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Mathematics in Bukit Timah | Separable Differential Equations, Growth, Decay and Equilibrium

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eduKate Asia

Did you know that an equation can describe how a quantity changes without directly telling you the quantity’s value? The statement dN/dt=0.2N says that the instantaneous rate of change is always 20% of the current amount per unit time. Given the starting value N(0)=100, calculus reveals the whole function: N(t)=100e^(0.2t). This is the central idea behind a differential equation.

This Mathematics in Bukit Timah article teaches rates of change, separable first-order differential equations, initial conditions, continuous exponential growth and decay, equilibrium, logistic models, Newton’s cooling law, direction fields, Euler’s numerical method and model limitations. Every quantity and time measurement below belongs to a fictional classroom process. The equations are not measurements of Bukit Timah pupil growth, tuition outcomes, local temperature, health, environmental conditions or real equipment performance.

Quick answer: A separable differential equation has a form such as dy/dt=g(t)h(y), allowing the variables to be arranged as dy/h(y)=g(t)dt, wherever h(y)≠0. Integrate both sides, solve for y, and use an initial condition to determine constants. Crucially, also check any equilibrium solutions h(y)=0 that were excluded by dividing through by h(y).

Article contents

Original calculus revision illustration supporting fictional differential equations, logistic growth, cooling and numerical Euler examples.

Why differential equations describe processes

An ordinary algebraic equation relates unknown numbers.

A differential equation instead relates an unknown function to one or more of its derivatives. The derivative records local change.

This allows Mathematics to describe a rule such as “the faster the quantity grows, the larger it already is,” then derive the function consistent with that rule.

Bukit Timah is the educational context, not empirical data

A family near King Albert Park, Beauty World or Sixth Avenue may encounter changing-temperature or growth diagrams in a school lesson.

But the quantities in this article are chosen for exact mathematical calculation. They do not describe an actual local institution, biological population or physical instrument.

A differential equation supplies a conditional model. Its solution is exact given the stated equation and initial value, not proof that the equation matches reality.

Rate of change is a derivative

For a differentiable quantity y(t), the derivative dy/dt measures instantaneous change in y per unit change of t.

A positive derivative means local increase; a negative derivative means local decrease.

A derivative of zero means no instantaneous change at that moment. It does not necessarily mean the whole function stays constant forever unless the differential equation and conditions support that conclusion.

Worked example 1: constant rate

Suppose dy/dt=4 and y(0)=3.

Integrate to obtain y(t)=4t+C. The initial condition gives C=3.

Thus y(t)=4t+3. At t=2, y=11. Differentiating this solution returns four, independently checking the rate rule.

A differential equation needs an initial value for one specific trajectory

The equation dy/dt=4 has infinitely many solutions y=4t+C, one for each constant C.

The initial condition y(0)=3 selects one of them.

This distinction appears throughout differential equations: the governing rule can define a family of solutions, while an initial condition selects the particular solution under appropriate existence and uniqueness assumptions.

Worked example 2: two different starts

For dy/dt=4, initial value y(0)=3 gives y=4t+3.

Initial value y(0)=10 gives y=4t+10.

Both satisfy the same derivative equation. Their values differ by seven at every time because they begin from different starting conditions.

What makes an equation separable?

Consider dy/dt=g(t)h(y).

When h(y)≠0, rearrange to put y terms with dy and t terms with dt:

[1/h(y)]dy=g(t)dt.

Integrating each side is the standard separation method. The manipulation is shorthand supported by the chain rule and integral calculus.

A separation step can lose solutions

Dividing by h(y) is invalid at values where h(y)=0.

Those values may define constant equilibrium solutions y(t)=c satisfying h(c)=0.

Therefore a complete solution process identifies such constants before dividing. Otherwise the final list may omit legitimate solutions.

Worked example 3: separate a simple product rule

Solve dy/dt=t y for y≠0.

Rearrange as (1/y)dy=t dt. Integrate:

ln|y|=t²/2+C.

Exponentiate to obtain y=K e^(t²/2), where K can be any nonzero real constant. The constant solution y=0 must also be included, so the final family permits K=0.

Worked example 4: apply an initial value

For dy/dt=t y with y(0)=2, use y=K e^(t²/2).

