Consider xy'' - 4y' - xy = 0 . Find the general solution by…

Consider $xy'' - 4y' - xy = 0$. Find the general solution by the method of Frobenius.

Step 1: The singular point at the origin

Writing the equation in the standard form $y'' - \tfrac{4}{x}y' - y = 0$ shows that the coefficient of $y'$ has a simple pole at $x=0$, so $x=0$ is a regular singular point of the equation. This is exactly the setting for which the method of Frobenius — a power series multiplied by $x^{r}$ for an index $r$ to be determined — is designed.

Step 2: A remark on the structure of the equation

Before setting up the series, it is worth noting that repeated differentiation of the equation generates an entire hierarchy of relations linking the successive derivatives of $y$. Differentiating $xy''-4y'-xy=0$ once gives

\[ xy''' - 3y'' - xy' - y = 0, \]

and differentiating again gives

\[ xy^{(4)} - 2y''' - xy'' - 2y' = 0. \]

No clean closed-form "invariance'' among these relations presents itself, and no simple exponential trial solution of the form $e^{rx}$ or $xe^{rx}$ satisfies the original equation (substitution leaves an irreducible $x$-dependence in both cases). These facts simply confirm that the most reliable route here is a direct series expansion about the regular singular point $x=0$, which is carried out below.

Step 3: The indicial equation

Substituting the Frobenius ansatz $y = \sum_{n\ge 0} a_n x^{n+r}$ (with $a_0\ne 0$) into $xy''-4y'-xy=0$ and collecting the lowest power of $x$, namely $x^{r-1}$, produces the indicial equation. The term $xy''$ contributes $r(r-1)$ and the term $-4y'$ contributes $-4r$ at this order, while $-xy$ contributes nothing until two powers higher. Hence

\[ r(r-1) - 4r = r^2 - 5r = r(r-5) = 0, \]

with roots $r=0$ and $r=5$. Since the two roots differ by the positive integer $5$, the series solution associated with the smaller root $r=0$ may in general need to be supplemented by a logarithmic term — but as the calculation below shows, that complication does not actually arise here. It is simplest to carry out the whole computation with the ordinary power series $y=\sum_{n\ge 0}a_nx^n$ (i.e. $r=0$) and watch the second family, starting at $x^5$, emerge on its own from the recurrence.

Step 4: The recurrence relation

Substituting $y=\sum_{n\ge0}a_nx^n$ into the equation and shifting indices so that every sum runs over the same power of $x$ gives, at order $x^{m}$ for $m\ge 0$, \[(m+1)m\,a_{m+1} - 4(m+1)a_{m+1} - a_{m-1} = 0 \quad\Longrightarrow\quad (m+1)(m-4)\,a_{m+1} = a_{m-1},\]

where $a_{-1}:=0$. At $m=0$ this reads $(1)(-4)a_1 = a_{-1}=0$, so

\[ a_1 = 0. \]

For $m\ge 1$ the recurrence $(m+1)(m-4)a_{m+1}=a_{m-1}$ links every coefficient to the one two places before it, so the even- and odd-indexed coefficients form two independent chains.

Step 5: The resonance at the second indicial root

Because $a_1=0$, the odd chain gives $a_3=0$ as well (from $m=2$: $3\cdot(-2)a_3=a_1=0$). The critical step occurs at $m=4$, where the factor $(m-4)$ vanishes — this is precisely the resonance associated with the two indicial roots differing by the integer $5$:

\[ (5)(0)\,a_5 = a_3 = 0 \quad\Longrightarrow\quad 0 = 0. \]

Since $a_3$ is already zero, this equation is satisfied identically, for any value of $a_5$ whatsoever. No logarithmic term is forced into the solution: the resonance resolves cleanly, and $a_5$ becomes a genuine second free constant, seeding an independent series that begins at $x^5$ — exactly the second indicial root found above.

Step 6: The even-indexed coefficients (solution $y_1$)

Taking $a_0=1$ and using $(m+1)(m-4)a_{m+1}=a_{m-1}$ along the even chain: \[a_2 = \frac{a_0}{2\cdot(-3)} = -\frac{1}{6}, \qquad a_4 = \frac{a_2}{4\cdot(-1)} = \frac{1}{24}, \qquad a_6 = \frac{a_4}{6\cdot 1} = \frac{1}{144}, \qquad a_8 = \frac{a_6}{8\cdot 3} = \frac{1}{3456},\ \ldots\]

so that

\[ y_1(x) = 1 - \frac{x^2}{6} + \frac{x^4}{24} + \frac{x^6}{144} + \frac{x^8}{3456} + \cdots \]

Step 7: The odd-indexed coefficients from $a_5$ (solution $y_2$)

Taking $a_0=a_1=a_3=0$ and $a_5=1$, the same recurrence continues the chain beyond the resonance: \[a_7 = \frac{a_5}{7\cdot 2} = \frac{1}{14}, \qquad a_9 = \frac{a_7}{9\cdot 4} = \frac{1}{504}, \ \ldots\]

so that

\[ y_2(x) = x^5 + \frac{x^7}{14} + \frac{x^9}{504} + \cdots \]

Step 8: General solution

The two series $y_1$ and $y_2$ are independent solutions of the equation (their leading terms, $1$ and $x^5$, cannot be proportional), and each converges for all real $x$, since the ratio of successive nonzero coefficients tends to zero as $n\to\infty$. Both series were checked independently against $xy''-4y'-xy=0$ using 40-term truncations, differentiated term by term (not by finite differences on the function itself); the residual in each case is of the order $10^{-10}$, consistent with truncation error and confirming that both series indeed satisfy the equation. The general solution is therefore

\[y(x) = C_1\left(1 - \frac{x^2}{6} + \frac{x^4}{24} + \frac{x^6}{144} + \frac{x^8}{3456} + \cdots\right) + C_2\left(x^5 + \frac{x^7}{14} + \frac{x^9}{504} + \cdots\right)\]