Recipe

Solve a higher-order homogeneous equation with real roots

Use when: you have a constant-coefficient homogeneous equation $a_ny^{(n)}+\cdots+a_1y'+a_0y=0$ of order 3 or higher, and its characteristic polynomial factors entirely into real roots

Goal: the general solution $y(x)$ , built as one linear combination of $n$ independent functions, one set per root

Taught in The characteristic polynomial and its roots

Quick recipe

Substitute $y=e^{rx}$ to get the characteristic polynomial
Factor out a common power of $r$ , then find the remaining roots
Determine each root's multiplicity
Build one fundamental-set function per root, per its multiplicity
Assemble the general solution

1
Replace $y^{(k)}$ with $r^k$ for every derivative (and $y$ itself with $r^0=1$ ), giving $a_nr^n+\cdots+a_1r+a_0=0$ .

Example

For $y''''-3y''-2y'=0$ : the derivatives present are $y''''$ , $y''$ , and $y'$ , so substituting gives $r^4-3r^2-2r=0$

Why: substituting $y=e^{rx}$ makes every derivative a power of $r$ times the same never-zero exponential factor, so the equation holds exactly when this polynomial in $r$ vanishes.
2
Pull out any common power of $r$ first, then test small integers (the rational root theorem) against whatever's left.

Example

$r^4-3r^2-2r=r(r^3-3r-2)$ . Testing $r=-1$ in the cubic: $(-1)^3-3(-1)-2=-1+3-2=0$ , so $(r+1)$ divides out, leaving $r^2-r-2=(r-2)(r+1)$ . Altogether: $r(r^3-3r-2)=r(r+1)(r^2-r-2)=r(r+1)(r-2)(r+1)=r(r-2)(r+1)^2$

Why: every root of this polynomial is, by construction, a value of $r$ that makes $e^{rx}$ solve the equation, so the roots are exactly the exponents the fundamental set is built from.
3
Multiplicity is how many times a factor $(r-r_i)$ appears in the fully factored polynomial; count it directly from the factored form.

Example

From $r(r-2)(r+1)^2=0$ : $r=0$ appears once (multiplicity $1$ ), $r=2$ appears once (multiplicity $1$ ), and $r=-1$ appears twice (multiplicity $2$ ).
- a root appears once — it contributes one function to the fundamental set, in Step 4.
- a root appears more than once (multiplicity $m$ ) — it contributes $m$ functions to the fundamental set, in Step 4.
Why: a root's multiplicity is exactly how many independent solutions that one root is responsible for — the fundamental set needs $n$ functions total to match the equation's order.
4
A simple root $r$ contributes $e^{rx}$ . A root repeated $m$ times contributes $e^{rx}, xe^{rx}, \ldots, x^{m-1}e^{rx}$ — $m$ independent functions from that one root.

Example

$r=0$ (multiplicity $1$ ) contributes $e^{0x}=1$ . $r=2$ (multiplicity $1$ ) contributes $e^{2x}$ . $r=-1$ (multiplicity $2$ ) contributes $e^{-x}$ and $xe^{-x}$ .

Why: this guarantees exactly $n$ linearly independent solutions in total — one for each root counted with multiplicity — matching the order of the equation.
5
Write $y$ as $C_1,\ldots,C_n$ times each function found in Step 4, added together.

Example

$y = C_1(1) + C_2e^{-x} + C_3xe^{-x} + C_4e^{2x}$

Why: superposition guarantees any linear combination of solutions to a linear homogeneous equation is itself a solution, and $n$ independent ones span every solution there is.

Common mistakes

Forgetting to pull out a common factor of $r$ before hunting for rational roots, and missing $r=0$ as a root entirely.
Reporting a repeated root's multiplicity as $1$ , and so missing the extra $xe^{rx}$ (or $x^2e^{rx}$ , and so on) factor it contributes.
Writing $x^{m-1}e^{rx}$ with the wrong power of $x$ — the exponents run $0,1,\ldots,m-1$ , so a multiplicity- $2$ root contributes exactly $e^{rx}$ and $xe^{rx}$ , not $xe^{rx}$ and $x^2e^{rx}$ .
Stopping once the polynomial is factored without actually assembling the $n$ -term general solution.