Mathematics Tutor Singapore | JC H2 Maths differential equations become easier when students understand what a rate says about an unknown quantity. Separation of variables is not merely moving dy and dx symbols: it reflects a particular mathematical relationship and has restrictions that affect possible solutions.
Did you know that dividing by a y-dependent factor can lose an entire constant solution? Or that exponential growth and decay differ by the sign of a rate parameter but can be solved through the same separation technique? This guide explains the steps and the checks.
The worked teaching examples are aligned to the 2026 Singapore-Cambridge H2 Mathematics 9758 separable differential-equation topic, including mathematical models, initial conditions and a clearly identified enrichment comparison with logistic growth.
Browse the JC H2 differential-equations chapters
- A differential equation describes how a quantity changes
- What separable means in H2 Mathematics
- A reliable six-stage solution method
- Worked separable example: dy/dx=2xy
- General solutions and particular solutions
- A y-dependent denominator and branch restriction
- An example whose integral uses arctangent
- Why logarithmic absolute values matter
- Exponential growth from a proportional rate
- Find the doubling time of the growth model
- Exponential decay and half-life
- Newton-style cooling toward a surrounding temperature
- When does the cooled object pass a threshold?
- Constant solutions lost by dividing a factor
- Optional extension: bounded logistic growth
- A given substitution can make a differential equation separable
- Interpreting the sign and units of a rate
- Where a mathematical solution may stop applying
- The first incorrect step tells the tutor what to repair
A differential equation describes how a quantity changes
A differential equation connects an unknown function to one or more of its derivatives. Instead of being given y explicitly as a function of x, we may be given dy/dx in terms of x and y. The task is to recover functions that satisfy this rate relationship.
For example dy/dx=2x says that the rate of change of y with x is twice x. Integrating gives y=x²+C. Many curves share the same derivative, so the arbitrary constant matters until an additional condition fixes one member of the family.
In real applications, x may represent time and y temperature, population or quantity remaining. The mathematical equation alone does not establish a valid physical model; its assumptions and units must be stated.
What separable means in H2 Mathematics
The 2026 H2 Mathematics 9758 syllabus includes differential equations of the form dy/dx=f(x)g(y), including cases reduced to separable form by a given substitution. This form means the derivative can be written as a product of one factor depending on x and another depending on y.
When g(y) is nonzero on the solution interval, separate variables as [1/g(y)]dy=f(x)dx, then integrate both sides. Each integration requires care with domain, logarithms and arbitrary constants.
The symbolic dy and dx notation is a practical guide to separation, but the justification relies on the chain rule and differentiability. Dividing by a y-dependent factor can exclude constant solutions where that factor is zero, so such cases must be checked separately.
A reliable six-stage solution method
Begin by identifying the unknown function y(x) and which variable is independent. Next state the separable equation and any restrictions on the values of x or y. Then move the y-dependent factor to one side, leaving the x-dependent factor on the other.
Integrate both sides, including a constant of integration. Simplify only where the steps remain valid and check the original domain restrictions. If an initial condition is given, substitute it to determine the constant and obtain a particular solution.
Finally differentiate the proposed solution and insert it back into the original differential equation. A correct differentiation check detects a lost factor or sign much faster than trusting an attractive exponential answer.
For a physical model, the final stage is interpreting the units, sign of the derivative and sensible time domain. A calculated mathematical function need not remain physically meaningful outside the assumptions of the word problem.
| Stage | Question |
|---|---|
| Identify | What is y and what does x or t represent? |
| Separate | Can dy/dx=f(x)g(y) be rearranged safely? |
| Integrate | What is the general antiderivative? |
| Constant | Is an initial condition supplied? |
| Check | Does differentiation recreate the original rate? |
| Interpret | Are signs, units and domain sensible? |
Worked separable example: dy/dx=2xy
Solve dy/dx=2xy with initial condition y(0)=3. When y≠0 we can write dy/y=2x dx. Integrating gives ln|y|=x²+C. Exponentiating gives |y|=e^C e^(x²), so nonzero solutions have the form y=Ae^(x²) for a nonzero signed constant A.
The missing equilibrium solution y=0 also satisfies the differential equation and can be included by allowing A=0 in the final expression. We therefore write the full family as y=Ae^(x²), with real A.