At t=0, e⁰=1, so K=2.

Therefore y(t)=2e^(t²/2). At t=2, it gives y(2)=2e²≈14.778. Differentiation recovers dy/dt=ty.

The exponential function is its own derivative

The function e^(kt) differentiates to k e^(kt) for constant k.

That is why rate laws proportional to a quantity often lead to exponential solutions.

A coefficient k may be positive, negative or zero. Its sign determines growth versus decay when the starting quantity is positive.

The proportional growth equation

Suppose a fictional index N(t) satisfies

dN/dt=kN, with k>0 and N(0)=N₀>0.

Separate variables: dN/N=k dt. Integrate to obtain ln N=kt+C. Exponentiating and using the start gives

N(t)=N₀e^(kt).

Worked example 5: a twenty-percent instantaneous rate

Let k=0.2 per time unit and N₀=100.

Then N(t)=100e^(0.2t).

At t=5, N(5)=100e¹≈271.83. This result is not 100×(1.2)⁵, which would be a different discrete-growth model.

Continuous and discrete growth differ

A discrete rule may update once per step as Nₙ₊₁=1.2Nₙ.

The continuous differential equation dN/dt=0.2N instead gives a growth factor e^0.2 per time unit, approximately 1.2214.

The two formulas have related ideas but do not represent identical assumptions. The rate parameter must be interpreted correctly.

Worked example 6: compare five-step values

Discrete growth at exactly 20% per step from 100 gives 100(1.2)⁵=248.832.

Continuous growth with instantaneous coefficient 0.2 per unit time gives 100e≈271.828 at t=5.

Neither number is an error. They answer different models whose update rules must not be conflated.

Doubling time follows from logarithms

For continuous proportional growth N(t)=N₀e^(kt), doubling means N(t)=2N₀.

Cancel N₀>0 to obtain e^(kt)=2. Taking natural logarithms gives

t_double=ln2/k, for k>0.

This time does not depend on the starting size in the simple constant-k model.

Worked example 7: calculate doubling time

Let k=0.2.

Then t_double=ln2/0.2≈0.693147/0.2=3.4657 time units.

Substituting t into 100e^(0.2t) gives 100e^(ln2)=200. The doubling formula is verified exactly by the exponential identity.

Decay uses a negative proportional rate

A simple fictional decay process has equation

dC/dt=−kC, with k>0 and C(0)=C₀.

Its solution is C(t)=C₀e^(−kt). For positive starting amount, the derivative is negative and the quantity approaches zero as time increases.

In the mathematical model it remains positive for every finite real t when C₀>0.

Worked example 8: a decay example

Let C₀=80 and k=0.3 per time unit.

Then C(t)=80e^(−0.3t). At t=4,

C(4)=80e^(−1.2)≈24.10. The amount has decreased but is still positive under the continuous exponential rule.

Half-life is an exact model property

For C(t)=C₀e^(−kt), half-life is the time required for the quantity to become C₀/2.

Solve e^(−kt)=1/2. Taking natural logarithms gives

t_half=ln2/k.

The formula resembles doubling time because both involve the same logarithm of two, with opposite growth and decay signs.

Worked example 9: decay half-life

Let k=0.3.

Then t_half=ln2/0.3≈2.3105 time units.

At that time, C=40 when C₀=80. After another half-life, the model gives C=20, not zero.

Exponential decay never reaches zero at finite time

For C₀>0 and finite t, the value C₀e^(−kt) is strictly positive.

As t→∞, it tends toward zero. This is a limit, not a claim that the quantity becomes exactly zero after some finite number of steps.

If a physical process has a detection threshold or discrete particles, the simple continuous model may cease to be appropriate.

The logarithm step requires care about sign

From ln|y|=kt+C, the general nonzero solution can be written y=Ke^(kt), with K nonzero and potentially negative.

The zero solution must be considered separately when the original rate law allows it.

If the quantity represents a nonnegative physical amount, an initial-value condition can restrict the permitted branch to positive or zero values.

Worked example 10: the missing zero solution

The equation dy/dt=0.2y is satisfied by y(t)=0 for every t, because both sides equal zero.

Separating by dividing through by y would be invalid for this solution.

