Differential-Equations · Unit 5 · Video 3 · Interactive Practice

Interest Minus a Thief: dy/dt = ry - w Has One Critical Balance, and Every Other Solution Runs Away From It

IKey Formulas

Formula	Name	What it says
$\dfrac{dy}{dt} = ry - w$	Interest minus a fixed theft	interest $ry$ grows with the balance $y$ ; the theft rate $w$ has no $y$ attached
$ry - w = 0 \iff y = \dfrac{w}{r}$	Step 1: the one critical point	at $y = \tfrac{w}{r}$ the interest $r \cdot \tfrac{w}{r} = w$ exactly pays the thief
$f(y) = ry - w = r\left(y - \dfrac{w}{r}\right)$	Step 2: the sign of the line	slope $r > 0$ : $f > 0$ for $y > \tfrac{w}{r}$ (rises), $f < 0$ for $y < \tfrac{w}{r}$ (falls)
$y(t) = \dfrac{w}{r} + \left(y_0 - \dfrac{w}{r}\right)e^{rt}$	The separable solution, as a check	the gap from $\tfrac{w}{r}$ is multiplied by $e^{rt}$ , whatever its sign

Key Insight: Interest grows with the balance but the theft does not, so the gap $u = y - \tfrac{w}{r}$ obeys $\dfrac{du}{dt} = ru$ : a gap above or below the critical balance feeds on itself and runs away.

IIVisualization 1 — The Sign of One Straight Line

Right of $\tfrac{w}{r}$ the interest out-earns the thief and the balance rises; left of it the balance falls.

interest rate r 0.5 theft rate w 1.5 balance y 4

IIIVisualization 2 — From the Sign to the Solution Curves

A solution's slope is $f(y) = r\left(y - \tfrac{w}{r}\right)$ , so the farther it strays from $\tfrac{w}{r}$ , the faster it runs.

start y₀ 4 time t 0

IVVisualization 3 — Every Solution Is a Slid Copy

The slope depends on the height $y$ alone, so sliding a solution along $t$ gives another solution.

height y 5

VQuiz Questions

Problem 1 · The Critical Balance

Given: An account earns interest at $r = 0.04$ per year while a teller removes 600 dollars per year, so $\dfrac{dy}{dt} = 0.04y - 600$ with $t$ in years — find the balance at which the account stays put.

✅ Correct! At

y = 15000

the interest

0.04 \times 15000 = 600

per year exactly pays the thief, so

\dfrac{dy}{dt} = 0

❌ That is $r \cdot w$ .

0.04 \times 600 = 24

multiplies the two rates; the balance must satisfy

ry = w

, so divide:

y = \tfrac{w}{r}

❌ Sign slip.

y = -15000

solves

0.04y + 600 = 0

. The theft is subtracted, so the condition is

0.04y = +600

❌ That is the theft rate, not a balance. At

y = 600

the interest is only

0.04 \times 600 = 24

per year, far short of the

600

taken, so

\dfrac{dy}{dt} = -576

Show solution

The critical points are the zeros of the right-hand side:

0.04y - 600 = 0 \;\Longrightarrow\; 0.04y = 600 \;\Longrightarrow\; y = \frac{600}{0.04} = 15000.

Check: at $y = 15000$ the interest earned is $0.04 \times 15000 = 600$ per year, precisely what the teller removes, so the balance never moves. This is $y = \tfrac{w}{r}$ with $w = 600$ , $r = 0.04$ .

Problem 2 · A Different Thief

Given: Interest is still $4\%$ per year, but now the teller removes a fixed fraction, $1\%$ of the balance per year: $\dfrac{dy}{dt} = 0.04y - 0.01y$ — which describes the critical points and the other solutions?

✅ Correct!

0.04y - 0.01y = (0.04 - 0.01)y = 0.03y

is the interest equation again with a smaller rate: its one critical point is

y = 0

and its solutions are plain exponentials

y_0e^{0.03t}

❌ That formula belongs to the other thief.

y = \tfrac{w}{r}

comes from a theft with no

y

attached. Here the theft is

0.01y

, and

0.03y

vanishes only at

y = 0

: at

y = 0.25

it equals

0.03 \times 0.25 = 0.0075 \ne 0

❌ Check the sign of the net rate.

0.04 - 0.01 = +0.03 > 0

: interest still beats a thief who takes only

1\%

, so a positive balance grows.

❌ An empty account stays empty. At

y = 0

the right-hand side is

0.03 \cdot 0 = 0

, so

y = 0

is a critical point: the constant solution

y(t) = 0

Show solution

Factor the right-hand side:

\frac{dy}{dt} = 0.04y - 0.01y = (0.04 - 0.01)\,y = 0.03y.

