\prime\prime}+ \frac{x}{x^2-2x-3}y^{\prime} + \frac{4}{x^2-2x-3}y = 0 $$
02

Analyze the Singularity of the Differential Equation

In the above equation, we can find singularities by determining the values of \(x\) for which the coefficients of \(y'\) and \(y\) become undefined. This occurs when the denominator of each coefficient, namely \((x^2-2x-3)\), is equal to zero: $$ x^2-2x-3 = 0 $$ Solve this quadratic equation for \(x\) to find the singularities. $$ x = \frac{-(-2)\pm \sqrt{(-2)^2- 4(1)(-3)}}{2(1)} $$ $$ x = \frac{2\pm \sqrt{16}}{2} $$ $$ x = 1\pm 2 \Rightarrow x = 3, -1 $$ So the singularities are located at \(x = -1\) and \(x = 3\).
03

Find the Lower Bound for the Radius of Convergence

Now, we will find the distance from each given point, \(x_0\), to the nearest singularity: 1. For \(x_0 = 4\), the nearest singularity is at \(x = 3\). The distance is \(4-3 = 1\). So, the lower bound for the radius of convergence is \(1\). 2. For \(x_0 = -4\), the nearest singularity is at \(x = -1\). The distance is \(|-4 - (-1)| = |-4+1|= 3\). So, the lower bound for the radius of convergence is \(3\). 3. For \(x_0 = 0\), the nearest singularity is at \(x=-1\). The distance is \(|0 - (-1)| = |1|=1\). So, the lower bound for the radius of convergence is \(1\). Thus, the lower bounds for the radii of convergence for the given points are as follows: - \(x_0 = 4: \text{Radius} \ge 1\) - \(x_0 = -4: \text{Radius} \ge 3\) - \(x_0 = 0: \text{Radius} \ge 1\)

Key Concepts

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

Differential Equations
A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. These equations are fundamentally used in a wide array of scientific and engineering disciplines.

For instance, the differential equation provided in the exercise, \[\begin{equation}\left(x^{2}-2 x-3\right) y^{\' \'}+x y^{\' }+4 y=0; \quad x_{0}=4, \quad x_{0}=-4, \quad x_{0}=0 \end{equation}\], is a second-order linear homogeneous differential equation. Its solutions can reveal the behavior of a physical system, like the motion of a particle in a potential field or the distribution of temperatures along a rod over time. Understanding how to work with differential equations, such as by finding series solutions, is crucial in the mathematical modeling of real-world situations.
Series Solutions
When solving differential equations, series solutions are a powerful tool, especially when the equations cannot be solved with simple algebraic methods. Series solutions involve expressing the solution to a differential equation as an infinite sum of terms, usually power series. These series are centered around a point, often denoted as \( x_0 \), and have coefficients that can be determined systematically.

With series solutions, one typically looks for a recurrence relation that the coefficients must satisfy. In the exercise, a series solution approach could help find a meaningful solution about the point \( x_0 \). The radius of convergence for these series solutions is essential because it indicates the interval around \( x_0 \) within which the series converges to an actual solution of the differential equation. When the series converges, it means that as you add more terms of the series, the sum gets closer and closer to a specific value, which represents the solution to the differential equation.
Singular Point Analysis
Singular point analysis is vital when working with differential equations, particularly when finding series solutions. A singular point of a differential equation is a value of \( x \) at which the behavior of the solution can change dramatically or where a solution in the form of a power series may not exist. More technically, it is a point where the coefficients of the lowest order derivatives in the differential equation have poles or become unbounded.

In this exercise, after rewriting the equation in standard form, it becomes possible to find the singular points by setting the denominator of the coefficients to zero, leading to the quadratic equation \( x^2-2x-3 = 0 \). Solving for \( x \), you can find the singular points at \( x = 3 \) and \( x = -1 \). By identifying singular points, you can then perform a radius of convergence analysis. It measures the distance from the point \( x_0 \) to the nearest singularity, which helps determine where the series solutions will be valid. In physical terms, the radius of convergence gives information about the range of conditions for which the system's model holds true.

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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 the I.egendre equation can also be written as $$ \left[\left(1-x^{2}\right) y^{\prime}\right]=-\alpha(\alpha+1) y $$ Then it follows that \(\left[\left(1-x^{2}\right) P_{n}^{\prime}(x)\right]^{\prime}=-n(n+1) P_{n}(x)\) and \(\left[\left(1-x^{2}\right) P_{m}^{\prime}(x)\right]^{\prime}=\) \(-m(m+1) P_{m}(x) .\) By multiplying the first equation by \(P_{m}(x)\) and the second equation by \(P_{n}(x),\) and then integrating by parts, show that $$ \int_{-1}^{1} P_{n}(x) P_{m}(x) d x=0 \quad \text { if } \quad n \neq m $$ This property of the Legendre polynomials is known as the orthogonality property. If \(m=n,\) it can be shown that the value of the preceding integral is \(2 /(2 n+1) .\)

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 \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) The Legendre polynomial \(P_{n}(x)\) is defined as the polynomial solution of the Legendre equation with \(\alpha=n\) that also satisfies the condition \(P_{n}(1)=1\). (a) Using the results of Problem 23 , find the Legendre polynomials \(P_{0}(x), \ldots . P_{5}(x) .\) (b) Plot the graphs of \(P_{0}(x), \ldots, P_{5}(x)\) for \(-1 \leq x \leq 1 .\) (c) Find the zeros of \(P_{0}(x), \ldots, P_{5}(x)\).

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}+2 x y^{\prime}+4 y=0\)

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x y^{\prime \prime}+e^{x} y^{\prime}+(3 \cos x) y=0\)

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 \(x \rightarrow \infty\).
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