Including K=0 in the final expression y=Ke^(0.2t) recovers it, but a complete derivation should acknowledge why division originally excluded that case.

Not all differential equations are proportional growth

Some equations include a constant input, a changing external condition or a carrying-capacity term.

For example dX/dt=5−0.2X describes an abstract quantity with constant incoming contribution five units per time and a removal rate proportional to the current X.

Its solution tends toward a nonzero equilibrium rather than growing indefinitely or decaying to zero.

An equilibrium makes the derivative zero

For dX/dt=5−0.2X, set the right-hand side equal to zero:

5−0.2X=0.

The equilibrium is **X*=25**. If the system begins at 25, its derivative is zero and it stays at 25 under the model.

Worked example 11: solve toward equilibrium

Rewrite the equation as dX/dt=−0.2(X−25).

Separate, or recognise the shifted exponential solution:

X(t)=25+(X₀−25)e^(−0.2t).

For X₀=10, this becomes X(t)=25−15e^(−0.2t). The starting value and long-run limit are both correct.

Worked example 12: evaluate the equilibrium approach

For X₀=10, at t=5 we have

X(5)=25−15e^(−1)≈25−5.5182=19.4818.

The quantity lies between its start ten and equilibrium 25. Its derivative is positive while X<25, so the direction of movement is consistent.

An initial value above equilibrium descends

If X₀=30, the same equation gives

X(t)=25+5e^(−0.2t).

The exponential term is positive and decays with time. Therefore X(t)>25 for finite t and approaches 25 from above.

The equilibrium is attractive from both sides under this simple linear model.

Worked example 13: test equilibrium direction

At X=10, dX/dt=5−2=3, so the quantity increases.

At X=30, dX/dt=5−6=−1, so it decreases.

At X=25, dX/dt=0. A small sign table explains stability without solving the full equation first.

Direction fields show the slope rule

A direction field places short line segments at selected points (t,y), each with slope prescribed by dy/dt=f(t,y).

A solution curve follows those slopes as it passes through the plane.

For an autonomous equation dy/dt=f(y), slope depends on y but not directly on t, so points at the same y-level have the same local slope.

Worked example 14: a direction field for growth

For dy/dt=0.2y:

At y=0, slope zero. At y=5, slope one. At y=10, slope two.

At y=−5, slope −1. These local slopes describe possible solutions at those levels, not separate measured data points.

A phase line focuses on equilibrium values

For an autonomous one-dimensional equation dy/dt=f(y), mark zeros of f and the sign of f in intervals between them.

If f>0, solutions move toward larger y as time increases. If f<0, they move toward smaller y.

The sign pattern can identify stable and unstable equilibria under suitable regularity conditions, even without an explicit formula.

Logistic growth adds a capacity term

A classic nonlinear equation is

dP/dt=rP(1−P/K),

where r>0 and K>0 in the stipulated model.

For 0<P<K, growth is positive. At P=K or P=0, the derivative vanishes. For P>K, the derivative is negative, pushing the quantity downward toward K in the positive-state regime.

Why logistic is not unlimited exponential growth

When P is very small compared with K, the factor (1−P/K) is near one, so growth resembles rP.

As P approaches K from below, the factor approaches zero, slowing growth.

This is an exact consequence of the chosen equation, not evidence that an actual classroom or ecosystem has a fixed carrying capacity.

Worked example 15: logistic slope at three values

Let r=0.4 and K=100.

At P=25, derivative is 0.4×25×(1−0.25)=7.5.

At P=50, derivative is 0.4×50×0.5=10.

At P=75, derivative is 0.4×75×0.25=7.5. The growth rate peaks in the middle of the idealised curve.

Find the logistic maximum growth rate

Treat the rate as a function g(P)=rP(1−P/K)=rP−(r/K)P².

This is a downward-opening quadratic in P for r,K>0. Its vertex occurs at P=K/2.

The maximum rate is rK/4. The quantity itself has not peaked there; its rate of increase has peaked.

Worked example 16: the rate maximum

For r=0.4,K=100, the midpoint is P=50.

The maximum derivative is rK/4=0.4×100/4=10 units per time unit.

At P=100, the quantity is larger but its derivative is zero. Distinguish the level of a quantity from the speed at which it changes.

