Differential-Equations · Unit 9 · Video 4 · Interactive Practice

Equal Roots and the Missing Solution: Reduction of Order Finds $t\,e^{-at}$

IKey Formulas

Formula	Name	Where it comes from
$(r + a)^2 = r^2 + 2ar + a^2 = 0$	Double root $r = -a$ , $a > 0$	$y'' + 2a\,y' + a^2 y = 0$ : $A = 2a$ , $B = a^2$ , so $A = 2\sqrt{B}$
$y = y_1\,u$	Reduction of order	One known solution $y_1$ of $y'' + p\,y' + q\,y = 0$ ; $p$ , $q$ may depend on $t$
$e^{-at}u'' = 0 \;\Rightarrow\; u = c_1 t + c_2$	Condition on $u$	Substitute $y = e^{-at}u$ ; the $u$ and $u'$ columns cancel
$y = (c_1 + c_2 t)\,e^{-at}$	Critically damped general solution	$y_1 = e^{-at}$ , $y_2 = t\,e^{-at}$ , independent since $y_2/y_1 = t$

Key Insight: A double root gives the exponential trial only one function, $e^{-at}$ . Writing $y = e^{-at}u$ turns the equation into $u'' = 0$ , and the simplest non-constant choice $u = t$ supplies the factor of $t$ no exponential could produce.

IIVisualization 1 — The Case Between Over- and Underdamping

For $y'' + Ay' + y = 0$ , so $B = 1$ , only $A = 2\sqrt{B} = 2$ merges the two roots into a double root.

Damping A 1.0

IIIVisualization 2 — Which $u$ Makes $e^{-at}u$ a Solution?

Substituting $y = e^{-at}u$ leaves only $e^{-at}u''$ , so $u$ works exactly when $u'' = 0$ .

multiply by		$u$ column	$u'$ column	$u''$ column
$a^2$	$y =$	$e^{-at}u$
$2a$	$y' =$	$-a\,e^{-at}u$	$+\,e^{-at}u'$
$1$	$y'' =$	$a^2 e^{-at}u$	$-2a\,e^{-at}u'$	$+\,e^{-at}u''$
sum	$y'' + 2a\,y' + a^2 y =$	$(a^2 - 2a^2 + a^2)\,e^{-at}u = 0$	$(2a - 2a)\,e^{-at}u' = 0$	$e^{-at}u''$

Try u(t) =

IVVisualization 3 — Two Constants Match Any Start

For $y'' + 2y' + y = 0$ , one constant times $e^{-t}$ fixes $y'(0) = -y(0)$ ; adding $t\,e^{-t}$ frees the slope.

VQuiz Questions

Problem 1 · The Critical Damping Value

Given: $y'' + Ay' + 16y = 0$ with $A > 0$ . Find the value of $A$ that makes the system critically damped, and the resulting double root.

What is A?

What is the double root?

✅ Correct!

A = 2\sqrt{16} = 8

, and

r^2 + 8r + 16 = (r + 4)^2

has the double root

r = -4

, so

a = 4

❌ That is $\sqrt{B}$ , which is $a$ . The damping coefficient is

A = 2a = 2\sqrt{B}

❌ That is $2B$ . Matching

(r + a)^2 = r^2 + 2ar + a^2

gives

B = a^2

, so

a = \sqrt{B}

, not

B

❌ That is $A^2$ , not $A$ . A double root needs

A^2 - 4B = 0

, so

A^2 = 64

and

A = 8

❌ Check $A$ . A double root means the discriminant

A^2 - 4B

is zero.

❌ Check the sign.

(r + 4)^2 = 0

r = -4

; with damping present the root is negative.

❌ That is $-A$ . The double root of

r^2 + Ar + B

-A/2

❌ Check the root. Factor

r^2 + 8r + 16

as a perfect square.

Show solution

The characteristic equation is $r^2 + Ar + 16 = 0$ . It has a double root when the discriminant vanishes:

A^2 - 4 \cdot 16 = 0 \;\Rightarrow\; A^2 = 64 \;\Rightarrow\; A = 8 \quad (A > 0).

Equivalently, $A = 2\sqrt{B} = 2\sqrt{16} = 8$ . Then

r^2 + 8r + 16 = (r + 4)^2 = 0 \;\Rightarrow\; r = -4 \text{ (double)},

so $a = 4$ and the general solution is $y = (c_1 + c_2 t)\,e^{-4t}$ .

Problem 2 · The Same Exponential Twice

Given: $y'' + 4y' + 4y = 0$ with $y(0) = 1$ , $y'(0) = 0$ . A student tries $y = c_1 e^{-2t} + c_2 e^{-2t}$ . Which statement is true?

✅ Correct!

y = (c_1 + c_2)e^{-2t}

has one effective constant;

y(0) = 1

forces

y'(0) = -2(c_1 + c_2) = -2 \neq 0

❌ That pair gets $y(0) = 1$ but not the slope. Any pair with

c_1 + c_2 = 1

gives

y = e^{-2t}

, whose slope at

0

-2

, not

0

❌ $c_1 + c_2 = 1$ handles only $y(0)$ . All those pairs give the same function

e^{-2t}

, and its slope at

0

-2

, so

y'(0) = 0

still fails.

