7.4 Series Solutions of Differential Equations

Big idea

Constant coefficients were a gift. The characteristic equation turned a differential equation into a quadratic, and exponentials did the rest. Put a variable coefficient in front of $y'$ or $y$ and that machinery stops: $y'' + xy = 0$ has no solution built from exponentials, sines and polynomials, yet it is an ordinary equation describing real physical systems.

The response is to widen the search. Instead of guessing a specific function, guess an infinite polynomial: a power series with unknown coefficients $a_0, a_1, a_2, \dots$, and let the equation decide what they must be. Differentiating term by term is legal inside the interval of convergence, so substitution produces a series set equal to zero, and a power series is zero only when every coefficient is zero. That single fact converts a differential equation into infinitely many algebraic equations, one per power of $x$.

Those equations turn out to be tidy. Each one relates a coefficient to an earlier coefficient, producing a recurrence relation such as $a_{k+2} = -a_k/((k+2)(k+1))$. Two coefficients, $a_0$ and $a_1$, are never determined by the equation. They are exactly the two free constants a second-order equation must have, and they are the values of $y(0)$ and $y'(0)$. Setting $a_0 = 1, a_1 = 0$ and then $a_0 = 0, a_1 = 1$ produces the two independent solutions.

One step in this process is where nearly everyone loses the thread: the powers of $x$ coming out of $y''$, $y'$ and $y$ do not line up, so the sums cannot be combined until the indices are shifted. The shift is bookkeeping, not mathematics, but it has to be exact. Done slowly and checked against the first two terms, it becomes routine.

Decoder

If the leading coefficient of a linear second-order equation is nonzero at a point, and all coefficients are analytic there, then the equation has a power series solution centered at that point, and the series converges at least out to the nearest point where the leading coefficient vanishes.

The condition is about the leading coefficient only, and it is checked before any computation. For $P(x)y'' + Q(x)y' + R(x)y = 0$ with polynomial coefficients, a point $a$ is ordinary when $P(a) \ne 0$ and singular when $P(a) = 0$. Series methods centered at an ordinary point are guaranteed to work; centered at a singular point they need heavier tools.

The second half gives the radius for free. For $(x^2 - 4)y'' + xy' + y = 0$ the leading coefficient vanishes at $x = \pm 2$, so a series centered at $0$ converges at least for $|x| < 2$ - an estimate obtained without summing anything.

The classic mistake happens one line into the substitution: combining $\sum_{n \ge 2} n(n-1)a_n x^{n-2}$ with $\sum_{n \ge 0} a_n x^n$ as though matching $n$ means matching powers. It does not. The first sum contributes $x^{n-2}$ and the second contributes $x^n$, so the coefficient of $x^k$ in one comes from $a_{k+2}$ and in the other from $a_k$.

Definitions and results

Ordinary and singular points. For $P(x)y'' + Q(x)y' + R(x)y = 0$, a point $a$ is ordinary if $P(a) \ne 0$ and $Q/P$, $R/P$ are analytic there; otherwise it is singular. With polynomial coefficients the singular points are just the roots of $P$.

The assumed form. At an ordinary point $a$, write

$$ y = \sum_{n=0}^{\infty} a_n (x - a)^n $$

Every example below is centered at $a = 0$, so the series is $\sum a_n x^n$. The coefficients are unknowns, and $a_0 = y(0)$, $a_1 = y'(0)$.

Term-by-term differentiation. Inside the interval of convergence,

$$ y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \qquad y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} $$

The starting index rises each time because differentiating kills the constant term.

Shifting the index. To read the coefficient of $x^k$ out of $y''$, substitute $n = k + 2$. The condition $n \ge 2$ becomes $k \ge 0$, and

$$ y'' = \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2}\, x^{k} $$

Check the first terms against the original: at $n = 2$ the old sum gives $2a_2$, and at $k = 0$ the new one gives $(2)(1)a_2$; at $n = 3$ the old gives $6a_3 x$, and at $k = 1$ the new gives $(3)(2)a_3 x$. A shift never changes the sum, only how it is addressed.

Multiplying by a power of $x$. A term like $xy'$ becomes $\sum_{n \ge 1} n a_n x^{n}$, already in powers $x^k$ with $k = n$. Since the $k = 0$ term would be $0 \cdot a_0$, the sum may start at $k = 0$ at no cost, which helps when combining it with the others.

Matching coefficients. Once every sum runs over the same index and the same power, add them into a single series and set the coefficient of $x^k$ to zero for each $k \ge 0$. This is valid because a power series that vanishes on an interval has every coefficient equal to zero.

The recurrence relation. The resulting equation expresses a later coefficient in terms of earlier ones. Solve it for the highest-index coefficient and iterate from $a_0$ and $a_1$. A relation linking $a_{k+2}$ to $a_k$ splits into two independent chains, one over even indices and one over odd.

The two independent solutions. Collect the terms containing $a_0$ into one series and those containing $a_1$ into another, so $y = a_0 y_1(x) + a_1 y_2(x)$. Since $y_1(0) = 1$ with $y_1'(0) = 0$ and $y_2(0) = 0$ with $y_2'(0) = 1$, their Wronskian at $0$ equals $1$ and the pair is independent.

