Differential-Equations · Unit 9 · Video 2 · Interactive Practice

The Characteristic Equation and the Overdamped Case (Real Distinct Roots)

IKey Formulas

Formula	Name	What it says
$y = e^{rt},\; y' = re^{rt},\; y'' = r^2e^{rt}$	Trial exponential	Each derivative brings out one factor of $r$
$y'' + Ay' + By = 0 \;\to\; r^2 + Ar + B = 0$	Characteristic equation	Cancel $e^{rt}$ , which is positive for every $t$
$y = c_1e^{r_1t} + c_2e^{r_2t}$	Case 1: real roots $r_1 \neq r_2$	Two independent exponentials
$r_1r_2 = B > 0,\quad r_1 + r_2 = -A < 0$	Overdamped signs	$m, c, k > 0$ forces both roots negative

Key Insight: At $t = 0$ every exponential equals $1$ , so $y(0)$ and $y'(0)$ become two linear equations for $c_1$ and $c_2$ .

IITrying $e^{rt}$

Substituting $y = e^{rt}$ into $y'' + 4y' + 3y$ gives $(r^2 + 4r + 3)e^{rt}$ , which vanishes only at a root.

Exponent r −2

IIIThree Cases from One Quadratic

For a spring $A = c/m$ and $B = k/m$ ; the vertex of $r^2 + Ar + B$ decides which root case occurs.

A 4 B 3

IVFitting $c_1$ and $c_2$ : $y'' + 4y' + 3y = 0$

Two initial conditions fix $c_1$ and $c_2$ in $y = c_1e^{-3t} + c_2e^{-t}$ , and the motion crosses equilibrium at most once.

y(0) 1 y′(0) 0

VQuiz Questions

Problem 1 · Write the General Solution

Given: $y'' + 5y' + 6y = 0$ — find the general solution.

✅ Correct!

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

has the distinct real roots

-2

and

-3

❌ Check the signs. The factors

(r + 2)(r + 3)

vanish at

r = -2

and

r = -3

, not at

2

and

3

❌ Not quite.

(r + 1)(r + 6) = r^2 + 7r + 6

: the product is right but the sum is

7

, not

5

❌ Not quite. The coefficients

5

and

6

are not the roots; the roots are the values of

r

that make

r^2 + 5r + 6 = 0

Show solution

Write the characteristic equation directly from the coefficients:

r^2 + 5r + 6 = 0 \quad\Longrightarrow\quad (r + 2)(r + 3) = 0 \quad\Longrightarrow\quad r = -2,\; r = -3.

The roots are real and different (Case 1), so $e^{-2t}$ and $e^{-3t}$ are independent solutions and

y = c_1e^{-2t} + c_2e^{-3t}.

Problem 2 · Starting at Equilibrium

Given: $y'' + 4y' + 3y = 0$ with $y(0) = 0$ and $y'(0) = 2$ (the mass sits at equilibrium and is pushed to the right) — find $y(t)$ .

✅ Correct!

y(0) = 1 - 1 = 0

and

y'(0) = -1 + 3 = 2

❌ Check the sign of the velocity. Here

y'(0) = -3 + 1 = -2

: this mass moves left, not right.

❌ Not quite. Here

y'(0) = -2 + 6 = 4

, twice the given velocity: from

-2c_1 = 2

you get

c_1 = -1

, not

-2

❌ Not quite. This starts at

y(0) = 1 + 1 = 2

, not at equilibrium.

Show solution

From the video, the general solution is $y = c_1e^{-3t} + c_2e^{-t}$ , so $y' = -3c_1e^{-3t} - c_2e^{-t}$ . At $t = 0$ every exponential equals $1$ :

c_1 + c_2 = 0, \qquad -3c_1 - c_2 = 2.

Add the equations: $-2c_1 = 2$ , so $c_1 = -1$ and $c_2 = -c_1 = 1$ .

y = -e^{-3t} + e^{-t} = e^{-t} - e^{-3t}.

Check: $y(0) = 1 - 1 = 0$ , $y'(0) = -1 + 3 = 2$ ✓. Since $e^{-t} > e^{-3t}$ for $t > 0$ , the mass moves right, then returns toward equilibrium without crossing it.

