Differential Equations Notes (JC1 H2 Math) — Free PDF

Q: How do you know if a differential equation is separable?

It is separable if the right-hand side factorises into a function of x only multiplied by a function of y only, dy/dx = f(x)g(y). For example, dy/dx = x²y³ is separable, but dy/dx = x + y is not, because a sum cannot be split into a product. Such equations come with a substitution given in the question.

Q: Do I have to come up with the substitution myself in H2 Maths?

No. At H2 level the substitution is always given in the question, such as y = vx or u = x + y. Differentiate it with respect to x, substitute to eliminate y and dy/dx, solve the separable DE, then substitute back so the final answer is in x and y.

Q: How do you find the constant k in a differential equation word problem?

Write the DE with k still unknown, then substitute the one snapshot the question gives: a value of the quantity and its rate of change, with the correct sign. For example, if dV/dt = −kV and the volume is decreasing at 20 cm³ s⁻¹ when V = 400, then −20 = −400k, so k = 1 ÷ 20.

Q: How do you find the long-term behaviour of a differential equation solution?

For a solution V = Ae^(−kt) + B with k > 0, e^(−kt) → 0 as t → ∞, so V → B. State the limit in context with units, for example “the volume of water in the tank approaches 160 cm³ in the long run”. You can check B quickly by setting dV/dt = 0 in the original DE.

A differential equation (DE) is an equation that involves an unknown function and at least one of its derivatives. Solving one means finding a function, usually y in terms of x, so that the equation holds for every value of x.

Differential equations sit at the end of the H2 calculus strand (9758 topic 5.5) and draw on both differentiation and integration. They appear in almost every A-Level paper, usually as a modelling question on populations, cooling or tanks of water.

What these notes cover

•What is a differential equation?
•General and particular solutions
•Separation of variables
•The modulus when integrating 1 ÷ y
•Reduction by a given substitution
•Formulating DEs from a problem situation
•Finding k from a data point
•Long-term behaviour as t → ∞
•Sketching solution curves and families of curves
•Commenting on the suitability of a model
•Common mistakes and exam technique

Key definitions

Differential equation (DE): An equation that involves an unknown function and at least one of its derivatives. Its solution is a function, not a number.
Order: The order of the highest derivative in the equation. dy/dx = 2 − y is first-order; d²s/dt² = −10 is second-order.
General solution: A solution containing an arbitrary constant (C, or A after simplification). It represents an infinite family of solution curves, one for each value of the constant.
Particular solution: The single member of the family picked out by an initial or boundary condition, such as y = 5 when x = 1. Substituting the condition into the general solution fixes the constant.
Steady-state value: For V = Ae^(−kt) + B with k > 0, the value B that V approaches as t → ∞. V never equals B at any finite time, so V = B is a horizontal asymptote.

Key formulas

Separation of variables

$\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$

For a separable DE dy/dx = f(x)g(y), with g(y) ≠ 0. Add one constant C, on the x-side, immediately after integrating.

Log integral

$\int \frac{1}{y}\,dy = \ln|y| + C$

The modulus is compulsory. Similarly ∫ (1 ÷ (a − y)) dy = −ln|a − y| + C, with a minus sign from the chain rule.

Derivative of the substitution y = vx

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

Product rule, because v is a function of x, not a constant. The same result holds for z = y ÷ x, i.e. y = zx.

The modelling equation

$\frac{dQ}{dt} = \text{rate in} - \text{rate out}$

Q is the changing quantity (volume, mass of solute, population, temperature). Each rate is either a constant or proportional to Q.

Newton's law of cooling

$\frac{d\theta}{dt} = -k(\theta - \theta_{\text{room}})$

θroom is room temperature. The minus sign makes θ move towards θroom from either side: a hot object cools and a cold object warms.

Long-term behaviour of an exponential solution

$V = Ae^{-kt} + B,\ k > 0 \;\Rightarrow\; V \to B \text{ as } t \to \infty$

B is the steady-state value and V = B is the horizontal asymptote. Check B by setting dV/dt = 0 in the original DE.

Worked example

Water flows into a tank at a constant rate of 24 cm³ s⁻¹ and leaks out at a rate proportional to the volume, V cm³, of water in the tank at time t seconds. When the volume reaches 600 cm³, it is decreasing at a rate of 12 cm³ s⁻¹. (a) Show that 50(dV/dt) = 1200 − 3V. (b) Given that V = 1000 when t = 0, find V in terms of t.