❌ Not quite. Factor out

e^{-2t}

and count the free constants.

Show solution

The characteristic equation $r^2 + 4r + 4 = (r + 2)^2 = 0$ has the double root $r = -2$ . The student's combination collapses:

y = c_1 e^{-2t} + c_2 e^{-2t} = (c_1 + c_2)\,e^{-2t} = C e^{-2t}.

Then $y(0) = C = 1$ , and $y'(t) = -2C e^{-2t}$ gives $y'(0) = -2C = -2 \neq 0$ .

One free constant cannot meet two conditions. The fix is the second solution $t\,e^{-2t}$ : $y = (1 + 2t)\,e^{-2t}$ has $y(0) = 1$ and $y'(0) = 2 - 2 = 0$ .

Problem 3 · Solving a Critically Damped Initial Value Problem

Given: $y'' + 6y' + 9y = 0$ , $y(0) = 2$ , $y'(0) = -1$ . Find $y(t)$ .

✅ Correct!

c_1 = y(0) = 2

and

y'(0) = c_2 - 3c_1 = -1

give

c_2 = 5

❌ $c_2$ is not $y'(0)$ . Differentiating

(c_1 + c_2 t)e^{-3t}

also hits the exponential:

y'(0) = c_2 - 3c_1

❌ The constants are in the wrong slots. At

t = 0

the factor is

c_1 + c_2 \cdot 0

, so

c_1 = y(0) = 2

❌ Check the sign of the exponential's derivative.

\frac{d}{dt}e^{-3t} = -3e^{-3t}

, so

y'(0) = c_2 - 3c_1

, not

c_2 + 3c_1

❌ Not quite. Use

y(0)

for

c_1

, then the product rule for

y'(0)

Show solution

$r^2 + 6r + 9 = (r + 3)^2 = 0$ gives the double root $r = -3$ , so

y = (c_1 + c_2 t)\,e^{-3t}, \qquad y' = c_2 e^{-3t} - 3(c_1 + c_2 t)\,e^{-3t}.

At $t = 0$ :

y(0) = c_1 = 2, \qquad y'(0) = c_2 - 3c_1 = c_2 - 6 = -1 \;\Rightarrow\; c_2 = 5.

y = (2 + 5t)\,e^{-3t}.

Check: $y'(0) = 5 - 3 \cdot 2 = -1$ ✓.

Problem 4 · Reduction of Order with Variable Coefficients

Given: For $t > 0$ , $y_1 = t^2$ solves $t^2 y'' - 3t\,y' + 4y = 0$ . Substitute $y = t^2 u$ . Find the condition on $u$ and a second solution $y_2$ .

What equation must u satisfy?

Which is a second solution?

✅ Correct! The

u

column cancels, leaving

t^4u'' + t^3u' = 0

, so

(t\,u')' = 0

u = c_1 \ln t + c_2

, and

y_2 = t^2\ln t

❌ That was the constant-coefficient case. Here the

u'

column does not cancel: it leaves

4t^3u' - 3t^3u' = t^3u'

❌ Check the condition on $u$ . Write

y

y'

y''

u

u'

u''

columns, multiply by

4

-3t

t^2

, and add each column.

❌ Multiplying by $t$ was special to $e^{-at}$ . Test it:

t^2(6t) - 3t(3t^2) + 4t^3 = t^3 \neq 0

❌ That is $u$ , not $y$ . The second solution is

y_2 = y_1 u = t^2\ln t

❌ Check $y_2$ . Solve the equation for

u

, then multiply by

y_1 = t^2

Show solution

With $y = t^2 u$ , arrange the derivatives in columns and multiply each row by its coefficient:

\begin{aligned} 4 \;]\quad y &= t^2 u \\ -3t \;]\quad y' &= 2t\,u + t^2 u' \\ t^2 \;]\quad y'' &= 2u + 4t\,u' + t^2 u'' \end{aligned}

Column sums: $u$ : $4t^2 - 6t^2 + 2t^2 = 0$ ; $u'$ : $-3t^3 + 4t^3 = t^3$ ; $u''$ : $t^4$ . So

t^4 u'' + t^3 u' = 0 \;\Rightarrow\; t\,u'' + u' = 0 \;\Rightarrow\; (t\,u')' = 0.

Hence $t\,u' = c_1$ , $u' = c_1/t$ , and $u = c_1 \ln t + c_2$ . The simplest non-constant choice $u = \ln t$ gives

y_2 = t^2 \ln t, \qquad \frac{y_2}{y_1} = \ln t \text{ (not constant)}.

As the video promised, reduction of order works even when the coefficients depend on $t$ .

Solved: 0 / 4

Equal Roots and the Missing Solution: Reduction of Order Finds t e−att\,e^{-at}te−at

IKey Formulas

IIVisualization 1 — The Case Between Over- and Underdamping

IIIVisualization 2 — Which uuu Makes e−atue^{-at}ue−atu a Solution?

IVVisualization 3 — Two Constants Match Any Start

VQuiz Questions

Equal Roots and the Missing Solution: Reduction of Order Finds $t\,e^{-at}$

IIIVisualization 2 — Which $u$ Makes $e^{-at}u$ a Solution?