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Calculus Volume 3

7.4 Series Solutions of Differential Equations

Calculus Volume 37.4 Series Solutions of Differential Equations
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 7.4.1. Use power series to solve first-order and second-order differential equations.

In Introduction to Power Series, we studied how functions can be represented as power series, y(x)=n=0anxn.y(x)=n=0anxn. We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives y(x)=n=1nanxn1y(x)=n=1nanxn1 and y(x)=n=2n(n1)anxn2.y(x)=n=2n(n1)anxn2. In some cases, these power series representations can be used to find solutions to differential equations.

Be aware that this subject is given only a very brief treatment in this text. Most introductory differential equations textbooks include an entire chapter on power series solutions. This text has only a single section on the topic, so several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises in this section were chosen for which power solutions exist. However, it is not always the case that power solutions exist. Those of you interested in a more rigorous treatment of this topic should consult a differential equations text.

Problem-Solving Strategy: Finding Power Series Solutions to Differential Equations
  1. Assume the differential equation has a solution of the form y(x)=n=0anxn.y(x)=n=0anxn.
  2. Differentiate the power series term by term to get y(x)=n=1nanxn1y(x)=n=1nanxn1 and y(x)=n=2n(n1)anxn2.y(x)=n=2n(n1)anxn2.
  3. Substitute the power series expressions into the differential equation.
  4. Re-index sums as necessary to combine terms and simplify the expression.
  5. Equate coefficients of like powers of xx to determine values for the coefficients anan in the power series.
  6. Substitute the coefficients back into the power series and write the solution.

Example 7.25

Series Solutions to Differential Equations

Find a power series solution for the following differential equations.

  1. yy=0yy=0
  2. (x21)y+6xy+4y=−4(x21)y+6xy+4y=−4

Solution

  1. Assume y(x)=n=0anxny(x)=n=0anxn (step 1). Then, y(x)=n=1nanxn1y(x)=n=1nanxn1 and y(x)=n=2n(n1)anxn2y(x)=n=2n(n1)anxn2 (step 2). We want to find values for the coefficients anan such that
    yy=0n=2n(n1)anxn2n=0anxn=0(step 3).yy=0n=2n(n1)anxn2n=0anxn=0(step 3).

    We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with n=0.n=0.
    To re-index the first term, replace n with n+2n+2 inside the sum, and change the lower summation limit to n=0.n=0. We get
    n=2n(n1)anxn2=n=0(n+2)(n+1)an+2xn.n=2n(n1)anxn2=n=0(n+2)(n+1)an+2xn.

    This gives
    n=0(n+2)(n+1)an+2xnn=0anxn=0n=0[(n+2)(n+1)an+2an]xn=0(step 4).n=0(n+2)(n+1)an+2xnn=0anxn=0n=0[(n+2)(n+1)an+2an]xn=0(step 4).

    Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of x are zero. So we have
    (n+2)(n+1)an+2an=0forn=0,1,2,….(n+2)(n+1)an+2an=0forn=0,1,2,….

    This recurrence relationship allows us to express each coefficient anan in terms of the coefficient two terms earlier. This yields one expression for even values of n and another expression for odd values of n. Looking first at the equations involving even values of n, we see that
    a2=a02a4=a243=a04!a6=a465=a06!⋮.a2=a02a4=a243=a04!a6=a465=a06!⋮.

    Thus, in general, when n is even, an=a0n!an=a0n! (step 5).
    For the equations involving odd values of n, we see that
    a3=a132=a13!a5=a354=a15!a7=a576=a17!⋮.a3=a132=a13!a5=a354=a15!a7=a576=a17!⋮.

    Therefore, in general, when n is odd, an=a1n!an=a1n! (step 5 continued).
    Putting this together, we have
    y(x)=n=0anxn=a0+a1x+a02x2+a13!x3+a04!x4+a15!x5+⋯.y(x)=n=0anxn=a0+a1x+a02x2+a13!x3+a04!x4+a15!x5+⋯.

    Re-indexing the sums to account for the even and odd values of n separately, we obtain
    y(x)=a0k=01(2k)!x2k+a1k=01(2k+1)!x2k+1(step 6).y(x)=a0k=01(2k)!x2k+a1k=01(2k+1)!x2k+1(step 6).

