A more direct approach to series solutions

In the previous post we found a solution to

y'' + 7y' + 12 = x^2.

using operator calculus, i.e. treating the differential operator D like a number and doing tricks with it. See the earlier post for a justification of why we can get away with unorthodox manipulations.

We can generalize the method of the previous post to say that a solution to any differential equation of the form

p(D) y = f(x)

is given by

y = \frac{1}{p(D)} f(x).

In the previous post we had

 p(D) = D^2 + 7D + 12


f(x) = x^2

but the method works more generally.

We then find a power series for 1/p(D), most likely by partial fraction decomposition, and apply the result to f(x). There may be a fair amount of labor left, but it’s purely calculus; all the differential equation work is done.

Conceptually, this method subsumes other differential equation techniques such as undetermined coefficients and power series solutions.

Let’s use this method to find an approximate solution to

y'' + 7y' + 12 = \zeta(x)

where ζ(x) is the Riemann zeta function.

In the previous post we worked out that

 \frac{1}{D^2 + 7D + 12} = \frac{1}{D+3} - \frac{1}{D+4} = \frac{1}{12} - \frac{7}{144}D + \frac{37}{1728}D^2 + {\cal O}(D^3)

and so an approximate solution to our differential equation near 0 is

 y(x) = \frac{1}{12}\zeta(0) - \frac{7}{144} \zeta'(0) + \frac{37}{1728}\zeta''(0) + {\cal O}(x^3).

Numerically this works out to

y(x) = - 0.0416667 + 0.0446706 x - 0.0429602 x^2.

If you want more terms, carry the series for 1/(D² + 7D + 12) out to more terms.

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5 thoughts on “A more direct approach to series solutions

  1. OK, this is so simple, it’s like stealing. Mathematica makes the whole process dead easy, e.g. Series[Apart[1/(D^2 + 7 D + 12)], {D, 0, 2}] does all the heavy lifting.

  2. You can also expand the series for 1/p(D) using long division, since p(D) is polynomial.

    (Typo? You have in the second-to-last formula y(z) = … but no z’s in the right-hand side. Maybe you meant y(x) and meant to include some powers of x next to the coefficients.)

  3. Thanks Michael. I automatically think “z” when writing power series but “x” or “t” when doing differential equations. Worlds collide. :)

  4. Actually, I didn’t read carefully enough (sorry about that). The operator solution would be y(x) = (1/12 – 7/144 D + 37/1728 D^2 + …) ζ(x), but I doubt that series converges.

    In any case, I get a different solution, namely, y = -0.0416667 + 0.0446706 x – 0.156347 x^2, to the IVP with y(0) = ζ(0)/12, y'(0) = -7 ζ'(0)/144. Mathematica code:

    AsymptoticDSolveValue[{y”[x] + 7 y'[x] + 12 y[x] == Zeta[x],
    y[0] == 1/12 Zeta[0], y'[0] == -7/144 Zeta'[0]},
    y[x], {x, 0, 2}] // N

  5. There must be some conditions under which such simple method works. As Michael commented, the series might be conditionally convergent only.

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