Differential-Equations · Unit 12 · Video 3 · Interactive Practice

Stability: Characteristic Roots with Negative Real Parts

IKey Formulas

Formula	Name	What it says
$y'' + A\,y' + B\,y = f(t)$ is stable $\iff c_1y_1 + c_2y_2 \to 0$ for all $c_1, c_2$	Stability	Then $y_c$ is the transient and $y_p$ the steady-state solution
$r_1 \neq r_2$ real: $\ r_1, r_2 < 0$ ; $\quad r_1 = r_2$ : $\ r_1 < 0$	Real roots	$e^{rt}$ and $t\,e^{rt}$ die away exactly when $r < 0$
$r = a \pm bi$ : $\ e^{at}(c_1\cos bt + c_2\sin bt) \to 0 \iff a < 0$	Complex roots	The oscillation repeats; only $e^{at}$ decides
stable $\iff$ every root of $r^2 + Ar + B = 0$ has $\operatorname{Re} r < 0$	Stability criterion	One sentence for all three cases, whatever $f(t)$ is

Key Insight: Stability is a property of the left side alone. For a stable equation, every solution is the steady state $y_p$ plus a transient that dies away, so in the long run the initial conditions stop mattering.

IIVisualization 1 — Do the Starting Values Fade?

Five solutions with the same input $10\cos t$ start from $y(0) = -3, -1, 1, 3, 5$ with slope $0$ .

Equation t 12

IIIVisualization 2 — Where the Roots Sit Decides Stability

Each characteristic root contributes a basic solution; the equation is stable exactly when every root lies left of the imaginary axis.

Roots

IVQuiz Questions

Problem 1 · Which Equation Is Stable?

Given: the four equations below — which one is stable?

✅ Correct!

r^2 + 5r + 6 = (r + 2)(r + 3)

: both roots,

-2

and

-3

, are negative. The input

\sin t

plays no part.

❌ Not quite.

r^2 - r - 6 = (r - 3)(r + 2)

has the root

3 > 0

, so

c_1e^{3t}

grows.

❌ Not quite. The roots are

\pm 2i

with real part

0

c_1\cos 2t + c_2\sin 2t

oscillates forever without dying away.

❌ Not quite.

r^2 + 3r = r(r + 3)

has the root

0

, whose solution

c_1

is a constant that does not go to

0

Show solution

Only the reduced equation matters. Roots of each characteristic polynomial:

(a) $r^2 - r - 6 = 0$ : $r = 3, -2$ . Not stable. (b) $r^2 + 4 = 0$ : $r = \pm 2i$ , real part $0$ . Not stable. (c) $r^2 + 5r + 6 = 0$ : $r = -2, -3$ . Stable. (d) $r^2 + 3r = 0$ : $r = 0, -3$ . Not stable.

Problem 2 · A Solution That Rises First

Given: $y'' + 2y' + y = 0$ has the solution $t\,e^{-t}$ , which rises from $0$ to $1/e$ at $t = 1$ — is the equation stable?

✅ Correct! Every solution is

(c_1 + c_2t)e^{-t}

, and the exponential beats the linear factor: all of them go to

0

❌ Not quite. Stability is about

t \to \infty

, not the start. After its peak,

t\,e^{-t}

decays to

0

❌ Not quite. A repeated root

r_1

gives

(c_1 + c_2t)e^{r_1t}

, which goes to

0

whenever

r_1 < 0

. Only

r_1 = 0

(linear solutions) fails.

❌ Not quite.

t\,e^{-t} \to 0

as well, so

c_2 \neq 0

is fine.

Show solution

$r^2 + 2r + 1 = (r + 1)^2$ , so $r_1 = r_2 = -1$ and $y = (c_1 + c_2t)\,e^{-t}$ .

\lim_{t\to\infty} t\,e^{-t} = \lim_{t\to\infty}\frac{t}{e^{t}} = 0,

so every solution goes to $0$ : the equation is stable, even though some solutions rise before they decay.

Problem 3 · The Long-Run Value

Given: $y'' + 2y' + 5y = 10$ — find the limit of every solution as $t \to \infty$ .

✅ Correct! The roots

-1 \pm 2i

have real part

-1 < 0

, so the transient dies and every solution tends to

y_p = 2

❌ Not quite. The steady state solves the equation: a constant

y_p

needs

5y_p = 10

, so

y_p = 2

, not

10

❌ Not quite. The transient

y_c

goes to

0

, but the steady state

y_p = 2

remains.

❌ Not here. The equation is stable (roots

-1 \pm 2i

), so the initial values only affect the transient, which dies away.

Show solution

Reduced equation: $r^2 + 2r + 5 = 0$ , $r = -1 \pm 2i$ . Real part $-1 < 0$ : stable.

A constant particular solution: $0 + 0 + 5y_p = 10$ , so $y_p = 2$ .

y = 2 + e^{-t}(c_1\cos 2t + c_2\sin 2t) \ \to\ 2.

Problem 4 · A Third-Order Equation

Given: $y''' + y'' + y' + y = 0$ , whose characteristic polynomial factors as $(r + 1)(r^2 + 1)$ — is it stable?

✅ Correct!

c_2\cos t + c_3\sin t

never dies away, so not every solution goes to

0

❌ Not quite. Every root must have negative real part. The roots

\pm i

give

\cos t

and

\sin t

, which keep oscillating.

❌ Not quite. The same criterion holds for constant-coefficient linear equations of any order: all roots need negative real parts.

❌ Not quite. Check the roots:

\pm i

have real part

0

, and

\cos t

is a solution that does not go to

0

Show solution

$r^3 + r^2 + r + 1 = r^2(r + 1) + (r + 1) = (r + 1)(r^2 + 1)$ , so the roots are $-1$ and $\pm i$ .

y = c_1e^{-t} + c_2\cos t + c_3\sin t.

With $c_2 = 1$ , $c_3 = 0$ the solution $\cos t$ does not go to $0$ : not stable.

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