Apply y(0)=3: 3=Ae^0=A, so the particular solution is y=3e^(x²). Differentiate: y’=3e^(x²)×2x=2x(3e^(x²))=2xy, as required.
At x=1, this solution gives 3e≈8.155. Its derivative is 6e, positive at x=1. The rate is not constant; it increases with x and y according to the equation.
General solutions and particular solutions
A general solution represents a family of functions containing at least one arbitrary constant in a first-order problem. A particular solution uses information such as y(0)=3 to select one function from that family.
If a student integrates dy/dx=4x and writes only y=2x², they have omitted the family constant C. That answer satisfies the differential equation, but it is not its entire general solution. The general solution is y=2x²+C.
Once an initial condition is supplied, it must be applied to the integrated relation, not inserted blindly into the derivative. For example if y(1)=7, then 7=2(1)²+C, so C=5 and y=2x²+5.
A good tutor checks whether the question asks for the general solution, the particular solution or a value at a specified time. The required final response changes with that instruction.
A y-dependent denominator and branch restriction
Consider dy/dx=2x/y with y>0 and y(0)=3. Separate to obtain y dy=2x dx. Integrating gives y²/2=x²+C, or y²=2x²+K for a renamed constant K.
At x=0, y=3, so K=9. The equation y²=2x²+9 has two algebraic square-root branches, but the given condition y>0 selects the positive branch: y=√(2x²+9).
Differentiate: y’=2x/√(2x²+9)=2x/y, verifying the equation. A student who writes both positive and negative branches despite the positive initial condition has not applied the stated domain.
Here y never equals zero along the positive branch, so the division by y is valid for the entire real x-domain of this particular solution.
An example whose integral uses arctangent
Solve dy/dx=3x²(1+y²) with initial condition y(0)=0. Separate: dy/(1+y²)=3x² dx. Integrating gives arctan y=x³+C.
At x=0 and y=0, C=0. Therefore y=tan(x³) on a continuous interval around zero where the tangent is defined and the inverse relationship corresponds to the selected branch.
Locally, the natural interval containing x=0 is |x³|<π/2. At its boundary, tan(x³) is undefined, so we should not claim one globally continuous solution across each singularity.
Differentiate y=tan(x³): dy/dx=sec²(x³)×3x²=3x²[1+tan²(x³)]=3x²(1+y²). The original relation is verified.
Why logarithmic absolute values matter
For a real nonzero y, the integral of 1/y with respect to y is ln|y|+C. The absolute value accommodates both positive and negative solution branches on intervals where y does not cross zero.
If a student writes ln y without a given condition y>0, they have implicitly excluded negative values that might satisfy the equation. That can matter in purely mathematical problems even if some physical models require positivity.
After exponentiating ln|y|=F(x)+C, the sign of y is captured in a signed multiplicative constant A. It is often useful to express the solution as y=Ae^{F(x)} and then check whether A=0 gives another permitted equilibrium solution.
Exponential growth from a proportional rate
Suppose a hypothetical quantity P(t) has rate of change proportional to its current amount: dP/dt=kP, with positive rate constant k. Separating and integrating gives P(t)=P₀e^(kt) when P(0)=P₀.
Take P₀=80 units and k=0.2 per hour. The model is P(t)=80e^(0.2t). After five hours it gives 80e≈217.46 units. The increase is not a fixed amount per hour: it is a fixed proportional instantaneous rate.
The units of k are reciprocal time, such as h⁻¹, because dP/dt has units of quantity per hour while P has units of quantity. A student who reports k in hours rather than inverse hours has lost dimensional meaning.
This is an idealised model. A real population or manufacturing process may not grow at the same proportional rate indefinitely. The interpretation should remain conditional on the assumptions.
Find the doubling time of the growth model
In the model P(t)=80e^(0.2t), the doubling time satisfies 160=80e^(0.2t). Divide by eighty to get 2=e^(0.2t), then take natural logs: ln2=0.2t.
Therefore t=ln2/0.2≈3.466 hours. This doubling time does not depend on the starting amount when k remains constant, because the initial amount cancels from the ratio.
Check by substitution: at that t, e^(0.2t)=2 and P(t)=160. Do not round the exponent too early when finding a threshold time, especially if the problem asks for a required precision.
