Chapter 5: Problem 12
Find the first four nonzero terms in each of two linearly independent power series solutions about the origin. What do you expect the radius of convergence to be for each solution? $$ e^{x} y^{\prime \prime}+x y=0 $$
Short Answer
Expert verified
Based on the steps, we found two linearly independent power series solutions for the given differential equation:
$$ y_1(x) = a_0\left(1-\frac{1}{2}x^2+\frac{1}{4}x^4-\cdots\right) \text{ and } y_2(x) = a_1\left(x-\frac{1}{3}x^3+\frac{1}{5}x^5-\cdots\right)$$
with the first four nonzero terms of each solution listed. The radius of convergence for both solutions is infinite.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Ordinary Differential Equations
Linear Ordinary Differential Equations (ODEs) form the backbone of many mathematical models used in science and engineering. They are equations that involve an unknown function, its derivatives, and variable 'x', typically written in the form:
\[\begin{equation}onumbera_n(x)\frac{d^ny}{dx^n} + a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \cdots + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x),ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline \cdots \cdots \cdots \cdots \cdots \ a_0(x)y = g(x), \end{equation}\]where \( a_i(x) \) are known functions, and \( g(x) \) is the nonhomogeneous term. If \( g(x) = 0 \), the equation is called homogeneous. Power series methods, like that used in the exercise, are particularly useful in finding solutions to linear ODEs around ordinary points. These methods involve assuming a solution that can be expressed as an infinite sum of powers of 'x', and systematically determining the coefficients to satisfy the ODE.
Understanding linear ODEs is essential because they arise in various physical problems, such as in modeling spring-mass systems, electrical circuits, and the motion of celestial bodies. Moreover, the behavior of their solutions can often be predicted by their coefficients, leading to a profound understanding of the underlying physical phenomenon.
\[\begin{equation}onumbera_n(x)\frac{d^ny}{dx^n} + a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \cdots + a_1(x)\frac{dy}{dx} + a_0(x)y = g(x),ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline \cdots \cdots \cdots \cdots \cdots \ a_0(x)y = g(x), \end{equation}\]where \( a_i(x) \) are known functions, and \( g(x) \) is the nonhomogeneous term. If \( g(x) = 0 \), the equation is called homogeneous. Power series methods, like that used in the exercise, are particularly useful in finding solutions to linear ODEs around ordinary points. These methods involve assuming a solution that can be expressed as an infinite sum of powers of 'x', and systematically determining the coefficients to satisfy the ODE.
Understanding linear ODEs is essential because they arise in various physical problems, such as in modeling spring-mass systems, electrical circuits, and the motion of celestial bodies. Moreover, the behavior of their solutions can often be predicted by their coefficients, leading to a profound understanding of the underlying physical phenomenon.
Radius of Convergence
The radius of convergence is a crucial concept when discussing power series solutions to differential equations. It refers to the range within which the power series converges to a finite value. Mathematically, a power series:
\[\begin{equation}onumber y(x) = \sum_{n=0}^{\infty} a_nx^n \end{equation}\]will converge on the interval \(( -R, R )\), where 'R' is the radius of convergence, and will diverge outside of it.
To determine this radius, one typically employs the ratio or root tests on the series' terms. For the differential equation from the exercise, which exhibits the exponential function \( e^x \), paired with solutions expressed as power series, the function \( e^x \) is known to converge for all 'x', indicating that the radius of convergence for the series solution is infinite (\( R = \infty \),). This is an exceptional case since most of the time, power series have a finite radius of convergence.
Knowing the radius of convergence is not only theoretical but has practical implications. For instance, it dictates the interval over which the power series is a valid representation of the solution to the differential equation. This is crucial information when applying these solutions to model real-world phenomena, as it tells you where you can trust the model.
\[\begin{equation}onumber y(x) = \sum_{n=0}^{\infty} a_nx^n \end{equation}\]will converge on the interval \(( -R, R )\), where 'R' is the radius of convergence, and will diverge outside of it.
To determine this radius, one typically employs the ratio or root tests on the series' terms. For the differential equation from the exercise, which exhibits the exponential function \( e^x \), paired with solutions expressed as power series, the function \( e^x \) is known to converge for all 'x', indicating that the radius of convergence for the series solution is infinite (\( R = \infty \),). This is an exceptional case since most of the time, power series have a finite radius of convergence.
Knowing the radius of convergence is not only theoretical but has practical implications. For instance, it dictates the interval over which the power series is a valid representation of the solution to the differential equation. This is crucial information when applying these solutions to model real-world phenomena, as it tells you where you can trust the model.
Recurrence Relation
A recurrence relation is an equation that expresses each term of a sequence as a function of its predecessors. In the context of solving differential equations using power series, a recurrence relation allows us to find the coefficients (\( a_n \),) of the series.
In the given exercise, the recurrence relation:\[\begin{equation}onumber(n+2)(n+1)a_{n+2} = -a_newlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline \end{equation}\]is fundamental as it sets up a relationship between the coefficients of different terms. Such recurrence relations are derived by substituting the assumed power series into the differential equation and equating coefficients of like powers of 'x'. The solution of this relation often requires setting initial conditions. In the example, the relation is used to calculate the first four non-zero terms of the series.
The power and utility of recurrence relations in solving differential equations cannot be overstated. They convert the problem of solving a differential equation into an algebraic task of finding the terms of a sequence, enabling one to express complex functions as infinite series. This is particularly useful when looking for patterns within the series or when a closed-form expression of the solution is not easily attainable.
In the given exercise, the recurrence relation:\[\begin{equation}onumber(n+2)(n+1)a_{n+2} = -a_newlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline \end{equation}\]is fundamental as it sets up a relationship between the coefficients of different terms. Such recurrence relations are derived by substituting the assumed power series into the differential equation and equating coefficients of like powers of 'x'. The solution of this relation often requires setting initial conditions. In the example, the relation is used to calculate the first four non-zero terms of the series.
The power and utility of recurrence relations in solving differential equations cannot be overstated. They convert the problem of solving a differential equation into an algebraic task of finding the terms of a sequence, enabling one to express complex functions as infinite series. This is particularly useful when looking for patterns within the series or when a closed-form expression of the solution is not easily attainable.
