A capacitor charging through a resistor, salt flowing through a well-mixed tank and a balance growing with interest while it is paid down are the same equation with different letters. In each, the steady state is b/a, and the sign of a decides whether it attracts or repels. The loan has a negative a, which is why the naive payoff time of balance divided by payment is wrong.
Core ideas
Three ideas: how a situation is turned into the equation, what the steady state means in each, and why the loan behaves differently.
Each equation starts as a sentence: the rate of change of the unknown is a gain minus a loss. A tank gains salt at a fixed rate and loses it in proportion to how much is there. A loan gains interest in proportion to the balance and loses a fixed payment. Collect the terms in y on the left and the equation is y′ + a·y = b, with a taking its sign from what is on the right.
At y = b/a the rate is zero: the tank loses salt exactly as fast as it gains it, the capacitor draws no current, and the balance grows by interest exactly as fast as the payment removes it. The steady state is the particular solution for a constant forcing, from “Homogeneous Plus Particular”.
For the tank and the capacitor, loss is in proportion to the unknown and a is positive, so every solution is pulled toward the steady state. For the loan the proportional term is a gain, so a is negative, and the steady state is where the balance stalls: start below it and the balance falls, start above it and the balance grows for ever.
The material
The three models with their numbers, the steady state in each, the loan in detail and the payoff time it gives, and the naive answer it corrects.
All three situations on this page have the same skeleton. There is an unknown quantity, a constant source and a loss or gain proportional to the unknown. Once the equation is written as y′ + a·y = b the solution and its steady state follow from “Constant Coefficients and the Steady State” with no new idea.
A capacitor with capacitance C is charged through a resistor R from a source of voltage V. The charge q satisfies R·q′ + q/C = V. With R = 1, C = 1/2 and V = 5 in consistent units, dividing by R gives q′ + 2q = 5. The steady state is C·V = 5/2, the charge at which the capacitor's voltage equals the source's. From an empty capacitor the charge approaches 5/2 and is below it at every finite time.
A tank holds 20 litres of well-mixed brine. Brine at 10 units of salt per litre flows in at one litre per minute, and the mixture flows out at the same rate. The salt y gains 10 per minute and loses y/20 per minute, since a twentieth of the tank leaves each minute. So y′ = 10 − y/20, which is y′ + y/20 = 10. The steady state is 200, the inflow concentration times the volume.
A balance of 100 earns interest at 5% a year, charged continuously, and is repaid at 6 a year. The balance gains B/20 and loses 6, so B′ = B/20 − 6. Moving the B term to the left changes its sign: B′ − B/20 = −6, so a = −1/20 and b = −6. The steady state is b/a = (−6)/(−1/20) = 120.
That number has a meaning. At a balance of 120 the interest is 120/20 = 6 a year, exactly the payment, so the balance neither falls nor grows. It is a knife edge, not a destination: a < 0 makes it repelling. The loan starts at 100, below it, so the gap is −20 and grows.
The payoff time is where the balance is zero. It needs a natural logarithm, ln 6 ≈ 1.79176, which is irrational, so the figure ≈ 35.8352 is rounded. The lab does not print a payoff time; it prints the closed form, and the number is yours to compute from it. You can check the sign change in the lab's Euler view with a step of one year and at least 36 steps: the rounded closed-form column falls through zero between year 35 and year 36.
Compare the naive estimate. A balance of 100 repaid at 6 a year would take 100/6 ≈ 16.6667 years if no interest were charged. At year 17 the closed form gives 120 − 20·e^(17/20) ≈ 73.2071, rounded: more than seventy are still owed. The real payoff is more than twice as long, because the balance gains interest all the way down.
Start the same loan at B(0) = 130 instead. The gap to 120 is +10, so B = 120 + 10·e^(t/20), and the balance grows without limit: the interest on 130 is 6.5 a year and the payment is only 6.
The knife edge is exactly B = 120. A payment of 6 a year pays off a loan of 100 but not a loan of 130, and the equation says so at once through the sign of the gap.
Interactive
μ, y_h, y_p and the constant, exactly
Method
Name the unknown, write gain and loss, collect into standard form with the signs checked, and read the steady state before solving.
A charge, an amount of salt or a balance, and its rate of change per unit of time. Fix the unit of time once.
y′ = (what comes in) − (what goes out). A flow in proportion to the unknown carries the unknown; a fixed flow does not.
Move every term in y to the left and change its sign as it crosses. A proportional gain, like interest, becomes a negative a.
a > 0 attracts and a < 0 repels. Say what the number means in the situation: a charge, an amount of salt, a balance.
Write y = b/a + (y₀ − b/a)·e^(−a·t). For a payoff, set it to zero and take a logarithm; the answer is irrational and printed rounded.
The balance gains B/20 a year in interest and loses 6 in payments. The equation is written, the steady state read, and the balance set to zero.
The payoff time is more than twice the naive estimate. The naive estimate assumes each payment of 6 takes 6 off the balance. It does not: interest of about 5 a year on a balance near 100 puts most of that back, so the balance falls by only about 1 a year at first.
For a rehearsal, run the tank. Write y′ = 10 − y/20, collect it to y′ + y/20 = 10, read the steady state before the lab does, and check that it is 200 and attracting.
You should be able to write the rate as gain minus loss, collect it into standard form with the signs checked, read the steady state b/a and say whether it attracts or repels, write the closed form, and find when a loan balance reaches zero by setting the closed form to zero, giving the answer rounded.
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