    Analysis for part a.
    As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since y(x)=c1ex+c2exy(x)=c1ex+c2ex is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
    Fortunately, after writing the power series representations of exex and ex,ex, and doing some algebra, we find that if we choose
    c0=(a0+a1)2,c1=(a0a1)2,c0=(a0+a1)2,c1=(a0a1)2,

    we then have a0=c0+c1a0=c0+c1 anda1=c0c1,a1=c0c1, and
    y(x)=a0+a1x+a02x2+a13!x3+a04!x4+a15!x5+=(c0+c1)+(c0c1)x+(c0+c1)2x2+(c0c1)3!x3+(c0+c1)4!x4+(c0c1)5!x5+=c0n=0xnn!+c1n=0(x)nn!=c0ex+c1ex.y(x)=a0+a1x+a02x2+a13!x3+a04!x4+a15!x5+=(c0+c1)+(c0c1)x+(c0+c1)2x2+(c0c1)3!x3+(c0+c1)4!x4+(c0c1)5!x5+=c0n=0xnn!+c1n=0(x)nn!=c0ex+c1ex.

    So we have, in fact, found the same general solution. Note that this choice of c1c1 and c2c2 is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.
  2. Assume y(x)=n=0anxny(x)=n=0anxn (step 1). Then, y(x)=n=1nanxn1y(x)=n=1nanxn1 and y(x)=n=2n(n1)anxn2y(x)=n=2n(n1)anxn2 (step 2). We want to find values for the coefficients anan such that
    (x21)y+6xy+4y=−4(x21)n=2n(n1)anxn2+6xn=1nanxn1+4n=0anxn=−4x2n=2n(n1)anxn2n=2n(n1)anxn2+6xn=1nanxn1+4n=0anxn=−4.(x21)y+6xy+4y=−4(x21)n=2n(n1)anxn2+6xn=1nanxn1+4n=0anxn=−4x2n=2n(n1)anxn2n=2n(n1)anxn2+6xn=1nanxn1+4n=0anxn=−4.

    Taking the external factors inside the summations, we get
    n=2n(n1)anxnn=2n(n1)anxn2+n=16nanxn+n=04anxn=−4(step 3).n=2n(n1)anxnn=2n(n1)anxn2+n=16nanxn+n=04anxn=−4(step 3).

    Now, in the first summation, we see that when n=0n=0 or n=1,n=1, the term evaluates to zero, so we can add these terms back into our sum to get
    n=2n(n1)anxn=n=0n(n1)anxn.n=2n(n1)anxn=n=0n(n1)anxn.

    Similarly, in the third term, we see that when n=0,n=0, the expression evaluates to zero, so we can add that term back in as well. We have
    n=16nanxn=n=06nanxn.n=16nanxn=n=06nanxn.

    Then, we need only shift the indices in our second term. We get
    n=2n(n1)anxn2=n=0(n+2)(n+1)an+2xn.n=2n(n1)anxn2=n=0(n+2)(n+1)an+2xn.

    Thus, we have
    n=0n(n1)anxnn=0(n+2)(n+1)an+2xn+n=06nanxn+n=04anxn=−4(step 4).n=0[n(n1)an(n+2)(n+1)an+2+6nan+4an]xn=−4n=0[(n2n)an+6nan+4an(n+2)(n+1)an+2]xn=−4n=0[n2an+5nan+4an(n+2)(n+1)an+2]xn=−4n=0[(n2+5n+4)an(n+2)(n+1)an+2]xn=−4n=0[(n+4)(n+1)an(n+2)(n+1)an+2]xn=−4n=0n(n1)anxnn=0(n+2)(n+1)an+2xn+n=06nanxn+n=04anxn=−4(step 4).n=0[n(n1)an(n+2)(n+1)an+2+6nan+4an]xn=−4n=0[(n2n)an+6nan+4an(n+2)(n+1)an+2]xn=−4n=0[n2an+5nan+4an(n+2)(n+1)an+2]xn=−4n=0[(n2+5n+4)an(n+2)(n+1)an+2]xn=−4n=0[(n+4)(n+1)an(n+2)(n+1)an+2]xn=−4

    Looking at the coefficients of each power of x, we see that the constant term must be equal to −4,−4, and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,
    4a02a2=−4a2=2a0+2(step 3).4a02a2=−4a2=2a0+2(step 3).

    For n1,n1, we have
    (n+4)(n+1)an(n+2)(n+1)an+2=0(n+1)[(n+4)an(n+2)an+2]=0.(n+4)(n+1)an(n+2)(n+1)an+2=0(n+1)[(n+4)an(n+2)an+2]=0.

    Since n1,n1, n+10,n+10, we see that
    (n+4)an(n+2)an+2=0(n+4)an(n+2)an+2=0

    and thus
    an+2=n+4n+2an.an+2=n+4n+2an.

    For even values of n, we have
    a4=64(2a0+2)=3a0+3a6=86(3a0+3)=4a0+4⋮.a4=64(2a0+2)=3a0+3a6=86(3a0+3)=4a0+4⋮.

