First-Order Differential Equations, Separation of Variables & Modelling
Differential equations are the language of mathematical modelling in science, engineering, and finance. Whenever a rate of change is described in words, such as 'the rate of cooling of a liquid is proportional to the temperature difference' or 'the rate of population growth is proportional to both the current population and the remaining carrying capacity' - it translates directly into a first-order differential equation. At A-Level, you must formulate differential equations from connected rates of change, solve them by separating variables (often using partial fractions or reverse chain rule), and apply boundary conditions to find particular solutions.
- Translating Word Statements into Differential Equations
- 'The rate of increase of P with respect to t is directly proportional to Q' translates to dP/dt = kQ (where k > 0). 'The rate of decrease of V is proportional to V' translates to dV/dt = -kV (where k > 0). When multiple effects act simultaneously, for instance, water flows into a tank at a constant rate of 20 cm³ s⁻¹ and leaks out at a rate proportional to the depth h - combine the rates with signs: dV/dt = 20 - kh.
- Connected Rates of Change (Chain Rule Formulation)
- Often a question gives the rate of change of volume dV/dt (e.g., for a sphere, cone, or hemisphere) and asks you to show that the radius r or depth h satisfies a differential equation in dr/dt or dh/dt. Express V purely in terms of r (or h using similar triangles for a cone where r / h = R / H), differentiate to find dV/dr, and apply the Chain Rule: dV/dt = (dV/dr) × (dr/dt), rearranging to isolate dr/dt.
- Separation of Variables
- A first-order differential equation of the form dy/dx = f(x)g(y) is solved by separating the variables so all y-terms are multiplied by dy on the left and all x-terms are multiplied by dx on the right: ∫ [1 / g(y)] dy = ∫ f(x) dx. You may only move y-factors across by multiplication and division, never by addition or subtraction. Include a single constant of integration (+ C) on the right-hand side immediately after integrating.
Definite Integral & Trapezium Rule Step-by-Step Visualizer
Choose a function, slide the limits a and b to see the area under the curve update live, and compare Exact Integration vs the Trapezium Rule.
Step-by-Step Working for ∫03 (1x²) dx
- Step 1, Find the antiderivative F(x):F(x) = [(1/3)x³] from x = 0 to x = 3
- Step 2: Substitute upper limit (b = 3) minus lower limit (a = 0):F(3) - F(0) = (9.0000) - (0.0000) = 9.0000
- Step 3, Compare with Trapezium Rule (4 strips, h = 0.750):Area ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)] = 9.2813
Difference from exact integral: 3.13% (increase
nstrips to make the approximation more accurate!).
+ C on indefinite integrals, and use radians whenever integrating trigonometric functions like sin(x) or cos(x)!Key Concepts & Exam Definitions (6)
- Translating Word Statements into Differential Equations
- 'The rate of increase of P with respect to t is directly proportional to Q' translates to dP/dt = kQ (where k > 0). 'The rate of decrease of V is proportional to V' translates to dV/dt = -kV (where k > 0). When multiple effects act simultaneously, for instance, water flows into a tank at a constant rate of 20 cm³ s⁻¹ and leaks out at a rate proportional to the depth h - combine the rates with signs: dV/dt = 20 - kh.
- Connected Rates of Change (Chain Rule Formulation)
- Often a question gives the rate of change of volume dV/dt (e.g., for a sphere, cone, or hemisphere) and asks you to show that the radius r or depth h satisfies a differential equation in dr/dt or dh/dt. Express V purely in terms of r (or h using similar triangles for a cone where r / h = R / H), differentiate to find dV/dr, and apply the Chain Rule: dV/dt = (dV/dr) × (dr/dt), rearranging to isolate dr/dt.
- Separation of Variables
- A first-order differential equation of the form dy/dx = f(x)g(y) is solved by separating the variables so all y-terms are multiplied by dy on the left and all x-terms are multiplied by dx on the right: ∫ [1 / g(y)] dy = ∫ f(x) dx. You may only move y-factors across by multiplication and division, never by addition or subtraction. Include a single constant of integration (+ C) on the right-hand side immediately after integrating.
