Differential-Equations · Unit 8 · Video 5 · Interactive Practice

RC Circuits, Decay Chains and Negative k

IKey Formulas

Formula	Name	Where it comes from
$R\,\frac{dq}{dt} + \frac{q}{C} = E(t) \;\Rightarrow\; q' + \frac{1}{RC}\,q = \frac{E(t)}{R}$	RC circuit, form (2)	Kirchhoff: drops $Ri$ and $q/C$ add to the EMF; $k = \frac{1}{RC}$
$B' + k_2 B = k_1 A_0\,e^{-k_1 t}$	Decay chain $A \to B$ , form (2)	Inflow $k_1 A$ minus outflow $k_2 B$ , with $A = A_0 e^{-k_1 t}$
$B = \frac{k_1 A_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right)$ , or $B = k_1 A_0\,t\,e^{-k_1 t}$ if $k_1 = k_2$	Amount of $B$ when $B(0) = 0$	Integrating factor $e^{k_2 t}$
$y' - ay = q(t) \;\Rightarrow\; y = e^{at}\!\int q(t)\,e^{-at}\,dt + c\,e^{at}$	Negative $k = -a$ , $a \gt 0$	Same method, integrating factor $e^{-at}$

Key Insight: The solution method never looks at the sign of $k$ . With $k \gt 0$ the term $c\,e^{-kt}$ is a transient and the initial condition fades; with $k = -a \lt 0$ the term $c\,e^{at}$ grows without bound and the initial condition decides everything.

IIVisualization 1 — Charging the Capacitor: $k = 1/RC$

The coefficient $k = 1/RC$ sets how fast the charge forgets $q(0)$ ; the source $E(t)/R$ sets where it settles.

R 2 C 1 q(0) 0 Source E(t)

IIIVisualization 2 — The Decay Chain $A \to B \to \cdots$

$B$ rises while its inflow $k_1 A$ exceeds its outflow $k_2 B$ , and peaks exactly when the two are equal.

t 1.20 k₁ 1 k₂ 0.5

💡 Challenge: set $k_1 = k_2$ . The general formula divides by $k_2 - k_1 = 0$ , yet $B$ stays finite: it becomes $k_1 A_0\,t\,e^{-k_1 t}$ , peaking at $t = 1/k_1$ .

IVVisualization 3 — When $k$ Turns Negative

The same starting value feeds $y' + y = 1$ and $y' - y = -1$ ; only one of them forgets it.

y(0) 1.05

VQuiz Questions

Problem 1 · The RC Circuit in Standard Form

Given: A series circuit with resistance $R = 4$ , capacitance $C = 0.5$ and a battery $E = 6$ obeys $R\,\frac{dq}{dt} + \frac{q}{C} = E$ . Which is its standard form $q' + kq = q(t)$ ?

✅ Correct!

k = \frac{1}{RC} = \frac{1}{2}

and the right-hand side is

\frac{E}{R} = \frac{3}{2}

, so the charge settles at

CE = 3

❌ That coefficient is $RC$ , not $1/RC$ . Dividing

R\,q' + q/C = E

R

leaves

\frac{1}{RC}

in front of

q

; here

RC = 2

, so

k = \frac{1}{2}

❌ The left side was divided by $R$ , the right side was not. Every term is divided by

R = 4

, so the right-hand side is

\frac{6}{4}

❌ $3 = CE$ is the steady-state charge, not the right-hand side. Form (1) would read

q' + kq = k\cdot CE = \frac{1}{2}\cdot 3

; the right-hand side itself is

\frac{E}{R}

❌ Not quite. Divide every term by

R

Show solution

Divide $R\,q' + \frac{q}{C} = E$ through by $R$ :

q' + \frac{1}{RC}\,q = \frac{E}{R} = \frac{6}{4}, \qquad RC = 4 \cdot 0.5 = 2.

q' + \tfrac{1}{2}\,q = \tfrac{3}{2}

So $k = \frac{1}{RC} = \frac{1}{2}$ and $q(t) = \frac{E}{R} = \frac{3}{2}$ . The solution is $q = 3 + c\,e^{-t/2}$ : the steady state is $CE = 3$ , and $c\,e^{-t/2}$ is the transient. Writing the right side as $k\cdot CE$ would put the equation in form (1), but nobody writes the circuit that way.

Problem 2 · Equal Decay Constants

Given: In the chain $A \to B$ , $A_0 = 5$ , $k_1 = k_2 = 0.5$ and $B(0) = 0$ , so $B' + 0.5B = 2.5\,e^{-0.5t}$ . Find $B(t)$ .

✅ Correct! With equal rates,

e^{0.5t}

cancels the forcing exponential completely and the integral produces a factor of

t

❌ This misses $B(0) = 0$ . At

t = 0

it gives

2.5

, and it solves

B' + 0.5B = 0

, not the forced equation.

❌ The general formula is $\frac{0}{0}$ here, not $0$ .

B

receives atoms at rate

k_1 A \gt 0

, so it cannot stay zero; solve the equation directly.

❌ The coefficient is $k_1 A_0$ , not $A_0$ . Check:

B' + 0.5B

for

5te^{-0.5t}

5e^{-0.5t}

, twice the right-hand side.

❌ Not quite. Multiply by the integrating factor

e^{0.5t}

Show solution

Multiply $B' + 0.5B = 2.5\,e^{-0.5t}$ by the integrating factor $e^{0.5t}$ :

\left(B\,e^{0.5t}\right)' = 2.5\,e^{-0.5t}\,e^{0.5t} = 2.5

B\,e^{0.5t} = 2.5\,t + c, \qquad B(0) = 0 \Rightarrow c = 0

B = 2.5\,t\,e^{-0.5t}

The general formula $\frac{k_1 A_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right)$ cannot be used when $k_1 = k_2$ : the forcing exponential then has the same rate as the integrating factor, and the integrand is a constant instead of an exponential. $B$ peaks at $t = 1/k_1 = 2$ with $B(2) = 5e^{-1} \approx 1.84$ .