Students should distinguish an exact expression ln2/0.2 from a decimal approximation. Both may be appropriate under different question instructions.
Exponential decay and half-life
A decaying quantity can satisfy dQ/dt=−kQ with k>0. Integration yields Q(t)=Q₀e^(−kt). The negative derivative means the quantity decreases over time while remaining positive for a positive starting value.
Take Q(0)=200 and k=0.3 h⁻¹. Then Q(t)=200e^(−0.3t). After two hours Q≈200e^(−0.6)≈109.76 units. It is smaller than the starting value, as required.
The half-life T solves Q(T)=Q₀/2, giving e^(−kT)=1/2 and T=ln2/k. With k=0.3, T≈2.3105 hours.
Half-life is the time required for the amount to halve under a constant proportional decay model. It is not a statement that a fixed number of units disappears every hour.
| Model | Initial quantity | Constant | Value at t |
|---|---|---|---|
| Growth | 80 | k=+0.2 h⁻¹ | 80e^(0.2t) |
| Decay | 200 | k=0.3 h⁻¹ | 200e^(−0.3t) |
| Cooling toward 20°C | 80°C | k=0.1 min⁻¹ | 20+60e^(−0.1t) |
Newton-style cooling toward a surrounding temperature
A common idealised model assumes the rate of temperature change is proportional to the difference from a fixed surrounding temperature Tₐ. Write dT/dt=−k(T−Tₐ), with positive k and temperatures measured consistently.
Suppose the surroundings are 20°C, the object starts at 80°C and k=0.1 min⁻¹. Separating yields T−20=Ce^(−0.1t). Using T(0)=80 gives C=60, so T(t)=20+60e^(−0.1t).
At t=10 minutes, T≈20+60/e≈42.07°C. The temperature approaches 20°C as t grows, but this idealised continuous model does not make it exactly equal to 20 at any finite positive time.
The model assumes constant surroundings and proportional heat transfer. Real heat flow can be more complex; this is an illustrative mathematical model, not a measurement claim about a particular physical object.
When does the cooled object pass a threshold?
Continue T(t)=20+60e^(−0.1t). Find the first time when T≤30°C under the model. The inequality 20+60e^(−0.1t)≤30 becomes e^(−0.1t)≤1/6.
Taking natural logs gives −0.1t≤−ln6. Dividing by the negative number −0.1 reverses the inequality, so t≥10ln6≈17.918 minutes.
This question combines differential equations, logarithms and inequality direction. The temperature starts above thirty and decreases toward twenty, so the threshold should be reached at one finite positive time and remain below it afterwards.
A student who reports t≤17.918 has mishandled the negative division even though their differentiation and integration may be correct.
Constant solutions lost by dividing a factor
Consider dy/dt=y(1−y/10). If y is identically zero, the right-hand side is zero and y’=0, so y=0 is a solution. Likewise the constant function y=10 satisfies the equation.
If we separate by dividing through by y(1−y/10), we have implicitly required y≠0 and y≠10. Those two constant equilibrium solutions must be identified separately.
This is a conceptual check especially useful before any advanced logistic integration. One may study such a model as enrichment; the main H2 syllabus technique is recognising separable equations and interpreting the resulting solutions.
Every zero of the y-dependent factor g(y) is a candidate constant solution because y’=f(t)g(y) vanishes for a constant y at that root. Check it in the original equation rather than assuming division preserved it.
Optional extension: bounded logistic growth
Suppose a purely illustrative bounded-growth model satisfies dP/dt=0.2P(1−P/100), with P(0)=20. Its equilibrium values are zero and one hundred. For an initial amount between zero and one hundred, the derivative is positive and the model approaches the upper limit.
Separating variables and using partial fractions gives ln[P/(100−P)]=0.2t+C on the branch 0
At t=5, P=100/(1+4/e)≈40.46. As t increases, the exponential factor tends to zero and P approaches one hundred, unlike unbounded P₀e^(kt) growth.
This is an enrichment comparison, not a claim that every Secondary or JC learner must memorise a logistic closed form for 9758. The useful conceptual contrast is whether the rate remains proportional to the amount alone or also decreases near a limiting capacity.