It is zero only when $y = 0$ , so that is the one critical point. The equation is the plain interest equation with rate $0.03$ , so every solution is $y(t) = y_0e^{0.03t}$ : a positive balance grows exponentially, a debt deepens exponentially, and nothing new happens. The fixed-sum thief $w$ is what creates the critical balance $\tfrac{w}{r} \ne 0$ .

Problem 3 · Starting Just Below

Given: The account of Problem 1, $\dfrac{dy}{dt} = 0.04y - 600$ , starts at $y(0) = 14000$ — find the initial rate of change and when the balance reaches zero.

The initial rate of change

The time to reach zero

✅ Correct! The account starts

1000

below the critical balance, so the gap grows like

1000e^{0.04t}

and the balance hits zero after about

67.7

years, far sooner than the

350

years a constant drain of

40

per year would take.

❌ The thief is missing.

0.04 \times 14000 = 560

is only the interest; subtract the

600

taken:

560 - 600 = -40

❌ Subtraction reversed. The rate is interest minus theft,

560 - 600 = -40

, not

600 - 560

. Below

15000

the line

0.04y - 600

is under the axis.

❌ The interest is missing.

-600

is the rate at a zero balance. At

y = 14000

the account also earns

0.04 \times 14000 = 560

, so

\dfrac{dy}{dt} = 560 - 600 = -40

❌ The drain does not stay at 40.

14000 / 40 = 350

assumes a constant rate, but

\dfrac{dy}{dt} = 0.04(y - 15000)

grows more negative as the balance falls: the gap below

15000

is multiplied by

e^{0.04t}

❌ Wrong ratio inside the log.

25\ln\tfrac{15000}{14000} \approx 1.7

solves

e^{0.04t} = \tfrac{15000}{14000}

. Setting

y = 15000 - 1000e^{0.04t} = 0

needs

e^{0.04t} = 15

❌ Nothing holds the balance at zero. At

y = 0

\dfrac{dy}{dt} = -600 < 0

, so the account passes straight through zero into debt. The only constant solution is

y = 15000

, and this solution runs away from it.

Show solution

Initial rate:

\left.\frac{dy}{dt}\right|_{t=0} = 0.04(14000) - 600 = 560 - 600 = -40.

The gap equation: $\dfrac{dy}{dt} = 0.04(y - 15000)$ , so the gap $u = y - 15000$ obeys $\dfrac{du}{dt} = 0.04u$ with $u(0) = -1000$ . Hence

y(t) = 15000 - 1000e^{0.04t}.

Zero balance:

1000e^{0.04t} = 15000 \;\Longrightarrow\; e^{0.04t} = 15 \;\Longrightarrow\; t = \frac{\ln 15}{0.04} = 25\ln 15 \approx 67.70 \text{ years}.

The drain speeds up from $40$ per year toward $600$ per year as the balance falls, which is why the account empties in about $68$ years rather than $350$ .

Problem 4 · A Deposit and a Fee

Given: A trust account earns no interest and is charged a fee of $2\%$ of the balance per year, while a relative deposits 300 dollars per year: $\dfrac{dy}{dt} = 300 - 0.02y$ , with $y(0) = 10000$ — what does the solution do?

✅ Correct! Same critical balance as Problem 1, opposite verdict: the line

300 - 0.02y

slopes down, so both arrows point toward

15000

and every solution approaches it.

❌ That copies this video's verdict without redrawing the line. Here

f(y) = 300 - 0.02y

has slope

-0.02 < 0

, so it is positive to the left of

15000

f(10000) = 300 - 200 = 100 > 0

, and the balance rises.

❌ It cannot cross the constant solution. The rate

f(y) = -0.02(y - 15000)

shrinks to

0

y

nears

15000

, and above

15000

it is negative, so both arrows point toward

15000

❌ $10000$ is not a critical point.

f(10000) = 300 - 0.02 \cdot 10000 = 100 \ne 0

; the only critical point is

y = 15000

Show solution

Step 1, critical point: $300 - 0.02y = 0 \Longrightarrow y = \dfrac{300}{0.02} = 15000$ .

Step 2, sign: $f(y) = 300 - 0.02y = -0.02\,(y - 15000)$ is a line of slope $-0.02$ . For $y < 15000$ , $f > 0$ (the balance rises); for $y > 15000$ , $f < 0$ (it falls). Both arrows point toward $15000$ , so it is a stable critical point.

Check by solving: the gap $u = y - 15000$ obeys $\dfrac{du}{dt} = -0.02u$ with $u(0) = -5000$ , so

y(t) = 15000 - 5000e^{-0.02t},

which rises toward $15000$ and never reaches it. The sign of the line's slope decides: slope $r > 0$ in $ry - w$ repels, slope $-0.02 < 0$ here attracts.

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