Characteristic Equation (Auxiliary Equation)

Why is it called the "characteristic equation"?

For a constant-coefficient linear ODE $a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0,$ the guess $y = e^{r x}$ reduces every derivative to a power of $r$, so the ODE collapses to $a_n r^{n} + a_{n-1} r^{n-1} + \cdots + a_1 r + a_0 = 0.$

This polynomial in $r$ is called the characteristic equation (also the auxiliary equation) because its roots characterise every solution of the ODE — they tell you what exponentials to build out of, and they determine every qualitative property (stability, oscillation frequency, damping).

The given equation

$y'' + 4 y' + 4 y = 0$

$a = 1$, $b = 4$, $c = 4$.

Step-by-step solution

Step 1 — Form the characteristic equation.

Replace $y'' \to r^2$, $y' \to r$, $y \to 1$: $r^{2} + 4 r + 4 = 0$

Step 2 — Solve it.

This factors as a perfect square: $(r + 2)^{2} = 0 \implies r = -2 \; \text{(repeated, multiplicity 2)}$

The discriminant is $\Delta = 16 - 16 = 0$ — exactly on the boundary between real-distinct and complex cases.

Step 3 — Build the general solution.

With a repeated root we can't just write $C_1 e^{-2x} + C_2 e^{-2x}$: that's $(C_1 + C_2) e^{-2x}$, which has only one independent constant. We need a second linearly independent solution. The rule for a repeated root of multiplicity $m$ is: $y_1 = e^{r x}, \quad y_2 = x \, e^{r x}, \quad y_3 = x^2 \, e^{r x}, \quad \ldots, \quad y_m = x^{m-1} \, e^{r x}.$

For multiplicity 2 that gives $\boxed{\, y(x) = (C_1 + C_2 \, x) \, e^{-2 x} \,}$

See #185 for the full derivation of why $x \, e^{r x}$ is the second solution (hint: reduction of order, or an $\epsilon$-limit of the distinct-root formula).

Verification

Let $y = (C_1 + C_2 x) e^{-2x}$. $y' = C_2 e^{-2x} + (C_1 + C_2 x)(-2) e^{-2x} = e^{-2x}\bigl(C_2 - 2 C_1 - 2 C_2 x\bigr)$ $y'' = -2 e^{-2x}\bigl(C_2 - 2 C_1 - 2 C_2 x\bigr) + e^{-2x}\bigl(-2 C_2\bigr)$ $\;\;= e^{-2x}\bigl(-2 C_2 + 4 C_1 + 4 C_2 x - 2 C_2\bigr) = e^{-2x}\bigl(4 C_1 - 4 C_2 + 4 C_2 x\bigr)$

Plug $y'' + 4 y' + 4 y$: $e^{-2x}\bigl[(4 C_1 - 4 C_2 + 4 C_2 x) + 4(C_2 - 2 C_1 - 2 C_2 x) + 4(C_1 + C_2 x)\bigr]$

Collect:

• $x$ coefficient (inside brackets): $4 C_2 - 8 C_2 + 4 C_2 = 0$ ✓
• constant: $4 C_1 - 4 C_2 + 4 C_2 - 8 C_1 + 4 C_1 = 0$ ✓

Whole bracket is $0$. $\checkmark$

How to read the characteristic polynomial

Roots tell you everything:

• Real roots $r_1, r_2, \ldots$ (distinct): each contributes an exponential $e^{r x}$.
• Repeated real root $r$ of multiplicity $m$: contributes $e^{r x}, \, x e^{r x}, \, \ldots, \, x^{m-1} e^{r x}$.
• Complex conjugate pair $\alpha \pm i \beta$: contributes $e^{\alpha x} \cos \beta x$ and $e^{\alpha x} \sin \beta x$ (one oscillating mode; the imaginary part $\beta$ is the frequency, the real part $\alpha$ is the damping rate). (See #184.)
• Repeated complex pair (higher-order ODEs): multiply by $x, x^2, \ldots$ as for repeated real roots.

The number of solutions always equals the ODE order, matching the fundamental theorem of algebra's count of roots with multiplicity.

Stability by sign of the roots

• All roots have negative real part ⇒ solutions decay to $0$ ⇒ asymptotically stable.
• Any root with positive real part ⇒ some solutions blow up ⇒ unstable.
• Roots on the imaginary axis with $\alpha = 0$ ⇒ neutrally stable (oscillations that neither grow nor decay).

For our equation: $r = -2$ (repeated, $\alpha = -2 < 0$) ⇒ asymptotically stable. Every solution decays exponentially to zero.

Initial value problem

With $y(0) = 1$, $y'(0) = 0$: from $y = (C_1 + C_2 x) e^{-2x}$, $y(0) = C_1 = 1$. And $y'(0) = C_2 - 2 C_1 = 0 \implies C_2 = 2$. $y(x) = (1 + 2 x) e^{-2 x}.$

Check: $y(0) = 1$ ✓. $y'(0) = 2 - 2 = 0$ ✓. As $x \to \infty$, the polynomial factor $1 + 2x$ is dominated by the exponential decay $e^{-2x}$, so $y \to 0$.

Physical interpretation — critical damping

A spring-mass-damper satisfies $m y'' + c y' + k y = 0$. The characteristic equation is $m r^2 + c r + k = 0$. The critically damped case is exactly $\Delta = c^2 - 4 m k = 0$ (repeated real root), and the solution $(C_1 + C_2 x) e^{r x}$ is the fastest non-oscillating return to equilibrium. One step less damping → oscillation; one step more → slow sluggish crawl.

Our equation $y'' + 4 y' + 4 y = 0$ models $m = 1, c = 4, k = 4$, which is exactly critically damped.

Common mistakes

• Writing $y = C_1 e^{r x} + C_2 e^{r x}$ for a repeated root. This has only one constant in disguise — it is $(C_1 + C_2) e^{rx}$, a one-parameter family, not a two-parameter one.
• Missing the $x$ factor in the second solution. Without $x e^{r x}$, you can't satisfy arbitrary initial conditions at a repeated root.
• Forgetting multiplicity. A triple root $r$ contributes $e^{r x}, x e^{r x}, x^2 e^{r x}$, not just the first two.
• Confusing discriminant signs. $\Delta = 0$ is repeated real; $\Delta < 0$ is complex. Don't swap them.

Try it in the visualization

Slide the coefficient $b$ through $4$, the critical-damping value. Watch the two roots collide on the real axis (becoming a double root), and see the solution shape morph smoothly from over-damped (two exponentials) through critical ($x e^{r x}$) to under-damped (oscillations).