A given substitution can make a differential equation separable
The 2026 H2 syllabus permits reducing a given differential equation to separable form by means of a given substitution. As an illustrative case, consider dy/dx=(x+y)/x for x>0 and let v=y/x be the given substitution.
Then y=xv and, by the product rule, dy/dx=v+x dv/dx. The original right-hand side is 1+v, so v+x dv/dx=1+v and therefore dv/dx=1/x.
Integrate: v=ln x+C. Substitute v=y/x back to obtain y=x(ln x+C) for x>0. If y(1)=2, then C=2 and y=x(ln x+2).
Differentiate the last expression to check: y’=ln x+3. Meanwhile (x+y)/x=1+ln x+2=ln x+3, so the proposed particular solution satisfies the original equation.
Interpreting the sign and units of a rate
A differential equation such as dQ/dt=−0.3Q says the quantity decreases whenever Q>0. A positive rate coefficient in dP/dt=0.2P gives increasing P for positive P. The sign of the derivative is part of the meaning, not a decoration.
If t is measured in hours, a proportionality constant multiplied by a quantity to produce its hourly rate has units h⁻¹. If t is minutes, the corresponding units are min⁻¹; mixing them without conversion changes the model’s rate by a factor of sixty.
A final quantity P(t) may have units of people, grams or arbitrary counts. The constant in an exponent such as kt must make the exponent dimensionless. This is a useful sense-check for real-context calculus.
Where a mathematical solution may stop applying
A differential equation can have a mathematical solution on an interval wider or narrower than the real story permits. In a cooling model time may be restricted to t≥0. In a population model the quantity may be interpreted as nonnegative and continuous even though real populations are discrete counts.
Some separated solutions have singularities. The solution tan(x³) cannot be continued through a point where the tangent becomes undefined while still remaining an ordinary differentiable function there. Restricting to a valid interval is part of a complete mathematical answer.
Other models use rates that change with season, external inputs or constraints, and may no longer fit one constant coefficient. The equation is only as useful as the assumptions under which it was formulated.
The first incorrect step tells the tutor what to repair
One student may write dy/y=2x dx and integrate ln y=x² but forget the constant. Another may find the correct general solution but fail to use the given initial value. A third may apply an exponent with units that do not match the stated time measurement.
Those are distinct difficulties: integration technique, constant selection and modelling interpretation. A good H2 Maths lesson should identify which occurred before assigning more questions.
Teachers should also check whether division by g(y) discarded a constant equilibrium solution. This is a mathematical reasoning issue rather than a careless algebraic slip.
Twenty original differential-equations questions with answers
1. Interpret a rate
dy/dx=2x says the derivative of y with respect to x is twice x.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
2. Recover a simple function
dy/dx=4x gives general solution y=2x²+C.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
3. Apply an initial condition
For y=2x²+C and y(1)=7, C=5.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
4. Separate a product
dy/dx=2xy becomes dy/y=2x dx when y≠0.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
5. General separable solution
dy/dx=2xy gives y=Ae^(x²) for real A, including A=0.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
6. Particular solution
With y(0)=3, solution to dy/dx=2xy is y=3e^(x²).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
7. Differentiate to check
Derivative of 3e^(x²) is 6xe^(x²)=2xy.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
8. Positive branch
dy/dx=2x/y, y(0)=3, y>0 gives y=√(2x²+9).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
9. Arctangent integration
dy/dx=3x²(1+y²), y(0)=0 gives local solution y=tan(x³).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
10. Constant equilibrium
For y’=y(1−y/10), y=0 and y=10 are constant solutions.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
11. Exponential growth
P’=0.2P, P(0)=80 gives P(t)=80e^(0.2t).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
12. Growth after five hours
80e^(0.2×5)=80e≈217.46 units.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
13. Doubling time
For constant rate 0.2 per hour, doubling time ln2/0.2≈3.466 h.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
14. Exponential decay
Q’=−0.3Q, Q(0)=200 gives 200e^(−0.3t).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
15. Decay after two hours
200e^(−0.6)≈109.76 units.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
16. Half-life
For proportional decay k=0.3 h⁻¹, half-life ln2/0.3≈2.3105 h.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
17. Cooling model
T’=−0.1(T−20), T(0)=80 gives T=20+60e^(−0.1t).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
18. Cooling at ten minutes
T(10)=20+60/e≈42.07°C.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
19. Cooling threshold
T≤30 in that model first occurs at t=10ln6≈17.918 minutes.