Separate the logistic equation

For 0<P<K, start with

dP/[P(1−P/K)]=r dt.

Because 1−P/K=(K−P)/K, the left integrand is K/[P(K−P)].

A partial-fraction identity gives K/[P(K−P)]=1/P+1/(K−P). Integrating both terms produces logarithms.

The logistic integration step

Integrate 1/P+1/(K−P) with respect to P.

The result is ln|P|−ln|K−P|=rt+C, because the derivative of K−P is −1.

Combine logarithms:

ln|P/(K−P)|=rt+C. The signs and absolute values must be handled before choosing a branch.

Solve the logistic initial-value problem

For initial P₀ with 0<P₀<K, the ratio P/(K−P) is positive.

Exponentiate to obtain P/(K−P)=A e^(rt), where A=P₀/(K−P₀)>0.

Solving for P gives

P(t)=K/[1+(K/P₀−1)e^(−rt)].

Worked example 17: logistic from twenty

Let K=100,r=0.4 and P₀=20.

Then K/P₀−1=100/20−1=5−1=4.

Thus the solution is P(t)=100/[1+4e^(−0.4t)].

At t=0, the denominator is five, giving P=20 exactly as required.

Worked example 18: time to reach fifty

Set the logistic solution equal to 50:

50=100/[1+4e^(−0.4t)].

Then 1+4e^(−0.4t)=2, so e^(−0.4t)=1/4.

Taking logarithms gives t=ln4/0.4≈3.4657 time units. The curve reaches half the capacity at the time its growth rate is maximal.

Equilibrium solutions can be lost during separation

The logistic separation divided by P and K−P.

Therefore it omitted the possible constant solutions P=0 and P=K, which make the original derivative zero.

Those equilibria must be stated separately. The explicit formula with 0<P₀<K describes the positive interior family, not all possible initial conditions.

The logistic equilibrium at zero is unstable to small positive starts

For 0<P<K, dP/dt>0, so positive solutions move away from zero as time increases.

For P close to K from below, growth remains positive but slows; for P>K, the derivative is negative and the quantity decreases.

Thus K is attracting in the positive-state setting, while zero is unstable to small positive perturbations. These are qualitative properties of the model.

Worked example 19: above-capacity derivative

Take P=120, r=0.4, K=100.

Then dP/dt=0.4×120×(1−1.2)=48×(−0.2)=−9.6.

The model predicts an instantaneous decrease back toward its positive equilibrium. It does not mean a real system necessarily allows a population above a fixed capacity.

Logistic growth is an idealisation

Actual growth systems may have changing capacities, delays, random events, seasonal effects or multiple interacting populations.

The logistic equation assumes constant positive parameters and one specific functional relation between rate and current amount.

Its exact solution therefore establishes what the hypothetical differential equation predicts, not that every real-world growth process follows it.

Newton’s law of cooling is another separable model

A simple cooling law says the temperature difference from a fixed environment decays proportionally to that difference:

dT/dt=−k(T−T_env) with k>0.

The environment temperature is assumed constant. This is distinct from Newton–Raphson root-finding, which uses tangents to solve f(x)=0.

The two topics share Newton’s name but are mathematically different methods and models.

Worked example 20: cooling toward a fixed environment

Let T_env=20, starting T(0)=80 and k=0.2 per minute.

The temperature difference obeys d(T−20)/dt=−0.2(T−20).

Thus T(t)=20+60e^(−0.2t). At t=0 it gives 80, and as t grows it approaches 20 from above.

Worked example 21: calculate temperature after five minutes

Substitute t=5:

T(5)=20+60e^(−1).

Since e^(−1)≈0.367879, the value is 42.07 in the fictional temperature units.

The number is conditional on ideal constant-environment cooling and does not describe a real Bukit Timah room or device.

How fast does the difference shrink?

The difference D=T−T_env obeys dD/dt=−kD, so D(t)=D₀e^(−kt).

The temperature itself need not decay to zero; it approaches the environment’s level.

This is why solving for the difference from equilibrium is often simpler than solving directly for the original quantity.

Worked example 22: time to a six-degree difference

Start with difference D₀=60, rate k=0.2.

