Series Solutions of Differential Equations
5.2 Series Solution to Differential Equations
Power Series Solutions to Linear Differential Equations
In earlier discussions, we primarily focused on homogeneous linear differential equations with constant coefficients. However, many physical applications lead to more complex second-order homogeneous linear differential equations of the form
[asciimath]P_0(x)y''+P_1(x)y'+P_2(x)y=0[/asciimath] (5.2.1)
where [asciimath]P_0, \ P_1,[/asciimath] and [asciimath]P_2[/asciimath] are polynomials with no common factor. Often, the solutions to Equation 5.2.1 cannot be expressed in terms of familiar functions, prompting the use of series solutions. We start by normalizing the equation, dividing by [asciimath]P_0(x)[/asciimath] to make the coefficient of [asciimath]y''[/asciimath] one.
[asciimath]y''+(P_1(x))/(P_0(x) )y'+(P_2(x))/(P_0(x) ) y=0[/asciimath]
Given the continuity of polynomials, [asciimath]P_1//P_0,[/asciimath] and [asciimath]P_2//P_0[/asciimath] are continuous except possibly where [asciimath]P_0(x_0)=0[/asciimath]. A point [asciimath]x_0[/asciimath] where [asciimath]P_0(x_0)!=0[/asciimath] is called an ordinary point of Equation 5.2.1; otherwise it is a singular point. Importantly, at ordinary points, [asciimath]P_1//P_0[/asciimath] and [asciimath]P_2//P_0[/asciimath] are analytic, allowing for power series representation.
Theorem. Suppose [asciimath]P_0, \ P_1,[/asciimath] and [asciimath]P_2[/asciimath] are polynomials with no common factor and [asciimath]P_0(x)!=0[/asciimath]. If [asciimath]x_0[/asciimath] is an ordinary point of Equation 5.2.1, then every solution of the equation can be represented by a power series
[asciimath]y(x)=sum_(n=0)^oo\ a_n(x-x_0)^n[/asciimath] (5.2.2)
Moreover, the radius of convergence [asciimath]R[/asciimath] of such a power series solution is at least as large as the distance from [asciimath]x_0[/asciimath] to the nearest singular point (real or complex) of the equation. If [asciimath]P_0[/asciimath] is constant, implying it is never zero, the radius of convergence will be infinity and the interval of convergence will be [asciimath](-oo,+oo)[/asciimath].
To find series solutions of Equation 5.2.1, we consider a power series converging near an ordinary point [asciimath]x_0[/asciimath]. We assume that the solution can be written as a power series 5.2.2, substitute [asciimath]y[/asciimath] and its derivatives in the given differential equation, and collect like powers of [asciimath]x-x_0[/asciimath]. Setting the coefficient of each power to zero, we can systematically solve for the [asciimath]a_n[/asciimath] coefficients, often resulting in a recursive relation.
How to Find a Series Solution to a Differential Equation
1. Determine the differential equation and choose the point [asciimath]x_0[/asciimath] around which to expand the series (typically an ordinary point)
2. Assume a power series solution (Equation 5.2.2) for [asciimath]y[/asciimath] and find its derivatives [asciimath]y'[/asciimath] , [asciimath]y''[/asciimath], etc., as required by the differential equation.
3. Substitute the series and its derivative into the differential equation.
4. Organize like powers of [asciimath]x-x_0[/asciimath] by aligning terms, ensuring all series are expressed from the same starting value of [asciimath]n[/asciimath].
5. Collect and group the coefficients of like powers of [asciimath]x-x_0[/asciimath].
6. Solve equations by equating coefficients of like powers of [asciimath]x-x_0[/asciimath] to find relations among [asciimath]a_n[/asciimath]‘s.
7. Use the given initial or boundary conditions to find specific [asciimath]a_n[/asciimath]‘s. Use the recursive relation to determine all coefficients.