Worked examples

Locating the singular points

For $(x^2 - 9)y'' + 2xy' - y = 0$ the leading coefficient vanishes at $x = \pm 3$, so those are the singular points and every other real number is ordinary. A series centered at $0$ converges at least on $|x| < 3$, and one centered at $1$ at least on $|x - 1| < 2$, the distance from $1$ to the nearer bad point.

Recovering the sine and cosine

Solve $y'' + y = 0$ with a power series about $0$. The point is ordinary, since the leading coefficient is the constant $1$. Substituting $y = \sum_{n \ge 0} a_n x^n$,

$$ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + \sum_{n=0}^{\infty} a_n x^n = 0 $$

The powers do not match, so shift the first sum with $n = k+2$ and rename the second index to $k$:

$$ \sum_{k=0}^{\infty} (k+2)(k+1) a_{k+2} x^{k} + \sum_{k=0}^{\infty} a_k x^k = \sum_{k=0}^{\infty}\left[(k+2)(k+1)a_{k+2} + a_k\right]x^k = 0 $$

Every bracket must vanish, so

$$ a_{k+2} = \frac{-a_k}{(k+2)(k+1)} $$

Run the even chain from $a_0$: $a_2 = -a_0/2$, then $a_4 = -a_2/12 = a_0/24$. Run the odd chain from $a_1$: $a_3 = -a_1/6$, then $a_5 = -a_3/20 = a_1/120$. Assembling,

$$ y = a_0\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right) + a_1\left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) $$

The two series are the Maclaurin series of $\cos x$ and $\sin x$, which is what the characteristic equation gives, so the method checks out on a case whose answer you already know.

A variable coefficient

Solve $y'' - 2xy' - 2y = 0$ about $0$. The second derivative shifts as before, $2xy' = \sum_{n \ge 1} 2n a_n x^n = \sum_{k \ge 0} 2k a_k x^k$, and $2y = \sum_{k \ge 0} 2a_k x^k$. Combining,

$$ \sum_{k=0}^{\infty}\left[(k+2)(k+1)a_{k+2} - 2k a_k - 2a_k\right]x^k = 0 $$

Since $2k a_k + 2a_k = 2(k+1)a_k$, the bracket gives $(k+2)(k+1)a_{k+2} = 2(k+1)a_k$, and one factor of $k+1$ cancels:

$$ a_{k+2} = \frac{2a_k}{k+2} $$

From $a_0$: $a_2 = a_0$, $a_4 = a_2/2 = a_0/2$, $a_6 = a_4/3 = a_0/6$. That is $a_0\sum_{m \ge 0} x^{2m}/m!$, which is $a_0 e^{x^2}$. Confirm directly: if $y = e^{x^2}$ then $y' = 2xe^{x^2}$ and $y'' = (2 + 4x^2)e^{x^2}$, so $y'' - 2xy' - 2y = (2 + 4x^2 - 4x^2 - 2)e^{x^2} = 0$.

From $a_1$: $a_3 = 2a_1/3$, $a_5 = 4a_1/15$, $a_7 = 8a_1/105$, so the second solution is $x + \frac{2x^3}{3} + \frac{4x^5}{15} + \cdots$, with no elementary closed form. That is the usual situation; the first solution was the lucky one.

An initial value problem

Solve $y'' + xy' + y = 0$ with $y(0) = 1$ and $y'(0) = 0$. The initial conditions set $a_0 = 1$ and $a_1 = 0$ immediately. Shifting as before, $xy' = \sum_{k \ge 0} k a_k x^k$ and $y = \sum_{k \ge 0} a_k x^k$, so the coefficient of $x^k$ is

$$ (k+2)(k+1)a_{k+2} + k a_k + a_k = (k+2)(k+1)a_{k+2} + (k+1)a_k = 0 $$

Cancelling $k+1$ leaves $a_{k+2} = -a_k/(k+2)$. Since $a_1 = 0$, every odd coefficient is zero. The even ones are $a_2 = -1/2$, $a_4 = -a_2/4 = 1/8$, and $a_6 = -a_4/6 = -1/48$, so

$$ y = 1 - \frac{x^2}{2} + \frac{x^4}{8} - \frac{x^6}{48} + \cdots $$

The pattern $a_{2m} = (-1)^m/(2^m m!)$ identifies the sum as $e^{-x^2/2}$. Check: that function has $y' = -xe^{-x^2/2}$ and $y'' = (x^2 - 1)e^{-x^2/2}$, so $y'' + xy' + y = (x^2 - 1 - x^2 + 1)e^{-x^2/2} = 0$, and $y(0) = 1$, $y'(0) = 0$.

Practice

These problems run the whole procedure: confirm the center is an ordinary point, substitute the series, shift indices until every sum is a multiple of $x^k$, read off the recurrence, and generate enough coefficients to show both solutions. Write the shift out explicitly, and check its first two terms against the unshifted sum before moving on.

Quiz

Five items on ordinary and singular points, index shifting, recurrence relations, and the two independent solutions.