Set 60e^(−0.2t)=6, so e^(−0.2t)=0.1.

Therefore t=ln10/0.2≈11.513 time units. This is a time to approach the environment within the specified difference, not the time to reach the exact environment temperature.

Warming from below also follows the same form

If the initial T is below a constant warmer environment, then T−T_env is negative.

The formula T(t)=T_env+(T₀−T_env)e^(−kt) remains valid. The negative difference gradually approaches zero, so the temperature increases toward the environment.

The sign of the derivative follows from the sign of the difference, not from memorising that the formula is “cooling only.”

Worked example 23: warming toward twenty

Let T_env=20,T₀=10,k=0.2.

Then T(t)=20−10e^(−0.2t).

At t=5, T≈20−3.6788=16.3212, higher than the start ten but still below twenty. The same mathematical law describes approach from either side.

A constant environment is an assumption

If T_env changes over time, the equation may become

dT/dt=−k[T−T_env(t)].

That is generally no longer the same simple constant-equilibrium separation model, though it may be solvable by another method.

Using the original exponential formula with one fixed environment temperature when that parameter is actually changing can give a misleading forecast.

Differential equations can be checked by differentiation

A proposed solution must satisfy the rate equation at every relevant point, not just the initial condition.

For N(t)=100e^(0.2t), differentiate to obtain N′=20e^(0.2t).

The right side 0.2N is 0.2×100e^(0.2t)=20e^(0.2t). Both match, verifying the governing equation.

Worked example 24: verify the initial condition too

For N(t)=100e^(0.2t), N(0)=100e⁰=100.

We have already checked N′=0.2N.

Both the differential equation and initial value are satisfied. A formula meeting one but not the other is not the required particular solution.

A rate law may have units in its coefficients

If N is measured in index units and t in hours, dN/dt has index units per hour.

In dN/dt=kN, k must have units reciprocal hours so that kN has the same units as the derivative.

Dimensional consistency is a useful check in applied mathematics, even for fictional data.

Worked example 25: label the coefficient

Suppose N uses points and t uses minutes.

The equation dN/dt=0.2N needs k=0.2 per minute, not just a bare 20%.

The dimensionless exponent kt combines reciprocal-minute with minutes, yielding a dimensionless number. Exponential functions require dimensionless arguments in the standard mathematical interpretation.

An equation needs a domain

For dN/dt=kN, the mathematical solution exists over all real t for finite k and starting values.

An applied interpretation may restrict t≥0 because it describes time after an experiment begins.

For logistic models, positivity and any other feasible bounds should be stated. A mathematical extension to negative time does not automatically describe a real historical process.

A separable equation may have a singularity

Consider dy/dt=y² with initial y(0)=1.

For y≠0, separate: dy/y²=dt. Integrate to get −1/y=t+C.

Using y(0)=1 gives C=−1, so y(t)=1/(1−t). This solution becomes unbounded as t approaches one from below.

Worked example 26: finite-time blow-up

For y(t)=1/(1−t), at t=0.5 the value is two.

At t=0.9, value is ten. As t→1 from below, the function grows without bound.

The solution cannot be continued as a finite differentiable function through t=1 using that formula. Differential equations need attention to their valid intervals.

The zero solution may coexist with nonzero solutions

For dy/dt=y², the constant function y(t)=0 also satisfies the equation.

It was excluded by dividing by y² during separation.

This reinforces the earlier rule: any value making a separated denominator zero should be tested in the original differential equation for a possible equilibrium solution.

Not every first-order equation is separable

An equation such as dy/dt=t+y has the independent and dependent variables combined additively.

It cannot be written directly as g(t)h(y) by ordinary algebra alone.

Other methods, including linear integrating factors, may solve such equations. Method selection starts by identifying the form rather than forcing separation into every equation.

Worked example 27: a separable versus nonseparable comparison

Equation dy/dt=t y is separable because the right side factors as t×y.

Equation dy/dt=t+y is a linear first-order equation but is not directly separable in its given variables.

Both may be solvable analytically, yet the appropriate algebraic methods differ.

Initial conditions and uniqueness require assumptions

For many smooth right-hand-side functions f(t,y), appropriate existence and uniqueness theorems give one local solution through a specified initial point.