8. Construct the solution with the coefficients found and discuss the radius and interval of convergence.
Determine a series solution for the differential equation
[asciimath]y''+y=0[/asciimath]
Show/Hide Solution
1. Notice that [asciimath]P_1(x)=1[/asciimath] and thus the coefficients are analytic at every point. We assume [asciimath]x_0=0[/asciimath] and that the solution can be written as a power series
[asciimath]y=sum_(n=0)^oo\ a_nx^n[/asciimath]
2. First, we need to find [asciimath]y''[/asciimath]:
[asciimath]y'=d/dxsum_(n=0)^oo\ a_nx^n[/asciimath] [asciimath]=sum_(n=1)^oo\ n a_n x^(n-1)[/asciimath]
[asciimath]y''=d/dxsum_(n=1)^oo\ n a_nx^(n-1)[/asciimath] [asciimath]=sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath]
3. Next, we substitute [asciimath]y[/asciimath] and [asciimath]y''[/asciimath] into the equation:
[asciimath]y''+y=0[/asciimath]
[asciimath]sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath] [asciimath]+sum_(n=0)^oo\ a_nx^n=0[/asciimath]
4. The next step is to align terms. To do this we need to shift the summation indices to start at the same value. Letting [asciimath]k=n-2[/asciimath] or equally [asciimath]n=k+2[/asciimath] in the first summation and [asciimath]n=k[/asciimath] in the second summation, we have
[asciimath]sum_(k=0)^oo\ (k+2)(k+1) a_(k+2) x^(k)[/asciimath] [asciimath]+sum_(k=0)^oo\ a_kx^k=0[/asciimath]
5. Adding the series yields
[asciimath]sum_(k=0)^oo\ [(k+2)(k+1) a_(k+2)+a_k] x^(k) =0[/asciimath]
6. From the Power Series Vanishing on an Interval property discussed in Section 5.1, we know that If a power series is zero for all [asciimath]x[/asciimath], then all its coefficients must be zero. Therefore, we conclude that
[asciimath](k+2)(k+1) a_(k+2)+a_k=0[/asciimath]
or
[asciimath]a_(k+2)=(-a_k)/((k+2)(k+1) ),[/asciimath] [asciimath]k>=0[/asciimath]
This is called the recurrence relation for the values of [asciimath]k[/asciimath] for which the relation is true.
7. Next, we write a few terms of the series to see if we can determine the trend and hopefully the explicit formula of the series. Setting [asciimath]k=0,1,2,...,5[/asciimath], we get
| [asciimath]k=0 ->[/asciimath] [asciimath]a_2=(-a_0)/((2)(1) )[/asciimath] | [asciimath]k=1 ->[/asciimath] [asciimath]a_3=(-a_1)/((3)(2) )[/asciimath] |
| [asciimath]k=2 ->[/asciimath] [asciimath]a_4=(-a_2)/((4)(3) )[/asciimath] [asciimath]= (a_0)/((4)(3)(2)(1) )[/asciimath] | [asciimath]k=3 ->[/asciimath] [asciimath]a_5=(-a_3)/((5)(4) )[/asciimath] [asciimath]=(a_1)/((5)(4)(3)(2) )[/asciimath] |
| [asciimath]k=4 ->[/asciimath] [asciimath]a_6=(-a_4)/((6)(5) )[/asciimath] [asciimath]= (-a_0)/((6)(5)(4)(3)(2)(1) )[/asciimath] | [asciimath]k=5 ->[/asciimath] [asciimath]a_7=(-a_5)/((7)(6) )[/asciimath] [asciimath]=(-a_1)/((7)(6)(5)(4)(3)(2) )[/asciimath] |
Notice that the term with even indices can be written in terms of the previous term and eventually in terms of [asciimath]a_0[/asciimath] and so can be the odd indices in terms of [asciimath]a_1[/asciimath]. Therefore, by writing the recurrence relation separately for odd ([asciimath]k=2m+1[/asciimath]) and even ([asciimath]k=2m[/asciimath]) indices, we get
[asciimath]a_(2m)=((-1)^ma_0)/((2m)![/asciimath] [asciimath],\ m>=0[/asciimath]
[asciimath]a_(2m+1)=((-1)^ma_1)/((2m+1)![/asciimath] [asciimath],\ m>=0[/asciimath]
8. Thus the general solution of the equation can be written as
[asciimath]y=sum_(n=0)^oo\ a_nx^n[/asciimath]
[asciimath]=sum_(m=0)^oo\ a_(2m)x^(2m)[/asciimath] [asciimath]+sum_(m=0)^oo\ a_(2m+1)x^(2m+1)[/asciimath]
[asciimath]=a_0sum_(m=0)^oo\ (-1)^m x^(2m) /((2m)![/asciimath] [asciimath]+a_1sum_(m=0)^oo\ (-1)^mx^(2m+1) /((2m+1)![/asciimath]
We recognize that the series in the solution are the Maclaurin series of [asciimath]cos(x)[/asciimath] and [asciimath]sin(x)[/asciimath], respectively.
[asciimath]cos(x)=sum_(m=0)^oo\ (-1)^m x^(2m) /((2m)![/asciimath] and [asciimath]sin(x)=sum_(m=0)^oo\ (-1)^mx^(2m+1) /((2m+1)![/asciimath]
Therefore, the general solution can be expressed as
[asciimath]y=a_0cos(x)+a_1sin(x)[/asciimath]
for some arbitrary constant [asciimath]a_0[/asciimath] and [asciimath]a_1[/asciimath]. This is the same solution we would obtain using the methods learned in previous sections.