Problem 3 · When Does B Peak?

Given: $A_0 = 4$ , $k_1 = 1$ , $k_2 = 0.5$ , $B(0) = 0$ — so $B = 8\left(e^{-t/2} - e^{-t}\right)$ . Find the time of the peak of $B$ and the peak amount.

When is $B$ largest?

What is the peak amount of $B$ ?

✅ Correct! At

t = 2\ln 2

e^{-t/2} = \frac{1}{2}

and

e^{-t} = \frac{1}{4}

, so

B = 8 \cdot \frac{1}{4} = 2

, and the inflow

k_1 A = 1

equals the outflow

k_2 B = 1

❌ $\ln 2$ is $\ln(k_1/k_2)$ before dividing by $k_1 - k_2 = 0.5$ . Setting

B' = 0

gives

e^{t/2} = 2

, so

t = 2\ln 2

❌ That is a single decay time, $1/k_2$ or $1/k_1$ . The peak depends on both constants:

t^* = \frac{\ln(k_1/k_2)}{k_1 - k_2}

❌ Check the peak time. Set

B' = 8\left(-\tfrac{1}{2}e^{-t/2} + e^{-t}\right) = 0

❌ $B$ never holds all of $A_0$ . While

A

is still feeding

B

, part of

B

has already decayed onward.

❌ $1$ is the amount of $A$ at the peak, not of $B$ . At

t = 2\ln 2

A = 4e^{-2\ln 2} = 1

and the inflow is

k_1 A = 1

;

B

itself is

8\left(\tfrac{1}{2} - \tfrac{1}{4}\right)

❌ $8$ is the prefactor $\frac{k_1 A_0}{k_2 - k_1}$ in absolute value, not a value of $B$ . Substitute the peak time into

8\left(e^{-t/2} - e^{-t}\right)

❌ Check the peak amount. Substitute

e^{-t/2} = \frac{1}{2}

and

e^{-t} = \frac{1}{4}

Show solution

From the formula, $B = \frac{k_1 A_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right) = \frac{4}{-0.5}\left(e^{-t} - e^{-t/2}\right) = 8\left(e^{-t/2} - e^{-t}\right)$ .

Step 1: peak time.

B' = 8\left(-\tfrac{1}{2}e^{-t/2} + e^{-t}\right) = 0 \;\Rightarrow\; e^{-t} = \tfrac{1}{2}e^{-t/2} \;\Rightarrow\; e^{-t/2} = \tfrac{1}{2} \;\Rightarrow\; t = 2\ln 2 \approx 1.39

This agrees with $t^* = \frac{\ln(k_1/k_2)}{k_1 - k_2} = \frac{\ln 2}{0.5}$ .

Step 2: peak amount.

B(2\ln 2) = 8\left(\tfrac{1}{2} - \tfrac{1}{4}\right) = 2

Check with the rates: $A = 4e^{-2\ln 2} = 1$ , so the inflow is $k_1 A = 1$ and the outflow is $k_2 B = 0.5 \cdot 2 = 1$ . Then $B' = k_1 A - k_2 B = 0$ , as it must be at a peak.

Problem 4 · Negative k Amplifies the Initial Condition

Given: $y' - 0.5y = -1$ . Two solutions start at $y(0) = 2.1$ and $y(0) = 1.9$ . How far apart are they at $t = 4$ ?

✅ Correct! Here

k = -0.5

, so the gap

0.2

is multiplied by

e^{0.5 t}

instead of shrinking: at

t = 4

it is

0.2e^2 \approx 1.48

❌ That treats $k$ as $+0.5$ . In standard form

y' + ky

, the equation

y' - 0.5y

has

k = -0.5

, so the free term is

c\,e^{+0.5t}

, which grows.

❌ The two solutions differ by $c\,e^{0.5t}$ , not by a constant. Both share the particular solution

y = 2

; only the

c

-term differs, and it grows.

❌ Each $c$ is measured from the line $y = 2$ . The two solutions have

c = 0.1

and

c = -0.1

, so the gap is

0.2\,e^{0.5t}

❌ Not quite. Write each solution as

2 + c\,e^{0.5t}

Show solution

In standard form $y' + ky = q$ the coefficient is $k = -0.5 = -a$ with $a = 0.5$ . The constant $y = 2$ is a particular solution, since $0 - 0.5 \cdot 2 = -1$ , so

y = 2 + c\,e^{0.5t}, \qquad c = y(0) - 2.

The two solutions have $c = 0.1$ and $c = -0.1$ :

y_1(4) = 2 + 0.1e^{2} \approx 2.739, \qquad y_2(4) = 2 - 0.1e^{2} \approx 1.261

y_1(4) - y_2(4) = 0.2\,e^{2} \approx 1.48

A starting difference of $0.2$ has grown more than sevenfold by $t = 4$ . The line $y = 2$ is not a steady state that solutions approach; they move away from it, and the initial condition decides where each one goes.

Solved: 0 / 4

RC Circuits, Decay Chains and Negative k

IKey Formulas

IIVisualization 1 — Charging the Capacitor: k=1/RCk = 1/RCk=1/RC

IIIVisualization 2 — The Decay Chain A→B→⋯A \to B \to \cdotsA→B→⋯

IVVisualization 3 — When kkk Turns Negative

VQuiz Questions

IIVisualization 1 — Charging the Capacitor: $k = 1/RC$

IIIVisualization 2 — The Decay Chain $A \to B \to \cdots$

IVVisualization 3 — When $k$ Turns Negative