/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Find a series solution for the d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find a series solution for the differential equation. $$ \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+2 y=0 $$

Short Answer

Expert verified
The series solution involves coefficients found using recurrence relation derived from aligning power series terms.

Step by step solution

01

Assume a Power Series Solution

Assume the solution can be expressed as a power series: \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]This is a typical form for solutions expressed in terms of power series.
02

Derive Series for Derivatives

Find the derivatives of the assumed power series:First derivative: \[ \frac{dy}{dx} = \sum_{n=1}^{\infty} a_n n x^{n-1} \]Second derivative: \[ \frac{d^2y}{dx^2} = \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} \]
03

Substitute Series into Differential Equation

Substitute the expressions for y, \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \) into the differential equation:\[ \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} - x \sum_{n=1}^{\infty} a_n n x^{n-1} + 2 \sum_{n=0}^{\infty} a_n x^n = 0 \]
04

Align Series for Combination

Re-index the series to express them with the same powers of x for easy comparison:\[ \sum_{n=0}^{\infty} a_{n+2} (n+2)(n+1) x^n - \sum_{n=0}^{\infty} a_{n} n x^{n} + 2 \sum_{n=0}^{\infty} a_n x^n = 0 \]Align each series so that they match by powers of x.
05

Collect Like Terms

Combine the terms by the same powers of x from the re-indexed series:\[ \sum_{n=0}^{\infty} \left( a_{n+2} (n+2)(n+1) - a_n n + 2a_n \right) x^n = 0 \]This requires each coefficient of x to be zero for the equation to hold.
06

Formulate Recurrence Relation

Set the coefficient of each power of x to zero to obtain the recurrence relation:\[ a_{n+2} = \frac{a_n (n - 2)}{(n+2)(n+1)} \]This relation allows you to calculate successive coefficients in terms of the previous ones.
07

Solve for Series Coefficients

Use the recurrence relation to compute the coefficients. Set initial conditions, usually \(a_0\) and \(a_1\), based on boundary or initial conditions. Assume \(a_0\) and \(a_1\) are arbitrary constants:\[ a_2 = \frac{a_0 (0 - 2)}{2 \cdot 1}, a_3 = \frac{a_1 (1 - 2)}{3 \cdot 2}, \ldots \]Continue this pattern.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Series Solution
Solving differential equations can sometimes be simplified by expressing solutions as a series. A series solution is often used to find an approximate solution when an exact one is hard to derive. In this process, we express the unknown function as an infinite sum, usually in the form of a power series. This approach provides a flexible way to deal with complex equations and is useful when dealing with varied and complex problems in mathematics and physics.

Using a series solution generally involves the following steps:
  • Assume a series form for the solution, as a starting point.
  • Derive the series forms for the derivatives involved in the equation.
  • Substitute these series into the original differential equation.
  • Ensure that each term's coefficient of power series sums to zero to satisfy the equation.
This series method is particularly powerful as it transforms a difficult differential equation into a more manageable set of algebraic equations.
Power Series
A power series is a specific type of series where each term is a constant coefficient times a power of a variable, typically written as \[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]where \(a_n\) are the coefficients. This form makes it easier to carry out mathematical operations such as differentiation and integration.

Power series are infinite, but in practice, they can be truncated to provide an approximation to the solution. They have the following properties:
  • They can be used to approximate functions that are difficult to express analytically, like transcendental functions.
  • They allow for systematic calculations of derivatives.
  • They often have a radius of convergence, beyond which they do not accurately represent the function.
In the context of solving differential equations, power series expand the solution into simpler, more manageable parts which can be aligned and manipulated using simple algebraic operations.
Recurrence Relation
Recurrence relations play a significant role in finding the coefficients of a power series. They express each term of a sequence as a function of the preceding terms, allowing one to build the solution incrementally. It's essentially like a recipe where the next step depends on the results of the previous one.

In the context of a differential equation, once the power series and derivatives have been substituted and aligned, a recurrence relation is formulated:
This relation looks like:
  • \( a_{n+2} = \frac{a_n (n - 2)}{(n+2)(n+1)} \)
The above relation allows for finding each term in the series successively. By expressing higher terms in terms of initial coefficients like \(a_0\) and \(a_1\), one can compute terms of the series iteratively. Recurrence relations are therefore crucial to determine the nature and form of the series solution.
Initial Conditions
When solving differential equations using series solutions, initial conditions are essential to determine the particular solution from a family of solutions. They provide specific values for the unknown function and its derivatives at a certain point. These conditions help in defining constants in the power series.

Initial conditions could typically be set as:
  • The value of the function at a particular point, e.g., \( y(0) = a_0 \)
  • The value of the function's derivative, e.g., \( y'(0) = a_1 \)
These initial values aid in determining the coefficients of the series, guiding the computation of terms in the power series. Without initial conditions, the series would remain in terms of arbitrary constants, leaving the problem open-ended. Therefore, they are pivotal in deriving a specific solution that matches given physical or problem-specific requirements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the differential equation by \(\mid\) a) using an integrating factor and (b) separating the variables. The differential equation $$ R \frac{d I}{d t}+\frac{I}{C}=\frac{d V}{d t} $$ describes an electrical circuit consisting of an electromotive force \(V\) with a resistance \(R\) and capacitance \(C\) connected in series. If \(V\) is constant and if \(I=I_{0}\) when \(t=0,\) express \(I\) as a function of \(t .\)

Solve the differential equation. $$ \sin y \cos x d x+\left(1+\sin ^{2} x\right) d y=0 $$

Solve the differential equation. $$ 2 y^{\prime \prime}-4 y^{\prime}+y=0 $$

A solution of a differential equation of the form \(y^{\prime \prime}=f(x, y)\) with initial conditions \(y=y_{0}\) at \(x=a\) and \(y=y_{-1}\) at \(x=a-h\) is sometimes approximated using the formula \(y_{k+1}=2 y_{k}-y_{k-1}+h^{2} f\left(x_{k}, y_{k}\right)\) for \(k=0,1,2, \ldots, n-1,\) where \(h=(b-a) / n\) and \(x_{k}=\) \(a+k h .\) If \(h \approx 0,\) then \(y_{k}\) is an approximation to \(y\) at \(x=x_{k} .\) Use this formula, with \(n=4\) and \(a=0,\) to approximate \(y\) at \(x=\frac{1}{2}\) for the given differential equation and initial conditions. $$ y^{\prime \prime}=y-x ; \quad y_{0}=1, \quad y_{-1}=0.882823 $$

As a jar containing 10 moles of gas \(A\) is heated, the velocity of the gas molecules increases and a second gas \(\mathrm{B}\) is formed. When two molecules of gas A collide, two molecules of gas \(\mathrm{B}\) are formed. The rate \(d y / d t\) at which gas \(\mathrm{B}\) is formed is proportional to \((10-y)^{2},\) the number of pairs of molecules of gas A. Find a formula for \(y\) if \(y=2\) moles after 30 seconds.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.