Topic 5 of 6

Series solutions

Step 20 · Differential equations topics

Solve differential equations by assuming the answer is a power series, finding its coefficients from a recurrence relation, and knowing where the series can be trusted.

Power series solutions

When an equation has no neat solution in familiar functions, assume y = Σ aₙxⁿ. Differentiate term by term, substitute, and match the coefficients of each power of x.

y = \sum_{n=0}^{\infty} a_n x^{n} \qquad y' = \sum_{n=1}^{\infty} n a_n x^{n-1}

Example: A first check

Solve y′ = y by series, with y(0) = 1.

Matching the coefficients of xⁿ gives (n + 1)aₙ₊₁ = aₙ.
With a₀ = 1: a₁ = 1, a₂ = 1/2, a₃ = 1/6, and so on.
So aₙ = 1 ÷ n!.

Answer $y = \sum \frac{x^{n}}{n!}$ , which is $e^{x}$

Recurrence relations

Matching coefficients usually gives a recurrence relation: a rule for each coefficient in terms of earlier ones. The first one or two coefficients are free, set by the starting values.

Example: y″ + y = 0

Find the recurrence for y″ + y = 0.

The coefficient of xⁿ in y″ is $(n + 2)(n + 1)a_{n+2}$ .
So $(n + 2)(n + 1)a_{n+2} + a_n = 0$ .

Answer $a_{n+2} = -\frac{a_n}{(n + 2)(n + 1)}$

Example: Building the solution

With a₀ = 1 and a₁ = 0, find the first terms.

a₂ = −1/2 and a₄ = 1/24, and every odd coefficient is 0.
So y = $1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} - \cdots$

Answercos x

Example: Airy's equation

Find the recurrence for y″ = xy.

On the right, xy raises every power by one, so the coefficient of xⁿ there is $a_{n-1}$ .
So $(n + 2)(n + 1)a_{n+2} = a_{n-1}$ , and $a_2 = 0$ .

Answer $a_{n+2} = \frac{a_{n-1}}{(n + 2)(n + 1)}$

Where series solutions work

For y″ + p(x)y′ + q(x)y = 0, a point where p and q are well behaved is an ordinary point. There, power series solutions exist, and they converge at least out to the nearest point where p or q blows up.

Example: A guaranteed radius

At least how far from 0 does a series solution of (1 − x²)y″ − 2xy′ + 6y = 0 converge?

Divide by 1 − x²: p and q blow up at x = 1 and x = −1.
The nearest trouble is 1 away from 0.

AnswerAt least for |x| < 1

Example: Singular points

Where are the singular points of x²y″ + y′ + y = 0?

Divide by x²: p = 1 ÷ x² blows up at 0.

AnswerOnly x = 0, where a plain power series about 0 may fail

Try it yourself

Work each one on paper first, then open the answer to check.

If y = Σ aₙxⁿ, what is the coefficient of x³ in y′?
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4a₄
Solve y′ = 2y by series with y(0) = 1. What is a₂?
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2
With $a_{n+2} = -\frac{a_n}{(n + 2)(n + 1)}$ and a₁ = 1, find a₃.
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$-\frac{1}{6}$
Which function does the series for y″ + y = 0 with a₀ = 0 and a₁ = 1 give?
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sin x
Is x = 0 an ordinary point of y″ + xy = 0?
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Yes
Where are the singular points of (x − 2)y″ + y = 0?
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At x = 2