LightMySky
به انگلیسی

Series Solutions About an Ordinary Point

When the coefficients are not constant, assume a power series, match coefficients and generate a recurrence. Named functions arise this way, such as the Airy functions, which solve y'' = xy.

No account needed. Progress saves in this browser.

What a learner can do afterwards

  • Substitute a power series and derive the recurrence relation
  • Generate the first several coefficients from initial conditions
  • Say what the radius of convergence means for the solution's range

1 · Read

When the coefficients are not constant, assume a power series and let it tell you its own terms. Write y as the sum of a sub n times x to the n, differentiate term by term, shift the indices so like powers line up, and match coefficients. That matching hands you a recurrence relation linking each coefficient to the earlier ones. Named functions come out of this: y double prime minus x y equals 0, solved below, is Airy's equation, and its solutions are the Airy functions.

Try it together

Solve y prime equals 3 y with a0 equals 2. Term by term differentiation turns the equation into n plus 1 times a sub n plus 1 equals 3 a sub n. At n equals 0 that reads 1 times a1 equals 3 times 2, so a1 is 6. At n equals 1 it reads 2 times a2 equals 3 times 6, so a2 is 9. Feed each answer into the next step and the series builds itself: 2 plus 6 x plus 9 x squared onward.

(n + 1) a(n+1) = 3 a(n)n = 0: a1 = 3 × 2 = 6n = 1: 2 a2 = 18, a2 = 9y = 2 + 6x + 9x² + ...

Check your series against expansions you already know. The series above is 2 times e to the 3 x expanded, and matching a known expansion catches algebra slips fast. Also mind where the series can be trusted: the radius of convergence says how far from the centre the sum still represents the solution. At an ordinary point the method runs smoothly, and when the coefficients are polynomials, as in y double prime minus x y equals 0, the series converges for every x.

Try it together

Now a variable coefficient. For y'' - xy = 0, write y = Σ a_n x^n. Then y'' = Σ (n + 2)(n + 1) a_(n+2) x^n, and xy = Σ a_(n-1) x^n starting at n = 1. The x^0 terms give 2a_2 = 0, so a_2 = 0. For n ≥ 1, (n + 2)(n + 1) a_(n+2) = a_(n-1). The initial conditions set the seeds: a_0 = y(0) and a_1 = y'(0). With y(0) = 1 and y'(0) = 0, a_3 = 1/6, a_4 = a_5 = 0 and a_6 = a_3/30 = 1/180, so y = 1 + x³/6 + x⁶/180 + ...

Substitute the series, shift indices, read the recurrence, and grow the coefficients step by step.

Readings 'Finding cos x from its equation', 'Where a power series works' and 'Check against a series you know' are original to this dataset. All 6 extra questions are original to this dataset.

2 · Watch

به انگلیسی

Take it off screen

Print a worksheetA4 with an answer key page for grown-ups. No screen, no internet.

Where it sits

Learn first

Optional connections

This opens up

Then practise

14 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.

Spotted a problem on this page? Tell us
Series Solutions About an Ordinary Point · Mathematics, ages 20 to 21 · LightMySky