Second-Order ODE with Repeated Roots

Why the repeated-root case is tricky

For two distinct roots $r_1 \ne r_2$ the general solution is $y = C_1 e^{r_1 x} + C_2 e^{r_2 x}.$ Each exponential is linearly independent — $\{e^{r_1 x}, e^{r_2 x}\}$ spans a 2D solution space.

When $r_1 = r_2 = r$, writing "$y = C_1 e^{r x} + C_2 e^{r x}$" collapses to $(C_1 + C_2) e^{r x}$ — a one-parameter family. We have lost a dimension of the solution space, and the ODE is still order 2, so there must be a second linearly independent solution we're missing. The question is how to find it.

Answer: $y_2(x) = x \, e^{r x}$. We'll derive this and verify it.

The given equation

$y'' - 6 y' + 9 y = 0$

Step 1 — Characteristic equation. $r^{2} - 6 r + 9 = 0 \implies (r - 3)^{2} = 0 \implies r = 3 \; \text{(double root)}$

Discriminant $\Delta = 36 - 36 = 0$. The two roots coincide.

Step 2 — Take $y_1 = e^{3 x}$ and find a second solution.

Derivation 1 — Reduction of order

Assume $y_2 = u(x) \, e^{3 x}$ for some unknown function $u$. (If $u$ is constant, we'd get a multiple of $y_1$; we need $u$ to be genuinely new.) Compute: $y_2' = u' e^{3x} + 3 u e^{3x} = (u' + 3 u) e^{3x}$ $y_2'' = (u'' + 3 u') e^{3x} + 3 (u' + 3 u) e^{3x} = (u'' + 6 u' + 9 u) e^{3x}$

Plug into $y'' - 6 y' + 9 y$: $e^{3x} \bigl[(u'' + 6 u' + 9 u) - 6 (u' + 3 u) + 9 u\bigr] = e^{3x}\bigl[u'' + 6 u' + 9 u - 6 u' - 18 u + 9 u\bigr]$ $= e^{3x} \, u''$

Setting this to $0$: $u'' = 0$, so $u(x) = a + b x$ for constants $a, b$.

Therefore $y_2 = (a + b x) e^{3 x}$. The $a \, e^{3x}$ piece is just a multiple of $y_1$, so the genuinely new part is $b \, x \, e^{3x}$. Take $y_2 = x \, e^{3 x}$. ✓

This is the reduction-of-order derivation. See #199 for the general method.

Derivation 2 — Limit from distinct roots

Perturb the ODE to have distinct roots $r$ and $r + \varepsilon$ ($\varepsilon \ne 0$), solution $y_\varepsilon = C_1 e^{r x} + C_2 e^{(r + \varepsilon) x}$. Choose constants smartly: $\frac{e^{(r + \varepsilon) x} - e^{r x}}{\varepsilon} \xrightarrow{\varepsilon \to 0} \frac{d}{d r} e^{r x} = x \, e^{r x}.$

So $x \, e^{r x}$ arises as the derivative with respect to $r$ of $e^{r x}$, and this recipe generalises: a root of multiplicity $m$ contributes $\{e^{rx}, x e^{rx}, x^2 e^{rx}, \ldots, x^{m-1} e^{rx}\}$ as a limit of nearby distinct roots.

Same answer, different viewpoint.

Step 3 — General solution. $\boxed{\, y(x) = (C_1 + C_2 \, x) \, e^{3 x} \,}$

Verification

$y = (C_1 + C_2 x) e^{3x}$ $y' = C_2 e^{3x} + 3 (C_1 + C_2 x) e^{3x} = e^{3x}(3 C_1 + C_2 + 3 C_2 x)$ $y'' = 3 e^{3x}(3 C_1 + C_2 + 3 C_2 x) + e^{3x}(3 C_2) = e^{3x}(9 C_1 + 6 C_2 + 9 C_2 x)$

Plug: $y'' - 6 y' + 9 y = e^{3x}\bigl[(9 C_1 + 6 C_2 + 9 C_2 x) - 6(3 C_1 + C_2 + 3 C_2 x) + 9 (C_1 + C_2 x)\bigr]$

• $x$-coefficient: $9 C_2 - 18 C_2 + 9 C_2 = 0$ ✓
• constant: $9 C_1 + 6 C_2 - 18 C_1 - 6 C_2 + 9 C_1 = 0$ ✓

So $y'' - 6 y' + 9 y = 0$. $\checkmark$

Initial value problem

$y(0) = 1$, $y'(0) = 5$.

$y(0) = C_1 = 1$. $y'(0) = 3 C_1 + C_2 = 5 \implies C_2 = 5 - 3 = 2$. $y(x) = (1 + 2 x) e^{3 x}.$

Long-term: both $x$ and $e^{3x}$ grow, so $y \to \infty$ exponentially. The linear factor $1 + 2x$ adds a mild overshoot to the fundamental exponential growth.

Physical interpretation — critical damping (revisited)

Spring-mass-damper $m y'' + c y' + k y = 0$ with $c^{2} = 4 m k$ (exactly critical) has a repeated root and solutions $(C_1 + C_2 t) e^{-\gamma t}$ with $\gamma = c/(2m)$. The $t \, e^{-\gamma t}$ term is the signature of critical damping — the system can cross the equilibrium once (the polynomial has one root) then decays. Any less damping and it oscillates; any more and it sluggishly drags to zero.

Higher multiplicity

A triple root $r$ contributes $\{e^{rx}, x e^{rx}, x^{2} e^{rx}\}$. A quadruple root contributes $\{e^{rx}, x e^{rx}, x^{2} e^{rx}, x^{3} e^{rx}\}$. The rule is clean: multiplicity $m$ → first $m$ polynomial multiples of $e^{rx}$.

This extends to complex repeated roots: a double pair $\alpha \pm i \beta$ contributes $\{e^{\alpha x} \cos \beta x, e^{\alpha x} \sin \beta x, x e^{\alpha x} \cos \beta x, x e^{\alpha x} \sin \beta x\}$ — four independent solutions.

Common mistakes

• Writing $y = C_1 e^{rx} + C_2 e^{rx}$ for a double root. That's a one-parameter family and fails to satisfy arbitrary initial conditions.
• Including $x^{m} e^{rx}$ for multiplicity $m$ (one too many). The rule is powers $0$ through $m - 1$, not up to $m$.
• Forgetting that $x e^{rx}$ is a separate solution. When applying ICs, treat $C_1 e^{rx}$ and $C_2 x e^{rx}$ as independent basis vectors, even though they share the exponential factor.

Try it in the visualization

Start with a distinct-root ODE and slide the discriminant to zero. Watch the two exponentials merge, and see the $x \, e^{r x}$ solution emerge as the limiting shape. Highlight how at $\Delta = 0$ exactly, the solution family is two-dimensional precisely because of the polynomial factor $x$.