- General Solutions vs Particular Solutions
- The General Solution of a first-order differential equation contains one arbitrary constant (such as + C, or a multiplicative constant A = e^C when simplifying logarithmic solutions: ln|y| = kx + C → y = e^(kx + C) = e^C e^(kx) = A e^(kx)). A Particular Solution is obtained by substituting a given initial or boundary condition (e.g., y = y₀ when x = x₀) to determine the numerical value of C or A.
- Newton's Law of Cooling and Exponential Models
- Newton's Law of Cooling states that the rate of change of temperature θ of an object is proportional to the difference between its temperature and the ambient room temperature θ₀: dθ/dt = -k(θ - θ₀) (where k > 0). Separating variables gives ∫ [1 / (θ - θ₀)] dθ = ∫ -k dt → ln|θ - θ₀| = -kt + C → θ - θ₀ = A e^(-kt) → θ = θ₀ + A e^(-kt).
- Logistic Population Growth (Partial Fractions in DEs)
- In a logistic growth model with carrying capacity M, the rate of growth is proportional to both P and (M - P): dP/dt = kP(M - P). Separating variables gives ∫ [1 / (P(M - P))] dP = ∫ k dt. To integrate the left-hand side, you must decompose 1 / [P(M - P)] into partial fractions: (1/M)[1/P + 1/(M - P)]. Integrating gives (1/M)[ln|P| - ln|M - P|] = kt + C (remembering the minus sign from integrating 1 / (M - P)!), which combines to ln|P / (M - P)| = Mkt + C'.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
How do you integrate ∫ [1 / (a - by)] dy when separating variables?
How do you rewrite ln|y| = 2x + c in the form y = f(x) with a multiplicative constant A?
Why can you not separate variables in dy/dx = x + y by writing dy - y dx = x dx?
In a conical water tank of total height H = 6 m and top radius R = 2 m, how do you eliminate r from V = (1/3)πr²h before differentiating?
What is the difference between a general solution and a particular solution of a differential equation?
In the logistic differential equation dP/dt = 0.02 P(500 - P), what happens to P as t → ∞ and for what value of P is the growth rate dP/dt maximum?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Find the particular solution of the differential equation (1 + x²) dy/dx = 2xy², given that y = 1 when x = 0. Give your answer in the form y = f(x). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Separate the variables so all y-terms are with dy and all x-terms are with dx: ∫ (1 / y²) dy = ∫ [2x / (1 + x²)] dx [1 mark]. 2. Integrate the left-hand side (writing 1 / y² as y^(-2)): ∫ y^(-2) dy = -1 / y [1 mark]. 3. Integrate the right-hand side by recognising the f'(x) / f(x) form: ∫ [2x / (1 + x²)] dx = ln(1 + x²) + C [1 mark]. 4. Substitute the initial condition x = 0, y = 1 to find C: -1 / 1 = ln(1 + 0) + C → C = -1 [1 mark]. 5. Rearrange -1 / y = ln(1 + x²) - 1 to make y the subject: 1 / y = 1 - ln(1 + x²) → y = 1 / [1 - ln(1 + x²)] [1 mark].
- Q2
A population of bacteria P (in thousands) at time t hours is modelled by the differential equation dP/dt = (1/2) P(4 - P). Initially, at t = 0, P = 1. Use partial fractions to solve the differential equation and show that P = 4 / (1 + 3e^(-2t)). State the limiting size of the population as t → ∞. (7 marks)
Click “Check Mark Scheme” to compare your answer, 1. Separate the variables: ∫ [2 / (P(4 - P))] dP = ∫ 1 dt (or ∫ [1 / (P(4 - P))] dP = ∫ (1/2) dt) [1 mark]. 2. Express 2 / [P(4 - P)] in partial fractions: 2 / [P(4 - P)] ≡ A / P + B / (4 - P) → A = 1/2, B = 1/2; multiplying both sides of the DE by 2 gives ∫ [1/P + 1/(4 - P)] dP = ∫ 2 dt [1 mark]. 3. Integrate both sides (remembering the minus sign from 4 - P): ln|P| - ln|4 - P| = 2t + C [1 mark]. 4. Combine logarithms using the division law: ln|P / (4 - P)| = 2t + C [1 mark]. 5. Substitute t = 0, P = 1: ln(1 / 3) = C, so ln[P / (4 - P)] - ln(1/3) = ln[3P / (4 - P)] = 2t [1 mark]. 6. Exponentiate and rearrange to isolate P: 3P / (4 - P) = e^(2t) → 3P = (4 - P)e^(2t) → P(3 + e^(2t)) = 4e^(2t) → divide top and bottom by e^(2t) to get P = 4 / (1 + 3e^(-2t)) [1 mark]. 7. As t → ∞, e^(-2t) → 0, so P → 4 (i.e. 4,000 bacteria) [1 mark].