1Rate in = 24 and rate out = kV, where k is a positive constant, so dV/dt = 24 − kV.
2When V = 600 the volume is decreasing at 12 cm³ s⁻¹, so dV/dt = −12 at that instant: −12 = 24 − 600k ⇒ 600k = 36 ⇒ k = 3 ÷ 50.
3Hence dV/dt = 24 − (3 ÷ 50)V = (1200 − 3V) ÷ 50, so 50(dV/dt) = 1200 − 3V (shown).
4Separate the variables: ∫ (50 ÷ (1200 − 3V)) dV = ∫ 1 dt, which gives −(50 ÷ 3) ln|1200 − 3V| = t + C.
5Using V = 1000 when t = 0: −(50 ÷ 3) ln|1200 − 3000| = C ⇒ C = −(50 ÷ 3) ln 1800.
6Substituting back and tidying: ln|1200 − 3V| = ln 1800 − 3t ÷ 50, so |1200 − 3V| = 1800e^(−3t ÷ 50).
7At t = 0, 1200 − 3V = −1800 < 0, and by continuity it stays negative, so |1200 − 3V| = 3V − 1200.
8Hence 3V − 1200 = 1800e^(−3t ÷ 50), giving V = 400 + 600e^(−3t ÷ 50).
9As t → ∞, V → 400, the steady state where rate in equals rate out: 24 = (3 ÷ 50) × 400.

Watch out: “Decreasing at 12 cm³ s⁻¹” means dV/dt = −12, not 12. Define k as a positive constant and let the subtraction carry the sign.

Common mistakes

Adding + C after exponentiating

Wrong: ln|y| = x², so y = e^(x²), then adding the constant: y = e^(x²) + C.

Right: Add C at the line of integration: ln|y| = x² + C ⇒ |y| = e^C e^(x²) ⇒ y = Ae^(x²). The two families are completely different sets of curves.

Dropping the modulus without a ± sign

Wrong: From |y| = e^C e^(3x), writing y = e^C e^(3x) and concluding the solution is always positive.

Right: y = ±e^C e^(3x) = Ae^(3x), where A = ±e^C. Since y = 0 is also a solution (A = 0), A may be any real number; the initial condition fixes its size and sign.

Losing the v term when differentiating y = vx

Wrong: dy/dx = x(dv/dx), treating v as a constant.

Right: v is a function of x, so use the product rule: dy/dx = v + x(dv/dx).

Questions students ask

How do you know if a differential equation is separable?

It is separable if the right-hand side factorises into a function of x only multiplied by a function of y only, dy/dx = f(x)g(y). For example, dy/dx = x²y³ is separable, but dy/dx = x + y is not, because a sum cannot be split into a product. Such equations come with a substitution given in the question.

Do I have to come up with the substitution myself in H2 Maths?

No. At H2 level the substitution is always given in the question, such as y = vx or u = x + y. Differentiate it with respect to x, substitute to eliminate y and dy/dx, solve the separable DE, then substitute back so the final answer is in x and y.

How do you find the constant k in a differential equation word problem?

Write the DE with k still unknown, then substitute the one snapshot the question gives: a value of the quantity and its rate of change, with the correct sign. For example, if dV/dt = −kV and the volume is decreasing at 20 cm³ s⁻¹ when V = 400, then −20 = −400k, so k = 1 ÷ 20.

How do you find the long-term behaviour of a differential equation solution?

For a solution V = Ae^(−kt) + B with k > 0, e^(−kt) → 0 as t → ∞, so V → B. State the limit in context with units, for example “the volume of water in the tank approaches 160 cm³ in the long run”. You can check B quickly by setting dV/dt = 0 in the original DE.

In the full PDF

•Figures of solution families and tank sketches with labelled asymptotes
•Further worked examples: direct integration, general solutions, modulus and sign resolution, and three given-substitution questions (y = vx, u = x + y, u = yx²)
•The full translation table from question wording to DEs, plus population, radioactive decay, falling-body and mixing models
•Worked examples on long-term behaviour, the sketching checklist and commenting on the suitability of a model
•All six most expensive errors with corrected working, exam technique and the 30-second final check
•End-of-chapter summary