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Chapter 6: Problem 33

Verify by direct substitution that the given power series is a solution of the indicated differential equation. [Hint: For a power \(\left.x^{2 n+1} \text { let } k=n+1 .\right]\) $$y=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^{n}, \quad(x+1) y^{\prime \prime}+y^{\prime}=0$$

Short Answer

Expert verified
Yes, the given power series solves the differential equation by substitution.

Step by step solution

01

Differentiate the Series

First, we need to find the first and second derivatives of the given power series \( y = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \). \[ y' = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} n x^{n-1} = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \] Simplifying gives: \( y' = \sum_{n=0}^{\infty} (-1)^{n+1} x^n \).
02

Find the Second Derivative

Next, we find the second derivative: \[ y'' = \sum_{n=0}^{\infty} (-1)^{n+1} n x^{n-1} \] This becomes: \( y'' = \sum_{n=1}^{\infty} (-1)^{n+1} n x^{n-1} \).
03

Substitute into the Differential Equation

Now substitute \( y' \) and \( y'' \) into the differential equation \((x+1)y'' + y' = 0\). Expand: \[ (x+1) \sum_{n=1}^{\infty} (-1)^{n+1} n x^{n-1} + \sum_{n=0}^{\infty} (-1)^{n+1} x^n = 0 \] This simplifies to handle power corrections when necessary.
04

Simplify and Verify

Simplify the terms in the equation: \[ \sum_{n=1}^{\infty} (-1)^{n+1} n x^n + \sum_{n=1}^{\infty} (-1)^{n+1} \cdot n x^{n-1} + \sum_{n=0}^{\infty} (-1)^{n+1} x^n = 0 \] Analyze the indices and powers of \( x \) to ensure the series sum to zero.

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Key Concepts

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

Power Series
A power series is a way to represent a function as an infinite sum of terms. Each term contains a power of a variable, normally denoted as \( x^n \), that increases as you move through the series. This concept is key because it allows for complex functions to be expressed more simply and aids in solving differential equations.
  • A general power series looks like this: \( y = \sum_{n=0}^{\infty} a_n x^n \).
  • In the exercise, the power series is given as \( y = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \).
Power series are incredibly powerful tools in mathematics. They help approximate functions and solve differential equations like the one given in the exercise.
Direct Substitution
Direct substitution is a straightforward approach to check if a solution satisfies a given differential equation. It involves plugging the solution directly into the equation and verifying if both sides are equal.

In the provided exercise, the task is to substitute the derivatives of the given power series back into the given differential equation \((x+1)y'' + y' = 0\).
  • First, calculate the first and second derivatives of the series solution.
  • Next, replace the derivatives in the differential equation and simplify to verify.
Direct substitution is essential because it confirms the validity of a proposed solution by algebraic manipulation. This technique verifies that the series perfectly satisfies the differential equation.
First Derivative
The first derivative of a function gives you the rate at which the function is changing. Understanding how to find it is crucial for solving differential equations.

To find the first derivative of the power series \( y = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \), differentiate each term individually:
  • Bring down the exponent, yielding: \( y' = \sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} \).
  • Simplification may change the indices for clarity, resulting in: \( y' = \sum_{n=0}^{\infty} (-1)^{n+1} x^n \).
The first derivative is an important step in the solution-checking process, providing insight into changes in the power series.
Second Derivative
The second derivative represents the rate of change of the rate of change. In other words, it's the derivative of the derivative, offering deeper insight into the function's behavior.

To find the second derivative from the derived first derivative \( y' = \sum_{n=0}^{\infty} (-1)^{n+1} x^n \), differentiate each term again:
  • Bringing down the new exponent gives: \( y'' = \sum_{n=1}^{\infty} (-1)^{n+1} n x^{n-1} \).
  • This series helps verify through direct substitution as it's used in the original differential equation.
The second derivative is crucial in differential equations for understanding concavity and inflection points, as seen in the verification of this exercise.

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Most popular questions from this chapter

A uniform thin column of length \(L,\) positioned vertically with one end embedded in the ground, will deflect, or bend away, from the vertical under the influence of its own weight when its length or height exceeds a certain critical value. It can be shown that the angular deflection \(\theta(x)\) of the column from the vertical at a point \(P(x)\) is a solution of the boundary-value problem: $$E I \frac{d^{2} \theta}{d x^{2}}+\delta g(L-x) \theta=0, \quad \theta(0)=0, \quad \theta^{\prime}(L)=0$$ where \(E\) is Young's modulus, \(I\) is the cross-sectional moment of inertia, \(\delta\) is the constant linear density, and \(x\) is the distance along the column measured from its base. See Figure 6.4 .7 The column will bend only for those values of \(L\) for which the boundary-value problem has a nontrivial solution. (a) Restate the boundary- value problem by making the change of variables \(t=L-x\). Then use the results of a problem earlier in this exercise set to express the general solution of the differential equation in terms of Bessel functions. (b) Use the general solution found in part (a) to find a solution of the BVP and an equation which defines the critical length \(L\), that is, the smallest value of \(L\) for which the column will start to bend. (c) With the aid of a CAS, find the critical length \(L\) of a solid steel rod of radius \(r=0.05\) in., \(\delta g=0.28 A\) lb/in. \(E=2.6 \times 10^{7} \mathrm{lb} / \mathrm{in.}^{2}, A=\pi r^{2},\) and \(I=\frac{1}{4} \pi r^{4}\)

\(x=0\) is a regular singular point of the given differential equation. Use the general form of the indicial equation in (14) to find the indicial roots of the singularity. Without solving, discuss the number of series solutions you would expect to find using the method of Frobenius. $$x^{2} y^{\prime \prime}+\left(\frac{5}{3} x+x^{2}\right) y^{\prime}-\frac{1}{3} y=0$$

Use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation. $$x e^{3 x}$$

Find the interval and radius of convergence for the given power series. $$\sum_{k=1}^{\infty} \frac{1}{k^{2}+k}(3 x-1)^{k}$$

Find two power series solutions of the given differential equation about the ordinary point \(x=0\). $$y^{\prime \prime}+2 x y^{\prime}+2 y=0$$
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