7.3: Series Solutions Near an Ordinary Point II
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In this section we continue to find series solutions
y=∞∑n=0an(x−x0)n
of initial value problems
P0(x)y″+P1(x)y′+P2(x)y=0,y(x0)=a0,y′(x0)=a1,
where P0,P1, and P2 are polynomials and P0(x0)≠0, so x0 is an ordinary point of Equation ???. However, here we consider cases where the differential equation in Equation ??? is not of the form
(1+α(x−x0)2)y″+β(x−x0)y′+γy=0,
so Theorem 7.2.2 does not apply, and the computation of the coefficients {an} is more complicated. For the equations considered here it is difficult or impossible to obtain an explicit formula for an in terms of n. Nevertheless, we can calculate as many coefficients as we wish. The next three examples illustrate this.
Example 7.4.1
Find the coefficients a0, …, a7 in the series solution y=∑∞n=0anxn of the initial value problem
(1+x+2x2)y″+(1+7x)y′+2y=0,y(0)=−1,y′(0)=−2.
Solution
Here
Ly=(1+x+2x2)y″+(1+7x)y′+2y.
The zeros (−1±i√7)/4 of P0(x)=1+x+2x2 have absolute value 1/√2, so Theorem 7.2.2 implies that the series solution converges to the solution of Equation ??? on (−1/√2,1/√2). Since
y=∞∑n=0anxn,y′=∞∑n=1nanxn−1andy″=∞∑n=2n(n−1)anxn−2,
Ly=∞∑n=2n(n−1)anxn−2+∞∑n=2n(n−1)anxn−1+2∞∑n=2n(n−1)anxn+∞∑n=1nanxn−1+7∞∑n=1nanxn+2∞∑n=0anxn.
Shifting indices so the general term in each series is a constant multiple of xn yields
Ly=∞∑n=0(n+2)(n+1)an+2xn+∞∑n=0(n+1)nan+1xn+2∞∑n=0n(n−1)anxn+∞∑n=0(n+1)an+1xn+7∞∑n=0nanxn+2∞∑n=0anxn=∞∑n=0bnxn,
where
bn=(n+2)(n+1)an+2+(n+1)2an+1+(n+2)(2n+1)an.
Therefore y=∑∞n=0anxn is a solution of Ly=0 if and only if
an+2=−n+1n+2an+1−2n+1n+1an,n≥0.
From the initial conditions in Equation ???, a0=y(0)=−1 and a1=y′(0)=−2. Setting n=0 in Equation ??? yields
a2=−12a1−a0=−12(−2)−(−1)=2.
Setting n=1 in Equation ??? yields
a3=−23a2−32a1=−23(2)−32(−2)=53.
We leave it to you to compute a4,a5,a6,a7 from Equation ??? and show that
y=−1−2x+2x2+53x3−5512x4+34x5+618x6−44356x7+⋯.
We also leave it to you (Exercise [exer:7.3.13}) to verify numerically that the Taylor polynomials TN(x)=∑Nn=0anxn converge to the solution of Equation ??? on (−1/√2,1/√2).
Example 7.4.2
Find the coefficients a0, …, a5 in the series solution
y=∞∑n=0an(x+1)n
of the initial value problem
(3+x)y″+(1+2x)y′−(2−x)y=0,y(−1)=2,y′(−1)=−3.
Since the desired series is in powers of x+1 we rewrite the differential equation in Equation ??? as Ly=0, with
Ly=(2+(x+1))y″−(1−2(x+1))y′−(3−(x+1))y.
Since
y=∞∑n=0an(x+1)n,y′=∞∑n=1nan(x+1)n−1andy″=∞∑n=2n(n−1)an(x+1)n−2,
Ly=2∞∑n=2n(n−1)an(x+1)n−2+∞∑n=2n(n−1)an(x+1)n−1−∞∑n=1nan(x+1)n−1+2∞∑n=1nan(x+1)n−3∞∑n=0an(x+1)n+∞∑n=0an(x+1)n+1.
Shifting indices so that the general term in each series is a constant multiple of (x+1)n yields
Ly=2∞∑n=0(n+2)(n+1)an+2(x+1)n+∞∑n=0(n+1)nan+1(x+1)n−∞∑n=0(n+1)an+1(x+1)n+∞∑n=0(2n−3)an(x+1)n+∞∑n=1an−1(x+1)n=∞∑n=0bn(x+1)n,
where
b0=4a2−a1−3a0
and
bn=2(n+2)(n+1)an+2+(n2−1)an+1+(2n−3)an+an−1,n≥1.
Therefore y=∑∞n=0an(x+1)n is a solution of Ly=0 if and only if
a2=14(a1+3a0)
and
an+2=−12(n+2)(n+1)[(n2−1)an+1+(2n−3)an+an−1],n≥1.
From the initial conditions in Equation ???, a0=y(−1)=2 and a1=y′(−1)=−3. We leave it to you to compute a2, …, a5 with Equation ??? and Equation ??? and show that the solution of Equation ??? is
y=−2−3(x+1)+34(x+1)2−512(x+1)3+748(x+1)4−160(x+1)5+⋯.
We also leave it to you (Exercise [exer:7.3.14}) to verify numerically that the Taylor polynomials TN(x)=∑Nn=0anxn converge to the solution of Equation ??? on the interval of convergence of the power series solution.
Example 7.4.3
Find the coefficients a0, …, a5 in the series solution y=∑∞n=0anxn of the initial value problem
y″+3xy′+(4+2x2)y=0,y(0)=2,y′(0)=−3.
Solution
Here
Ly=y″+3xy′+(4+2x2)y.
Since
y=∞∑n=0anxn,y′=∞∑n=1nanxn−1,andy″=∞∑n=2n(n−1)anxn−2,
Ly=∞∑n=2n(n−1)anxn−2+3∞∑n=1nanxn+4∞∑n=0anxn+2∞∑n=0anxn+2.
Shifting indices so that the general term in each series is a constant multiple of xn yields
Ly=∞∑n=0(n+2)(n+1)an+2xn+∞∑n=0(3n+4)anxn+2∞∑n=2an−2xn=∞∑n=0bnxn
where
b0=2a2+4a0,b1=6a3+7a1,
and
bn=(n+2)(n+1)an+2+(3n+4)an+2an−2,n≥2.
Therefore y=∑∞n=0anxn is a solution of Ly=0 if and only if
a2=−2a0,a3=−76a1,
and
an+2=−1(n+2)(n+1)[(3n+4)an+2an−2],n≥2.
From the initial conditions in Equation ???, a0=y(0)=2 and a1=y′(0)=−3. We leave it to you to compute a2, …, a5 with Equation ??? and Equation ??? and show that the solution of Equation ??? is
y=2−3x−4x2+72x3+3x4−7940x5+⋯.
We also leave it to you (Exercise [exer:7.3.15}) to verify numerically that the Taylor polynomials TN(x)=∑Nn=0anxn converge to the solution of Equation ??? on the interval of convergence of the power series solution.