- Q3
Water is poured into an inverted right circular cone of height 30 cm and top radius 10 cm at a constant rate of 8π cm³ s⁻¹. At the same time, water leaks out of a tiny hole at the vertex at a rate of (1/2)πh cm³ s⁻¹, where h cm is the depth of water at time t seconds. Show that dh/dt = (72 - 9h / 2) / h² (or (144 - 9h) / (2h²)). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Form a rate equation for the net rate of change of water volume V: dV/dt = Rate In - Rate Out = 8π - (1/2)πh [1 mark]. 2. Use similar triangles to express the water surface radius r in terms of depth h: r / h = 10 / 30 = 1 / 3 → r = h / 3 [1 mark]. 3. Substitute r = h / 3 into the volume of a cone formula V = (1/3)πr²h: V = (1/3)π(h / 3)²h = (1/27)πh³ [1 mark]. 4. Differentiate V with respect to h: dV/dh = (3/27)πh² = (1/9)πh² [1 mark]. 5. Apply the Chain Rule dV/dt = (dV/dh)(dh/dt): (1/9)πh² (dh/dt) = 8π - (1/2)πh → divide by (1/9)πh² to get dh/dt = (144 - 9h) / (2h²) [1 mark].
- Q4
Find the general solution of the differential equation cos(x) (dy/dx) + y sin(x) = cos³(x), or solve dy/dx = y tan(x) for 0 ≤ x < π/2, giving y in terms of x. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. For dy/dx = y tan(x), separate the variables: ∫ (1 / y) dy = ∫ tan(x) dx [1 mark]. 2. Integrate both sides (using ∫ tan(x) dx = ln|sec x| = -ln|cos x|): ln|y| = ln|sec x| + C [1 mark]. 3. Let C = ln(A) and combine logs: ln|y| = ln|A sec x| [1 mark]. 4. Exponentiate to obtain the general solution: y = A sec(x) (or y = A / cos x) [1 mark]. (Note: for cos(x) dy/dx + y sin(x) = cos³(x), dividing by cos²x gives d/dx(y sec x) = cos x → y sec x = sin x + C → y = sin(x)cos(x) + C cos(x)).
Common Exam Questions (4)
When should I substitute the initial condition (x₀, y₀) to find the constant C?
Usually it is quickest and least prone to algebra errors to substitute (x₀, y₀) immediately after integrating both sides (while the equation is still in logarithmic form), rather than waiting until after you have rearranged for y.
How do I get from P = 4e^(2t) / (3 + e^(2t)) to the required show-that form P = 4 / (1 + 3e^(-2t))?
Multiply both the numerator and the denominator of the fraction by e^(-2t) (which is equivalent to dividing top and bottom by e^(2t)). Since e^(2t) × e^(-2t) = e⁰ = 1, the top becomes 4 and the bottom becomes 3e^(-2t) + 1.
What does it mean if a question asks for the depth h when the water level is stationary (at equilibrium)?
You do not need to solve the differential equation. A stationary level (equilibrium) occurs when dh/dt = 0 (Rate In = Rate Out). Simply set the numerator of dh/dt equal to zero and solve for h.
Why did I lose a mark when integrating ∫ [1 / (50 - y)] dy?
Because the coefficient of y is -1, ∫ [1 / (50 - y)] dy = -ln|50 - y|, not +ln|50 - y|. Missing that minus sign turns a decaying exponential into an exploding one and ruins the rearrangement.
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