01 · A colder object
In the same surroundings, an object starts at 5°C. Give θ(t) using the same k.
Hint
The initial difference is negative.
Worked solution
θ(t) = 20 − 15e⁻ᵏᵗ. Its temperature increases towards 20°C.
Understand · explore · practise
Differential equations: cooling and limiting values
Solve cooling and limiting-value differential equations. Model approach to equilibrium, constant inflow with proportional removal, logistic growth and the difference between a limit and an attained value.
Before you startSeparation of variables, exponential decay, logarithms and partial fractions.
01 / Model the difference from equilibrium
The quantity approaches a target when its rate opposes the difference.
y′ = −k(y − L) → y = L + (y₀ − L)e⁻ᵏᵗ
For k > 0, L is a stable equilibrium and y₀ is the initial value.
Above L, the derivative is negative; below L, it is positive. At L, the derivative is zero. The difference y − L decays exponentially, while the whole quantity need not be an exponential tending to zero.
Approach the equilibrium from either sideExplore
At t = 0, y = 80; excess y − 20 = 60.
Rate y′ = −12.
The value decreases towards 20 from above.
The limit is 20, but a non-equilibrium solution never reaches it at finite time.
The dashed line is the equilibrium. Each equal time interval multiplies the difference from 20 by the same factor; it does not multiply the whole value by that factor.
02 / Use temperature difference in Newton’s cooling model
The surrounding temperature must be treated as constant.
An object starts at 80°C in surroundings at 20°C. Write and solve a cooling model.Worked example
dθ/dt = −k(θ − 20), k > 0
Rate is proportional to excess temperature, with a negative sign.
dθ/(θ − 20) = −k dt
Separate for θ ≠ 20.
ln|θ − 20| = −kt + C
Integrate.
θ = 20 + Ae⁻ᵏᵗ; A = 60
Apply θ(0) = 80.
θ = 20 + 60e⁻ᵏᵗ
The excess above room temperature decays.
Watch: the difference from equilibrium shrinks
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Use a later temperature to find k
Subtract the ambient temperature before taking a ratio.
The 80°C object reaches 50°C after 10 minutes. Find k.Worked example
50 − 20 = 60e⁻¹⁰ᵏ
Use the excess temperature 30, not the total 50.
e⁻¹⁰ᵏ = 1/2
The excess has halved.
k = ln 2 / 10 per minute
Take logarithms.
θ(t) = 20 + 60 × 2^(−t/10)
At 20 minutes, θ = 35°C.
02 · Another ten minutes
What temperature does this model predict at t = 30 minutes?
Hint
The excess has halved three times.
Worked solution
20 + 60/8 = 27.5°C.
03 · Temperature versus excess
Does θ = 50°C at 10 minutes mean the temperature halves every ten minutes?
Hint
Identify what decays exponentially.
Worked solution
No. The excess θ − 20 halves; the total temperature does not.
04 / Solve for a target above the limiting temperature
Check that the target is reachable within the model.
When does the cooling model first reach 25°C?Worked example
25 − 20 = 60 × 2^(−t/10)
The target excess is 5.
2^(−t/10) = 1/12
Divide by 60.
t = 10ln 12 / ln 2
About 35.85 minutes.
04 · An impossible target
Can this model reach 15°C for t ≥ 0?
Hint
Its excess above 20 stays positive.
Worked solution
No. It approaches 20°C from above.
05 · Equilibrium exactly
Does the object reach exactly 20°C at a finite time?
Hint
Can a positive exponential equal zero?
Worked solution
No. It tends to 20°C as t → ∞, unless it started at equilibrium.
05 / Combine constant input with proportional removal
The equilibrium balances input and output rates.
An amount M has constant input 6 units per minute and removal rate 0.2M units per minute. Initially M = 0.Worked example
M′ = 6 − 0.2M = −0.2(M − 30)
The equilibrium is 6/0.2 = 30 units.
M = 30 + Ae^(−0.2t)
Use the approach-to-equilibrium family.
A = −30
Apply the empty initial state.
M = 30(1 − e^(−0.2t))
It increases towards 30, with initial rate 6.
06 · Change the initial amount
With the same rates but M(0) = 50, find M(t).
Hint
The initial excess above 30 is 20.
Worked solution
M = 30 + 20e^(−0.2t). It decreases towards 30.
07 · Find an equilibrium
For Q′ = 12 − 0.3Q, find the equilibrium amount.
Hint
Set Q′ = 0.
Worked solution
Q = 40.
06 / Check constant solutions before dividing
An equilibrium is a real solution, not just a horizontal guide.
Solve y′ = −k(y − L), including the case y = L.Worked example
If y = L, both sides are zero
Check this before dividing by y − L.