The interval of convergence for both the cosine and sine series is all real numbers [asciimath](-oo,oo)[/asciimath].
For both series in the solution, the Ratio Test indicates that as [asciimath]m->oo[/asciimath] the limit [asciimath]L[/asciimath] approaches zero, which means the series converge for all real numbers. Therefore, without prior knowledge of the series representing sine and cosine, we would conclude that the interval of convergence for each series and hence the combined series solution is all real number [asciimath](-oo,oo)[/asciimath].
Try an Example
In practice, we are interested in finding the series solution for equations with nonconstant coefficients. This is because equations with constant coefficients can be easily solved using the technique outlined in Chapter 3 for homogeneous equations with constant coefficients. Let us do another example for an equation with nonconstant coefficients.
Find a series solution for the differential equation
[asciimath](1+x^2)y''+2xy'-2y=0[/asciimath]
Show/Hide Solution
1. Note that [asciimath]P_1(x)=1+x^2[/asciimath] has no root and thus every point for this equation is an ordinary point. We assume that the solution can be written as a power series
[asciimath]y=sum_(n=0)^oo\ a_nx^n[/asciimath]
2. Next, we find [asciimath]y'[/asciimath] and [asciimath]y''[/asciimath] :
[asciimath]y'=d/dxsum_(n=0)^oo\ a_nx^n[/asciimath] [asciimath]=sum_(n=1)^oo\ n a_n x^(n-1)[/asciimath]
[asciimath]y''=d/dxsum_(n=1)^oo\ n a_nx^(n-1)[/asciimath] [asciimath]=sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath]
3. Next, we substitute [asciimath]y,\ y',[/asciimath] and [asciimath]y''[/asciimath] into the equation:
[asciimath](1+x^2)sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath] [asciimath]+2xsum_(n=1)^oo\ n a_n x^(n-1)[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n=0[/asciimath]
Multiplying the coefficients by the series, we get
[asciimath]sum_(n=2)^oo\ n(n-1) a_n x^(n-2)[/asciimath] [asciimath]+sum_(n=2)^oo\ n(n-1) a_n x^n[/asciimath] [asciimath]+2sum_(n=1)^oo\ n a_n x^n[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n=0[/asciimath]
4. Note that the exponent of [asciimath]x[/asciimath] is the same in all but the first series. Therefore, we only need to shift the index of the first summations by 2:
[asciimath]sum_(n=0)^oo\ (n+2)(n+1) a_(n+2) x^n[/asciimath] [asciimath]+sum_(n=2)^oo\ n(n-1) a_n x^n[/asciimath] [asciimath]+2sum_(n=1)^oo\ n a_n x^n[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n=0[/asciimath]
Also, notice that the second series is zero at [asciimath]n=0,1[/asciimath] . So its index can start at [asciimath]n=0[/asciimath] . Likewise, the third series is zero at [asciimath]n=0[/asciimath] so it can too start at [asciimath]n=0[/asciimath] . Then we rewrite the indices and we have
[asciimath]sum_(n=0)^oo\ (n+2)(n+1) a_(n+2) x^n[/asciimath] [asciimath]+sum_(n=0)^oo\ n(n-1) a_n x^n[/asciimath] [asciimath]+2sum_(n=0)^oo\ n a_n x^n[/asciimath] [asciimath]-2sum_(n=0)^oo\ a_nx^n=0[/asciimath]
5. Combining the series yields
[asciimath]sum_(n=0)^oo\ [(n+2)(n+1) a_(n+2)+(n(n-1)+2n-2)a_n] x^n=0[/asciimath]
[asciimath]sum_(n=0)^oo\ [(n+2)(n+1) a_(n+2)+(n^2+n-2)a_n] x^n=0[/asciimath]
6. Now setting the coefficient to zero gives
[asciimath](n+2)(n+1) a_(n+2)+(n^2+n-2) a_n=0[/asciimath]
[asciimath]a_(n+2)=-(n^2+n-2) /((n+2)(n+1) )a_n[/asciimath]
[asciimath]=-((n+2)(n-1)) /((n+2)(n+1) )a_n[/asciimath]
7. So the recursive relation is simplified to