But not every differential equation satisfies the needed regularity conditions. Some may have multiple solutions through the same initial point.

A learner should not claim uniqueness from the mere presence of an initial condition without considering the equation’s properties.

A counterexample with nonunique solutions

Consider dy/dt=2√|y| with y(0)=0.

The zero function is a solution. For any waiting time a≥0, the piecewise function y(t)=0 for t≤a and y(t)=(t−a)² for t≥a is also a differentiable solution meeting y(0)=0.

Thus the same initial condition allows many solutions, illustrating the importance of uniqueness conditions.

Worked example 28: verify one delayed start

Choose a=1. Define y(t)=0 for t≤1 and y(t)=(t−1)² for t≥1.

For t<1 both derivative and right side are zero. For t>1, derivative 2(t−1) equals 2√[(t−1)²] because t−1≥0.

At t=1 both sides equal zero, and the derivative joins continuously. The piecewise function satisfies the equation.

Euler’s method approximates a solution numerically

When an exact solution is inconvenient, forward Euler advances by a short step h:

yₙ₊₁=yₙ+h f(tₙ,yₙ).

The current derivative supplies a straight-line estimate for the next value. Smaller steps often improve accuracy for suitable smooth problems, though they increase computation.

This is a numerical method, not the same as Euler’s polyhedron formula or the identity e^(iπ)+1=0.

Worked example 29: first Euler growth step

For dy/dt=0.2y with y(0)=100 and step h=1:

y₁=100+1×0.2×100=120.

The exact continuous solution at t=1 is 100e^0.2≈122.14. The Euler estimate is below the exact value in this increasing convex example.

Worked example 30: second Euler step

Apply the same one-unit step from y₁=120:

y₂=120+0.2×120=144.

The exact solution at t=2 is 100e^0.4≈149.18. The two-step Euler result has an absolute error of about 5.18 units.

Smaller steps change the Euler formula

With h=0.5 for dy/dt=0.2y, each step multiplies the current approximation by

1+0.2×0.5=1.1.

Four half-unit steps reach t=2. Thus the Euler estimate is 100(1.1)⁴=146.41, closer to the exact 149.18 than the two one-unit steps in this example.

A smaller step is not a universal safety guarantee

Numerical stability also depends on the equation and the method.

A step that is too large can produce oscillation or divergence in a problem whose true solution smoothly approaches equilibrium.

Reducing step size appropriately can help, but a numerical solver still needs checks on positivity, error tolerance, stiffness and valid domains.

Worked example 31: a decay step that goes negative

For dy/dt=−2y, starting y₀=1 and Euler step h=1:

y₁=1+1(−2×1)=−1.

But the exact solution y(t)=e^(−2t) remains positive. The numerical result is an unstable artefact of an overly large Euler step for this model.

Stability of forward Euler on exponential decay

For dy/dt=−ky with k>0, forward Euler gives yₙ₊₁=(1−kh)yₙ.

For the discrete approximations to decay in magnitude, we require |1−kh|<1, equivalently 0<kh<2.

For nonnegative iterates from positive starts without alternating signs, the stronger 0≤1−kh<1 gives 0<kh≤1.

Worked example 32: choose a smaller decay step

Let k=2 and h=0.25.

The Euler multiplier is 1−2×0.25=0.5.

Starting at one, the approximations are 1,0.5,0.25,0.125,…, all nonnegative and decreasing. They are not exact exponential values, but avoid the negative overshoot of h=1.

Global numerical error is distinct from local algebra error

Euler’s method approximates the continuous solution using a finite step.

Even if each arithmetic operation is performed exactly, the approximation has discretisation error because it follows tangent segments rather than the full curve.

Rounding adds another possible source of error. A careful report distinguishes mathematical method error from calculator precision.

A graph can check qualitative behaviour

For dX/dt=5−0.2X with X₀=10, the solution should rise toward equilibrium 25 without crossing it under the exact model.

A sketch showing the solution decrease away from 25 would contradict the derivative sign.

Qualitative checks can catch errors before a detailed integration is finished.

Worked example 33: verify a proposed equilibrium solution

For dX/dt=5−0.2X, test constant X(t)=25.

Its derivative is zero. The right side is 5−0.2×25=5−5=0.