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
20. Given substitution
For dy/dx=(x+y)/x, x>0, v=y/x given, the solution is y=x(ln x+C).
Check the initial value, relevant interval and sign or units of the rate. If dividing by a y-dependent expression, inspect possible constant solutions in the original differential equation. The final result should answer the precise quantity requested.
Nine H2 differential-equation misconceptions
| First wrong move | Underlying issue | Repair |
|---|---|---|
| Leaves out +C after integration | General family lost | Differentiate multiple candidates |
| Uses initial condition on y’ rather than y | Condition misinterpreted | State y(t₀) |
| Omits |y| for ln integration | Negative branch excluded | Discuss nonzero y |
| Divides by y and loses y=0 | Equilibrium excluded | Check g(y)=0 |
| Gets e^(−kt) for positive growth | Sign of rate confused | Predict direction first |
| Gives wrong exponent units | Rate parameter dimension misunderstood | Make kt dimensionless |
| Reports negative time for cooling threshold | Inequality sign or log mishandled | Check monotone physical trend |
| Uses principal square root despite negative initial y | Branch restriction overlooked | Use initial sign |
| Ignores domain near tangent singularity | Interval of validity missed | Locate undefined points |
Ten-session JC1–JC2 revision progression
| Session | Concept | Independent evidence |
|---|---|---|
| 1 | Meaning of dy/dx | Interpret rate |
| 2 | Separate product f(x)g(y) | Valid rearrangement |
| 3 | Integrate logarithmic and polynomial forms | Includes +C |
| 4 | Apply initial conditions | Find particular solution |
| 5 | Check by differentiation | Recovers rate |
| 6 | Exponential growth and doubling time | Units and log method |
| 7 | Decay and half-life | Correct negative exponent |
| 8 | Cooling toward an equilibrium | Correct ambient shift |
| 9 | Equilibrium and given substitution | Keeps domain |
| 10 | Delayed mixed modelling question | Method independent |
A student who can integrate but cannot identify a model’s initial condition needs a different lesson from one who never recognises a separable equation. A useful tutor can explain the choice of model, the integration, the constant and the final physical interpretation without assuming they are one skill.
Frequently asked questions
What is a differential equation?
An equation connecting an unknown function with its derivatives and possibly the independent variable.
What does separable mean?
The derivative can be written as f(x)g(y), allowing x-dependent and y-dependent factors to be integrated on different sides where division is valid.
Why is there a constant C?
Indefinite integration recovers a family of functions with the same derivative.
How is a particular solution obtained?
Use the given initial or boundary condition to determine the family constant.
Why does proportional growth produce an exponential?
Solving P’=kP gives P=P₀e^(kt), a function whose rate is proportional to its current value.
What is a half-life?
The time required for an exponentially decaying quantity to halve, ln2/k under constant positive decay parameter k.
Can division by y make a solution disappear?
Yes. The zero solution can be excluded when dividing by y, so it should be checked separately in the original equation.
Are logistic growth equations all compulsory in H2 Maths?
This is an enrichment comparison. The 2026 syllabus specifies the separable first-order form and may provide a substitution; it does not require every advanced growth model to be memorised.
Does the H2 syllabus include modelling?
Yes. Section 5.5 includes formulating and interpreting differential equations from problem situations.
Can learning differential equations guarantee an A-Level grade?
No. It develops one important calculus topic within a much broader H2 Mathematics syllabus.
Official 2026 syllabus and further Maths reading
The official 2026 H2 Mathematics 9758 syllabus lists differential equations in Section 5.5. It includes finding general and particular solutions of dy/dx=f(x)g(y), using a given substitution to reduce an equation to that form, formulating an equation from a problem and interpreting its solution. Students should consult the examination-year specification rather than assuming university-level methods are compulsory.
Continue through the Mathematics Learning Hub, JC H2 Calculus Optimisation and Related Rates, JC H2 Definite Integration, and JC H2 Differential Equations and Modelling in Bukit Timah.
The unchanged Secondary 1 Mathematics Tutor Clementi small-group reference remains the original teaching standard. For a parent or JC student discussion, contact eduKate Singapore with one correctly separated equation and one example where the model, constant or domain failed.