[asciimath]a_(n+2)=-(n-1) /(n+1)a_n[/asciimath]
Setting [asciimath]n=0,1,2,...,5[/asciimath] , we get
| [asciimath]n=0 ->[/asciimath] [asciimath]a_2=a_0[/asciimath] | [asciimath]n=1 ->[/asciimath] [asciimath]a_3=0a_1=0[/asciimath] |
| [asciimath]n=2 ->[/asciimath] [asciimath]a_4=-1/3 a_2[/asciimath] [asciimath]=-1/3a_0[/asciimath] | [asciimath]n=3 ->[/asciimath] [asciimath]a_5=-2/4 a_3[/asciimath] [asciimath]=0[/asciimath] |
| [asciimath]n=4 ->[/asciimath] [asciimath]a_6=-3/5 a_4[/asciimath] [asciimath]=1/5a_0[/asciimath] | [asciimath]n=5 ->[/asciimath] [asciimath]a_7=-4/6 a_5[/asciimath] [asciimath]=0[/asciimath] |
Notice that all the terms with odd indices are zero except [asciimath]a_1[/asciimath]. Therefore, by writing the recurrence relation separately for odd ([asciimath]n=2m+1[/asciimath]) and even ([asciimath]n=2m[/asciimath]) indices, we obtain
[asciimath]a_(2m)=(-1)^(m+1)1/(2m-1)a_0[/asciimath] [asciimath],\ m>=1[/asciimath]
[asciimath]a_(2m+1)=a_1[/asciimath] [asciimath],\ m=0[/asciimath]
8. Thus the general solution of the equation can be written as
[asciimath]y=sum_(n=0)^oo\ a_nx^n[/asciimath]
[asciimath]=a_0+a_1x^(2(0)+1) +sum_(m=1)^oo\ a_(2m)x^(2m)[/asciimath]
[asciimath]=a_0+a_1x +a_0sum_(m=1)^oo\(-1)^(m+1)x^(2m)/(2m-1)[/asciimath]
Find the first six terms in the series solution of the initial value problem
[asciimath](1+x^2)y''+2xy'-2y=0,[/asciimath] [asciimath]y(0)=2,\ y'(0)=3[/asciimath]
Show/Hide Solution
In Example 5.2.2, we found the general series solution to this differential equation.
[asciimath]=a_0+a_1x +a_0sum_(m=1)^oo\(-1)^(m+1)x^(2m)/(2m-1)[/asciimath]
To apply the initial conditions, we first recognize that [asciimath]a_0=y(0)=2[/asciimath] and [asciimath]a_1=y'(0)=3[/asciimath]. Then, we substitute [asciimath]a_0[/asciimath] and [asciimath]a_1[/asciimath] into the general solution to compute the other terms.
[asciimath]a_0=2[/asciimath]
[asciimath]a_1=3[/asciimath]
[asciimath]m=1\ ->[/asciimath] [asciimath]a_2=[/asciimath] [asciimath]a_0[(-1)^(1+1)x^(2(1))/(2(1)-1)][/asciimath] [asciimath]->\ a_2=2x^2[/asciimath]
[asciimath]a_3=0[/asciimath]
[asciimath]m=2 \ ->[/asciimath] [asciimath]a_4=a_0[(-1)^(2+1)x^(2(2))/(2(2)-1)][/asciimath] [asciimath]->\ a_4=-2/3x^4[/asciimath]
[asciimath]a_5=0[/asciimath]
[asciimath]m=3 \ ->[/asciimath] [asciimath]a_6=a_0[(-1)^(3+1)x^(2(3))/(2(3)-1)][/asciimath] [asciimath]->\ a_6=2/5x^6[/asciimath]
[asciimath]a_7=0[/asciimath]
[asciimath]m=4 \ ->[/asciimath] [asciimath]a_8=a_0[(-1)^(4+1)x^(2(4))/(2(4)-1)][/asciimath] [asciimath]->\ a_8=-2/7x^8[/asciimath]
Therefore, the solution to the initial value problem is
[asciimath]y(x)=2+3x+2x^2-2/3x^4+2/5x^6-2/7x^8+...[/asciimath] .
Try an Example
Section 5.2 Exercises
- Find the first five terms in the series solution of the initial value problem
[asciimath](1+x)y''+(1-3x)y'-y=0,[/asciimath] [asciimath]y(0)=3,\ y'(0)=2[/asciimath]
Show/Hide Answer
[asciimath]y = 3 + 2 x +1/2 x^2 + x^3 -11/24x^4+...[/asciimath]
- Find the first five terms in the series solution of the initial value problem
[asciimath]y''-2xy'+y=0,[/asciimath] [asciimath]y(0)=4,\ y'(0)=3[/asciimath]
Show/Hide Answer
[asciimath]y = 4 + 3 x -2 x^2 + 1/2 x^3 -1/2x^4+...[/asciimath]
- Find the first five terms in the series solution of the initial value problem
[asciimath](2+x)y''+(1-4x)y'+(2+5x)y=0,[/asciimath] [asciimath]y(0)=1,\ y'(0)=2[/asciimath]
Show/Hide Answer
[asciimath]y = 1 + 2 x - x^2 + 1/4x^3 -73/96x^4+...[/asciimath]