Both sides match for all t. Thus constant 25 is an equilibrium solution.

The derivative is a local rule, not a causal explanation by itself

A mathematical equation can encode the assumption that rate is proportional to current quantity.

But writing such a relation does not establish what physical mechanism produced it or whether a real process follows it.

Causal claims require evidence and a justified model. This lesson’s fictitious rates are designed to teach calculus.

A solved differential equation can still be the wrong model

Suppose an actual process is affected by sudden interventions, changing parameters or time delays.

A constant-k exponential equation may then predict poorly even though its integration is correct.

Mathematical correctness of the solution and validity of the original process assumptions are independent issues.

Comparing equilibrium, growth and decay models

For dN/dt=kN with k>0 and N₀>0, the solution grows without bound as t increases.

For dC/dt=−kC, C₀>0, the solution decays toward zero.

For dX/dt=a−bX with a,b>0, the solution approaches equilibrium a/b.

For logistic growth, the positive-state solution approaches K under the specified r,K conditions. The right-hand-side sign and structure determine the qualitative behaviour.

Worked example 34: identify an equilibrium

For dY/dt=12−3Y, set the derivative to zero:

12−3Y=0, so **Y*=4**.

If Y=2, derivative is six, positive; if Y=6, derivative is minus six, negative. Thus the equation pushes values toward four under the model.

Common error 1: forget the integration constant

Integrating dy/y=k dt gives ln|y|=kt+C, not just kt.

The constant allows the family of solutions, and an initial condition selects the particular member.

Omitting it can force the starting value to one and give a formula that fails the stated initial data.

Common error 2: divide by a quantity that might be zero

Separation of dy/dt=y² through division by y² excludes y=0.

The zero solution still satisfies the original equation.

Check equilibrium values and any zero denominators before declaring the solution set complete.

Common error 3: confuse discrete compounding with instantaneous growth

The law dN/dt=0.2N gives N(t)=N₀e^(0.2t).

The recurrence Nₙ₊₁=1.2Nₙ gives Nₙ=N₀(1.2)ⁿ.

They use different mathematical assumptions. The exponentials look similar but give different values.

Common error 4: assign the wrong sign to decay

A decay law with positive k should be dC/dt=−kC for positive C.

Writing dC/dt=kC predicts growth.

Check whether the solution should move up or down before integrating, then verify the derivative sign.

Common error 5: treat equilibrium as zero quantity

In dX/dt=5−0.2X, equilibrium means the derivative is zero at X=25.

It does not mean X=0. The incoming and outgoing rates balance at a nonzero level.

Distinguish a zero rate from a zero state.

Common error 6: lose the logistic equilibria

Separating logistic growth through division by P(K−P) excludes P=0 and P=K.

Both are constant solutions of the original differential equation.

A complete response identifies these equilibria independently, even if an initial condition inside 0<P<K selects a different branch.

Common error 7: overstate the meaning of a model

An exact logistic solution for invented parameters cannot establish real student achievement trends.

A cooling formula derived for a constant ambient temperature cannot confirm the temperature of a real object without measurements.

Keep the mathematical prediction explicitly conditional on the stipulated system.

Common error 8: accept every Euler output as physically meaningful

A large forward-Euler step can turn a positive decaying quantity negative.

The numerical method may have crossed a stability or positivity limit even though the true solution stays positive.

Check step sizes, qualitative signs and errors before interpreting a numerical trajectory.

Primary preparation: changing quantities

Primary pupils can read tables in which a quantity increases by a fixed amount each step or by a fixed proportion.

They can contrast “add four” with “increase by ten percent,” laying foundations for later rate models.

Formal derivatives are not needed to recognise that the rule for change matters.

Lower Secondary: graphs and proportional patterns

Secondary learners can study gradients, functions, exponentials and the shape of growing or decaying curves.

They can interpret an equilibrium informally as a level where competing changes balance.

Separation of variables should follow differentiation, integration and logarithmic foundations at the appropriate programme level.

Advanced Mathematics and official scope

Separable differential equations, logistic equations and Euler numerical approximations are advanced calculus and modelling topics.

The SEAB 2027 G3 Mathematics and Additional Mathematics syllabus directory provides relevant Secondary prerequisites. The 2027 A-Level syllabus directory provides higher-level pathways.