Problem 3 · A Hard Push Toward Equilibrium

Given: $m = 1$ , $c = 6$ , $k = 5$ , so $y'' + 6y' + 5y = 0$ , with $y(0) = 1$ and $y'(0) = -9$ — find the motion and the time it crosses equilibrium.

What is $y(t)$ ?

At what time does the mass cross $y = 0$ ?

✅ Correct!

y = -e^{-t} + 2e^{-5t}

crosses once, at

t = \tfrac14\ln 2 \approx 0.173

, then approaches

0

from below.

❌ Check $y(t)$ . Substitute

t = 0

into

y

and

y'

: you need

y(0) = 1

and

y'(0) = -9

❌ Not quite. Here

y'(0) = -2 + 5 = 3

: the constants are swapped between the two exponentials.

❌ Check the derivative. Here

y'(0) = -\tfrac72 + \tfrac{25}{2} = 9

: the factors

-1

and

-5

from differentiating lost their signs.

❌ Check the crossing time. Set

2e^{-5t} = e^{-t}

and divide by

e^{-t}

: the exponents differ by

4

❌ Almost.

e^{-4t} = \tfrac12

gives

4t = \ln 2

; divide by

4

❌ Not quite.

c_1

and

c_2

have opposite signs and

-c_1/c_2 = \tfrac12 < 1

, so there is exactly one crossing at a positive time.

Show solution

Roots: $r^2 + 6r + 5 = (r + 1)(r + 5) = 0$ , so $r = -1$ and $r = -5$ , and $y = c_1e^{-t} + c_2e^{-5t}$ with $y' = -c_1e^{-t} - 5c_2e^{-5t}$ .

Constants: at $t = 0$ ,

c_1 + c_2 = 1, \qquad -c_1 - 5c_2 = -9.

Adding gives $-4c_2 = -8$ , so $c_2 = 2$ and $c_1 = 1 - 2 = -1$ :

y = -e^{-t} + 2e^{-5t}.

Check: $y(0) = -1 + 2 = 1$ , $y'(0) = 1 - 10 = -9$ ✓.

Crossing: $y = 0$ means $2e^{-5t} = e^{-t}$ , i.e. $e^{-4t} = \tfrac12$ , so

t = \tfrac14\ln 2 \approx 0.173.

Because $e^{-4t}$ is strictly decreasing, this is the only crossing; afterwards $-e^{-t}$ dominates and the mass returns to equilibrium from the negative side.

Problem 4 · When $B$ Is Negative

Given: $y'' + y' - 6y = 0$ , where $B = -6 < 0$ (a "spring" that pushes the mass away). Which statement is true?

✅ Correct!

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

; the product of the roots is

B = -6 < 0

, so they have opposite signs.

❌ Not quite. The sign argument needs

B > 0

; here

r_1r_2 = B = -6 < 0

, so the roots have opposite signs.

❌ Not quite. A negative product

r_1r_2 = -6

is impossible for a complex conjugate pair, whose product

|r|^2

is positive.

❌ Check the sum.

-2 + 3 = 1

, but the sum of the roots must be

-A = -1

Show solution

The characteristic equation is

r^2 + r - 6 = (r - 2)(r + 3) = 0 \quad\Longrightarrow\quad r = 2,\; r = -3.

Check with sum and product: $r_1 + r_2 = -1 = -A$ and $r_1r_2 = -6 = B$ ✓. The roots are real and distinct, so

y = c_1e^{2t} + c_2e^{-3t}.

With $B < 0$ the product of the roots is negative, so one root is positive. Whenever $c_1 \neq 0$ the term $c_1e^{2t}$ grows without bound; only the special initial conditions with $c_1 = 0$ give a decaying solution. The video's conclusion "both roots negative" relies on $k > 0$ .

Solved: 0 / 4

The Characteristic Equation and the Overdamped Case (Real Distinct Roots)

IKey Formulas

IITrying erte^{rt}ert

IIIThree Cases from One Quadratic

IVFitting c1c_1c1​ and c2c_2c2​: y′′+4y′+3y=0y'' + 4y' + 3y = 0y′′+4y′+3y=0

VQuiz Questions

IITrying $e^{rt}$

IVFitting $c_1$ and $c_2$ : $y'' + 4y' + 3y = 0$