    In general, a2k=(k+1)(a0+1)a2k=(k+1)(a0+1) (step 5).
    For odd values of n, we have
    a3=53a1a5=75a3=73a1a7=97a5=93a1=3a1⋮.a3=53a1a5=75a3=73a1a7=97a5=93a1=3a1⋮.

    In general, a2k+1=2k+33a1a2k+1=2k+33a1 (step 5 continued).
    Putting this together, we have
    y(x)=k=0(k+1)(a0+1)x2k+k=0(2k+33)a1x2k+1(step 6).y(x)=k=0(k+1)(a0+1)x2k+k=0(2k+33)a1x2k+1(step 6).
Checkpoint 7.22

Find a power series solution for the following differential equations.

  1. y+2xy=0y+2xy=0
  2. (x+1)y=3y(x+1)y=3y

We close this section with a brief introduction to Bessel functions. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order n is given by

x2y+xy+(x2n2)y=0.x2y+xy+(x2n2)y=0.

This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.

Example 7.26

Power Series Solution to the Bessel Equation

Find a power series solution to the Bessel equation of order 0 and graph the solution.

Solution

The Bessel equation of order 0 is given by

x2y+xy+x2y=0.x2y+xy+x2y=0.

We assume a solution of the form y=n=0anxn.y=n=0anxn. Then y(x)=n=1nanxn1y(x)=n=1nanxn1 and y(x)=n=2n(n1)anxn2.y(x)=n=2n(n1)anxn2. Substituting this into the differential equation, we get

x2n=2n(n1)anxn2+xn=1nanxn1+x2n=0anxn=0Substitution.n=2n(n1)anxn+n=1nanxn+n=0anxn+2=0Bring external factors within sums.n=2n(n1)anxn+n=1nanxn+n=2an2xn=0Re-index third sum.n=2n(n1)anxn+a1x+n=2nanxn+n=2an2xn=0Separaten=1term from second sum.a1x+n=2[n(n1)an+nan+an2]xn=0Collect summation terms.a1x+n=2[(n2n)an+nan+an2]xn=0Multiply through in first term.a1x+n=2[n2an+an2]xn=0.Simplify.x2n=2n(n1)anxn2+xn=1nanxn1+x2n=0anxn=0Substitution.n=2n(n1)anxn+n=1nanxn+n=0anxn+2=0Bring external factors within sums.n=2n(n1)anxn+n=1nanxn+n=2an2xn=0Re-index third sum.n=2n(n1)anxn+a1x+n=2nanxn+n=2an2xn=0Separaten=1term from second sum.a1x+n=2[n(n1)an+nan+an2]xn=0Collect summation terms.a1x+n=2[(n2n)an+nan+an2]xn=0Multiply through in first term.a1x+n=2[n2an+an2]xn=0.Simplify.

Then, a1=0,a1=0, and for n2,n2,

n2an+an2=0an=1n2an2.n2an+an2=0an=1n2an2.

Because a1=0,a1=0, all odd terms are zero. Then, for even values of n, we have

a2=122a0a4=142a2=14222a0.a6=162a4=1624222a0a2=122a0a4=142a2=14222a0.a6=162a4=1624222a0

In general,

a2k=(−1)k(2)2k(k!)2a0.a2k=(−1)k(2)2k(k!)2a0.

Thus, we have

y(x)=a0k=0(−1)k(2)2k(k!)2x2k.y(x)=a0k=0(−1)k(2)2k(k!)2x2k.
7.10

The graph appears below.

This figure is the graph of a function. The graph is oscillating with the highest amplitude above the origin. The horizontal axis is labeled in increments of 2.5. The vertical axis is labeled in increments of 0.2.
Checkpoint 7.23

Verify that the expression found in Example 7.26 is a solution to the Bessel equation of order 0.

Section 7.4 Exercises

Find a power series solution for the following differential equations.

104.

y+6y=0y+6y=0

105.

5y+y=05y+y=0

106.

y+25y=0y+25y=0

107.

yy=0yy=0

108.

2y+y=02y+y=0

109.

y2xy=0y2xy=0

110.

(x7)y+2y=0(x7)y+2y=0

111.

yxyy=0yxyy=0

112.

(1+x2)y4xy+6y=0(1+x2)y4xy+6y=0

113.

x2yxy3y=0x2yxy3y=0

114.

y8y=0,y(0)=−2,y(0)=10y8y=0,y(0)=−2,y(0)=10

115.

y2xy=0,y(0)=1,y(0)=−3y2xy=0,y(0)=1,y(0)=−3

116.

The differential equation x2y+xy+(x21)y=0x2y+xy+(x21)y=0 is a Bessel equation of order 1. Use a power series of the form y=n=0anxny=n=0anxn to find the solution.

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