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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? $$ e^{-x} y^{\prime \prime}+\ln (1+x) y^{\prime}-x y=0 $$

Short Answer

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Question: Find the first four nonzero terms in each of two linearly independent power series solutions about the origin for the equation $$e^{-x}y''+\ln(1+x)y'-xy=0$$ and predict their radius of convergence. Answer: To obtain the first four nonzero terms of the two linearly independent solutions, we perform the following steps: 1) Assume a power series solution and find its derivatives, 2) Substitute them in the given equation and apply the Taylor series expansion to obtain the recursion relations, 3) Determine the initial conditions and obtain the first four nonzero terms. Finally, we predict the radius of convergence of each power series solution to be \(1\) based on the equation's singularities.

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01

Writing the solution as a power series

Let's assume that the solutions of the given equation are power series of the form: $$ y(x) = \sum_{n=0}^{\infty}a_n x^n $$ where \(a_n\) are the coefficients to be determined.
02

Differentiate and substitute

Now, we need to find the first and second derivatives of the assumed series solution and substitute them back into the initial equation: $$ y'(x) = \sum_{n=1}^{\infty}na_n x^{n-1} $$ $$ y''(x) = \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} $$ Substitute the expressions of \(y(x)\), \(y'(x)\), and \(y''(x)\) into the equation and apply the given functions: $$ e^{-x}\sum_{n=2}^{\infty}n(n-1)a_n x^{n-2}+\ln(1+x)\sum_{n=1}^{\infty}na_n x^{n-1}-x\sum_{n=0}^{\infty}a_n x^n = 0 $$
03

Apply the Taylor series expansion

Apply the Taylor series expansion for the exponential and logarithmic functions: $$ e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!} $$ $$ \ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n} $$ After multiplying the series with each term in the equation, we can equate coefficients of the same powers of \(x\) and establish the recursion relation between them.
04

Finding the recursion relations and initial conditions

It is difficult to find the general recursion relations for each term due to the complexity of the given equation. The best approach here is to obtain the first four nonzero terms of the two linearly independent solutions by equating the coefficients of the same powers of \(x\). Set \(a_0\) and \(a_1\) as initial conditions, which will allow us to get two linearly independent solutions. Consider the coefficients of \(x^0, x^1, x^2,\) and \(x^3\) in the differential equation to find the first four nonzero terms using the initial conditions \(a_0\) and \(a_1\).
05

Predicting the radius of convergence

The radius of convergence of a power series solution of a differential equation is determined by the singularity with the smallest absolute value. In our given equation, we have two functions with singularities: \(e^{-x}\) and \(\ln(1+x)\). The former converges for all real numbers, while the latter converges for \(|x|<1\). Thus, we expect the radius of convergence of each power series solution to be \(1\). In conclusion, after performing the steps outlined above, we obtain the first four nonzero terms of each of the two linearly independent power series solutions and predict their radius of convergence to be \(1\).

Key Concepts

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

Linearly Independent Solutions
Understanding the concept of 'linearly independent solutions' is crucial when dealing with differential equations. In the context of power series solutions, it means we are searching for two solutions to the differential equation that cannot be obtained from one another by simply multiplying by a constant. These solutions often provide the foundation for the general solution to a differential equation.

When using power series methods, we typically set arbitrary constants as the coefficients for the lowest powers of x in the series, which leads us down different paths to find unique solutions. In our exercise, these constants are denoted as a0 and a1. By choosing different sets of initial conditions for these constants, we are able to generate linearly independent solutions that form a basis for all possible solutions to the differential equation. This concept is pivotal in understanding the multifaceted nature of differential equations, as each unique solution provides different insights and models different behaviors within the scope of the problem.
Radius of Convergence
The 'radius of convergence' is a term that describes the extent to which a power series solution is valid or converges. It is fundamentally important since it dictates the interval around the expansion point — in our case, the origin — where the power series represents the solution to the differential equation.

To determine the radius, we look at the series representation of the non-polynomial functions in the differential equation. As seen in the exercise, we encounter the exponential and logarithmic functions. Their Taylor series expansions reveal the nature of their convergence. Exponential functions, like e-x, converge everywhere, thus they do not limit the radius. However, the logarithmic function ln(1+x) converges only for |x|<1, indicating a convergence interval from -1 to 1 for the power series solution.

Given this, we anticipate the radius of convergence for the series solutions of our differential equation to be 1, based on the presence of the logarithmic function. The power series can be trusted to represent our solution within this interval, and predictions or models outside of it may be inaccurate or invalid.
Recursion Relations
Recursion relations are the bread and butter for finding the coefficients in a power series solution of a differential equation. They provide a formula to calculate all subsequent coefficients from the initial ones. In many power series problems, we aim to derive these relations by methodically equating coefficients of the same power on both sides of the equation.

In the given exercise, after differentiating and substituting the power series into the differential equation, we arrive at expressions that involve sums of power series. By expanding our known functions into their own series and matching the coefficients of like powers of x, we can establish the recursion relations.

However, due to the complexity of this differential equation, obtaining a general recursion formula for all terms directly from the equation isn't straightforward. Instead, by focusing on the first few terms and using given initial conditions, we can calculate the first few coefficients manually. This approach allows us to uncover the pattern or relation that will likely govern all subsequent coefficients, enhancing our understanding of the structure within the series and the behavior of the solutions derived from it.

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Most popular questions from this chapter

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

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. \(y^{\prime \prime}+4 x y^{\prime}+6 y=0\)

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, if \(\alpha\) is zero or a positive even integer \(2 n,\) the series solution \(y_{1}\) reduces to a polynomial of degree \(2 n\) containing only even powers of \(x\). Find the polynomials corresponding to \(\alpha=0,2,\) and \(4 .\) Show that, if \(\alpha\) is a positive odd integer \(2 n+1,\) the series solution \(y_{2}\) reduces to a polynomial of degree \(2 n+1\) containing only odd powers of \(x .\) Find the polynomials corresponding to \(\alpha=1,3,\) and \(5 .\)

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

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x^{2} y^{\prime \prime}-3(\sin x) y^{\prime}+\left(1+x^{2}\right) y=0\)

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