For y ≠ L, integrate dy/(y − L) = −k dt
Obtain the non-equilibrium branches.
y = L + Ae⁻ᵏᵗ, A any real constant
A = 0 includes the checked equilibrium.
08 · Direction from the equation
For y′ = −0.4(y − 10), what is the sign of y′ when y = 6?
Hint
Substitute into the right-hand side.
Worked solution
y′ = 1.6 > 0, so the value increases.
09 · Stability
Why is L a stable equilibrium when k > 0?
Hint
Inspect the derivative on both sides.
Worked solution
Above L the value decreases; below L it increases. Both motions approach L.
07 / Let proportional growth slow near a capacity
Logistic growth includes two equilibrium amounts.
P′ = rP(1 − P/K)
Take r > 0 and K > 0. The equilibria are P = 0 and P = K.
For 0 < P < K, growth is positive but the per-unit rate r(1 − P/K) decreases as P rises. For P > K, the derivative is negative. K is the model’s limiting capacity for positive initial values, subject to its assumptions.
10 · Logistic rate signs
For K = 600, classify the rate at P = 0, 300, 600 and 900.
Hint
Use the signs of P and 1 − P/600.
Worked solution
Zero, positive, zero, negative respectively.
08 / Separate logistic growth using partial fractions
Retain P = 0 and P = K before dividing.
Solve P′ = rP(1 − P/K) for a non-equilibrium branch.Worked example
K/[P(K − P)] dP = r dt
Separate variables.
K/[P(K − P)] = 1/P + 1/(K − P)
Recombine to check the identity.
ln|P| − ln|K − P| = rt + C
The second integral has a minus sign.
P/(K − P) = Beʳᵗ on a chosen branch
Absorb its fixed sign into the constant.
P = K/(1 + Ae⁻ʳᵗ)
This describes positive initial-value solutions with the appropriate A; also record the excluded equilibrium P = 0.
11 · Apply a positive initial value
For P(0) = P₀ > 0, find A in P = K/(1 + Ae⁻ʳᵗ).
Hint
Substitute t = 0.
Worked solution
A = K/P₀ − 1. If P₀ = K, A = 0 recovers the checked equilibrium.
12 · Verify the decomposition
Combine 1/P + 1/(K − P).
Hint
Use denominator P(K − P).
Worked solution
[(K − P) + P]/[P(K − P)] = K/[P(K − P)].
09 / Calibrate a logistic initial state
The limit and initial amount play different roles.
Let K = 600, r = 0.3 per day and P(0) = 100.Worked example
A = 600/100 − 1 = 5
Apply the initial value.
P(t) = 600/(1 + 5e^(−0.3t))
For t ≥ 0 it remains positive and below 600.
P = 300 when 1 + 5e^(−0.3t) = 2
Set a target amount.
t = ln 5 / 0.3
About 5.36 days.
P → 600 as t → ∞
Starting below capacity, it does not reach exactly 600 at finite time.
13 · Start above capacity
If P₀ = 900 with K = 600, what is A and what happens for t ≥ 0?
Hint
Use K/P₀ − 1.
Worked solution
A = −1/3. The denominator remains positive, and the solution decreases from 900 towards 600.
10 / Interpret assumptions and asymptotes in context
A mathematical limit is not automatically a recorded physical endpoint.
The cooling model assumes fixed surroundings and a constant proportional coefficient. The input-removal model assumes fixed input and proportional removal throughout. Logistic growth assumes a fixed capacity and a particular dependence of rate on amount. Changing conditions, measurement resolution and discrete populations can limit these models.
14 · A changing room
Why can a large change in surrounding temperature invalidate the calibrated cooling formula?
Hint
The model held its equilibrium temperature fixed.
Worked solution
The target temperature and possibly the coefficient change, so the original fixed-parameter equation no longer describes the same process.
15 · Limit versus attainment
Does writing lim P(t) = K prove P(t) = K for some finite t?
Hint
A horizontal asymptote need not be reached.
Worked solution
No. The non-equilibrium logistic branches approach K without attaining it at finite time.
11 / Find equilibrium, solve the difference, interpret the limit
Check excluded constant solutions and physical assumptions.
- Use the difference from ambient temperature in cooling.
- Balance constant input against proportional removal.
- Record equilibrium solutions before separation.
- Calibrate constants from initial and later data.
- Use partial fractions for logistic separation.
- Distinguish a limiting value from finite-time attainment.
- Keep units and contextual assumptions explicit.
16 · A final limiting model
Solve y′ = 2(7 − y), with y(0) = 3, and state the limit.
Hint
The equilibrium is 7 and the initial difference is −4.
Worked solution
y = 7 − 4e⁻²ᵗ, tending to 7 from below as t → ∞.
Section 1 of 11 · Model the difference from equilibrium