Quick Reference: Series Solutions & Euler Equations
Section 5.3 — Series Near Ordinary Points
Ordinary Point Test: The point $x_0$ is ordinary if $P(x_0) \neq 0$, where the DE is written as
$$P(x) y'' + Q(x) y' + R(x) y = 0$$
Equivalently: $p(x) = \frac{Q(x)}{P(x)}$ and $q(x) = \frac{R(x)}{P(x)}$ are both analytic at $x_0$.
General Series Solution Form:
$$y = \sum_{n=0}^{\infty} a_n (x - x_0)^n = a_0 y_1(x) + a_1 y_2(x)$$
where $y_1$ and $y_2$ are two linearly independent solutions, and $a_0, a_1$ are arbitrary constants.
Taylor Coefficient Formula:
$$a_n = \frac{\phi^{(n)}(x_0)}{n!}$$
where $\phi(x) = y(x)$ is the solution function.
Recurrence Relation Shortcut:
Isolate $y'' = \ldots$ from the DE. Differentiate repeatedly:
$y' = \sum a_n n (x-x_0)^{n-1}$
$y'' = \sum a_n n(n-1) (x-x_0)^{n-2}$
Substitute into DE, collect like powers, solve for $a_n$
Key Insight: The coefficients $a_0 = y(x_0)$ and $a_1 = y'(x_0)$ are always free (undetermined). They are the initial conditions of the IVP. All subsequent $a_2, a_3, \ldots$ follow from the recurrence relation.
Radius of Convergence: The radius $R \geq \min\{ \rho_p, \rho_q \}$, where $\rho_p$ and $\rho_q$ are the distances from $x_0$ to the nearest singularities of $p(x)$ and $q(x)$. Not infinite by default — always check for finite radius.
Problem 5.3 #1: Solve $y'' + xy' + y = 0$, expand around $x_0 = 0$.
Step 1 — Check if $x_0 = 0$ is an ordinary point
Write the DE in standard form: $P(x)y'' + Q(x)y' + R(x)y = 0$ where $P(x)=1$, $Q(x)=x$, $R(x)=1$.
Since $P(0) = 1 \neq 0$, the point $x_0 = 0$ IS an ordinary point. We can use power series.
Step 2 — Assume a power series solution
Assume $y = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$
Why? At ordinary points, the solution is guaranteed to be analytic (smooth), so it can be written as a power series. The coefficients $a_0$ and $a_1$ will be arbitrary (they come from initial conditions $y(0)$ and $y'(0)$).
Step 3 — Compute the derivatives First derivative: $y' = \sum_{n=1}^{\infty} n a_n x^{n-1} = a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3 + \cdots$
Multiply by $x$: $xy' = \sum_{n=1}^{\infty} n a_n x^n = a_1 x + 2a_2 x^2 + 3a_3 x^3 + \cdots$
Second derivative: $y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = 2a_2 + 6a_3 x + 12a_4 x^2 + 20a_5 x^3 + \cdots$
Why compute these? We need to substitute $y$, $y'$, and $y''$ into the original DE.
Step 4 — Substitute into the DE
Substitute into $y'' + xy' + y = 0$:
$$\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=0}^{\infty} a_n x^n = 0$$
Group the last two sums (they both have $x^n$ powers):
$$\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + \sum_{n=0}^{\infty} (n+1) a_n x^n = 0$$
Step 5 — Shift indices to match powers
The first sum has powers $x^{n-2}$. Let $m = n-2$, so $n = m+2$:
$$\sum_{m=0}^{\infty} (m+2)(m+1)a_{m+2} x^m + \sum_{n=0}^{\infty} (n+1) a_n x^n = 0$$
Now both sums have $x^m$ (or $x^n$ — same index variable). Combine:
$$\sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n+1)a_n] x^n = 0$$
Step 6 — Equate all coefficients to zero
For this series to equal zero for all $x$, every coefficient must be zero:
$$(n+2)(n+1)a_{n+2} + (n+1)a_n = 0 \quad \text{for } n = 0, 1, 2, 3, \ldots$$
Factor out $(n+1)$:
$$(n+1)[(n+2)a_{n+2} + a_n] = 0$$
Since $(n+1) \neq 0$ for $n \geq 0$, we have:
$$(n+2)a_{n+2} + a_n = 0$$
Step 7 — Find the recurrence relation
Solve for $a_{n+2}$:
$$a_{n+2} = -\frac{a_n}{n+2} \quad \text{for } n = 0, 1, 2, 3, \ldots$$
What does this mean? Every coefficient depends on the coefficient two steps back. So:
Even indices ($a_0, a_2, a_4, a_6, \ldots$) all depend on $a_0$
Odd indices ($a_1, a_3, a_5, a_7, \ldots$) all depend on $a_1$
Step 8 — Compute the coefficients Even sequence (starting with $a_0$):
Case I — Distinct Real Roots $r_1 \neq r_2$:
$$y = c_1 x^{r_1} + c_2 x^{r_2}$$
Use this when the discriminant $\Delta = (\alpha - 1)^2 - 4\beta > 0$.
Case II — Repeated Real Root $r_1 = r_2 = r$:
$$y = c_1 x^r + c_2 x^r \ln x$$
Use this when $\Delta = 0$. The second solution is $x^r \ln x$ (verify by substitution).
Case III — Complex Conjugate Roots $r = \lambda \pm \mu i$:
$$y = c_1 x^{\lambda} \cos(\mu \ln x) + c_2 x^{\lambda} \sin(\mu \ln x)$$
Use this when $\Delta < 0$. Extract $\lambda = \text{Re}(r)$ and $\mu = |\text{Im}(r)|$.
Singular Point Classification (Frobenius Method Extension):
For $x^2 y'' + \alpha x y' + \beta y = 0$ near $x = x_0$:
Regular singular point: If $\alpha$ and $\beta$ are constants (or analytic at $x_0$), $x = 0$ is regular singular.
Irregular singular point: If coefficients blow up faster than $(x-x_0)^{-2}$.
Use indicial equation $r^2 + (\alpha-1)r + \beta = 0$ to find exponents for series $y = (x-x_0)^r \sum a_n (x-x_0)^n$.
Domain of Validity: Solutions $y = x^r$ and variants are valid for $x > 0$ OR $x < 0$ separately. The Euler equation is singular at $x = 0$. If the original domain is $x \in (-\infty, \infty)$, solve on $x > 0$ and $x < 0$ separately, then piece together if needed.
Example 5.4 #14: Classify singular points of $x^2(1-x) y'' + 2x y' - 2y = 0$.
Singular points occur where the leading coefficient $x^2(1-x) = 0$, i.e., at $x = 0$ and $x = 1$.
At $x = 0$: Rewrite as $y'' + \frac{2}{x(1-x)} y' - \frac{2}{x^2(1-x)} y = 0$. Near $x=0$, $\frac{2}{x(1-x)} \approx \frac{2}{x} \cdot 1 = \frac{2}{x}$ (pole of order 1) and $\frac{2}{x^2(1-x)} \approx \frac{2}{x^2}$ (pole of order 2). Since the coefficient of $y$ has a pole of order $\geq 2$, $x=0$ is irregular singular.
At $x = 1$: Near $x=1$, let $u = x-1$. The coefficient $\frac{2}{x(1-x)} \approx \frac{2}{1 \cdot (-u)} = -\frac{2}{u}$ (pole of order 1) and $\frac{2}{x^2(1-x)} \approx \frac{2}{1 \cdot (-u)} = -\frac{2}{u}$ (pole of order 1). Since both coefficients have poles of order $\leq 1$, $x=1$ is regular singular.