This article does not claim every logistic integration or numerical stability result is examinable in a general SEC G3 course.

A twenty-minute family diagnostic

Minutes 1–5: Solve dy/dt=4 with y(0)=3, obtaining y=4t+3. Explain the role of the constant.

Minutes 6–10: Separate dy/dt=0.2y and use y(0)=100 to get 100e^(0.2t).

Minutes 11–15: Find the equilibrium of dX/dt=5−0.2X, then calculate X(t)=25−15e^(−0.2t) for starting value ten.

Minutes 16–20: Identify both equilibria of the logistic equation and explain why forward Euler with h=1 can turn dy/dt=−2y falsely negative.

Seven days of practice and transfer

Day one revises derivatives as rates. Day two integrates a constant-rate rule.

Day three separates a product differential equation. Day four models exponential growth and decay.

Day five studies stable equilibria. Day six derives and interprets the logistic form. Day seven compares exact solutions with Euler approximations, checking initial conditions, qualitative signs and step size.

Practice questions with checked answers

A. Solve dy/dt=4, y(0)=3. Answer: y=4t+3.

B. Solve dN/dt=0.2N, N(0)=100. Answer: N=100e^(0.2t).

C. Doubling time for coefficient 0.2? Answer: ln2/0.2≈3.4657.

D. Half-life for decay coefficient 0.3? Answer: ln2/0.3≈2.3105.

E. Equilibrium of dX/dt=5−0.2X? Answer: 25.

F. Logistic equilibria for dP/dt=rP(1−P/K)? Answer: 0 and K.

G. Maximum logistic rate when r=0.4,K=100? Answer: 10 at P=50.

H. Forward Euler for dy/dt=−2y, y₀=1,h=1: y₁? Answer: −1, an unstable numerical artefact for positive decay.

Frequently asked questions

What is a differential equation?

An equation involving an unknown function and one or more derivatives of that function.

What makes it separable?

Its variables can be arranged into a product form allowing one variable’s terms to be integrated separately from the other’s, where the necessary denominators are nonzero.

Why is an initial condition useful?

It identifies the particular solution from a family, when the differential equation has suitable existence and uniqueness properties.

Why does proportional growth produce an exponential?

Because e^(kt) differentiates to k times itself, matching the rule that growth rate is proportional to current amount.

Is logistic growth the same as exponential growth?

No. Logistic growth includes a capacity factor that slows growth near the equilibrium K.

What is an equilibrium?

A constant state making the differential equation’s right-hand side zero.

Can dividing during separation lose valid solutions?

Yes. Equilibria that make the divided expression zero must be checked in the original equation.

What is Euler’s method?

A numerical approximation that takes short steps along the derivative rule, using the current slope to estimate the next value.

Do the equations predict real Bukit Timah pupil progress?

No. Every rate, value and process is fictional educational Mathematics.

How can parents recognise understanding?

The learner can derive a solution, verify its derivative and initial value, identify omitted equilibria and explain why mathematical models and numerical methods have limitations.

Use the SEAB 2027 G3 Mathematics syllabus directory and 2027 A-Level syllabus directory to check the appropriate school scope. All equations and quantities here are invented teaching models, not verified local observations.

Continue with Exponential Models and Logarithmic Scales, Differentiation, Tangents and Rates, Integration and Accumulated Areas, Newton’s Method and Convergence and Taylor Series and Local Approximations for connected but distinct mathematical topics.

The immutable Clementi Secondary 1 Mathematics small-group teaching reference anchors first-principles explanation, accurate working, careful error diagnosis and independent verification.

The lasting habit is to identify the rate law and its units, check separation conditions and equilibria, apply the initial value, verify the derivative and state exactly what the mathematical model does—and does not—establish.

Companion Mathematics in Bukit Timah articles in this new four-part batch: Bézier Curves, Control Points and Smooth Geometric Design · Chinese Remainder Theorem, Congruences and Repeating Cycles · Graph Isomorphism, Degree Sequences and Network Symmetry. These new lessons extend parametric curve design, simultaneous modular constraints, network structural equivalence and calculus models, with original worked examples and explicit assumptions.