Edexcel IAL Maths
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What a differential equation is; general solutions (one arbitrary constant) and particular solutions
Solving dy/dx=f(x)g(y) by separating the variables, including equations that give ln|y|
Separated integrals that need partial fractions, a standard result from the formulae booklet, or integration by parts
Forming differential equations from statements about rates: proportional growth and decay
Newton's law of cooling and draining tanks
Limited (logistic) growth
Finding constants from data, interpreting solutions, long-term behaviour and the limits of a model
1. A differential equation links a function and its derivative; the general solution of a first-order equation contains one arbitrary constant, and a given condition such as y=2 when x=1 fixes it to give a particular solution.
2. Separate: dy/dx=f(x)g(y)⇒∫1/g(y) dy=∫f(x) dx. Add one constant, on the x side, as soon as you integrate.
3. Log constants: ln|y|=f(x)+c⇒y=Aeᶠ⁽ˣ⁾, with A=±eᶜ. Never y=eᶠ⁽ˣ⁾+c.
4. Forming: "rate of change of N" is dN/dt; "proportional to" brings in a constant k; a decreasing quantity gets a minus sign with k>0. Newton's law of cooling: dθ/dt=-k(θ-θₛ).
5. Interpret: use two pieces of data to find c and k; for the long term let t→∞ (e⁻ᵏᵗ→0); and say when the model stops being valid.
First-order differential equation. An equation connecting x, y and the first derivative dy/dx. A solution is a relationship between x and y, with no derivative, that makes the equation true.
Solving means integrating once, so the general solution contains one arbitrary constant. Each value of the constant gives a different curve: a whole family. A boundary or initial condition ("y=2 when x=1", "initially the temperature is 85 °C") picks out one member, the particular solution.
The equation itself tells you the gradient at every point. Drawing a short segment with that gradient at many points gives a slope field, and every solution curve follows the segments. To verify a solution, differentiate it (implicitly if it is not in the form y=…) and show that the result matches the equation.
WORKED EXAMPLE · QUESTION
(a) Show that y³=3x³+C, where C is a constant, satisfies the differential equation dy/dx=3x²/y². (b) Find the particular solution for which y=2 when x=1. (c) Use your answer to (b) to find y when x=2, giving your answer to 3 significant figures.
SOLUTION
(a) Differentiate implicitly: 3y²dy/dx=9x², so dy/dx=9x²/3y²=3x²/y² ✓.
(b) Substitute x=1, y=2: 8=3+C, so C=5 and y³=3x³+5.
(c) y³=24+5=29, so y=∛29=3.0723…= 3.07.
An equation is separable if the right-hand side factorises into a function of x times a function of y. Then each variable can be moved to its own side and integrated on its own.
dy/dx=f(x) g(y) ⟹ ∫1/g(y) dy=∫f(x) dx
Divide by g(y), so every y is on the same side as dy and every x goes with dx. Integrate both sides and add one constant c (on the x side) straight away; then use the condition, then rearrange if asked. Not in the formulae booklet: learn the method.
Factorise first: dy/dx=xy+x=x(y+1) is separable. dy/dx=x+y is not, and cannot be solved this way.
WORKED EXAMPLE · QUESTION
Given that dy/dx=(2x+1)e⁻ʸ and that y=0 when x=0, (a) find y in terms of x; (b) find the exact value of y when x=2, and give it to 3 significant figures.
SOLUTION
(a) e⁻ʸ=1÷eʸ, so multiply both sides by eʸ: ∫eʸ dy=∫(2x+1) dx, giving eʸ=x²+x+c. Use x=0, y=0: e⁰=1=c. So eʸ=x²+x+1 and y=ln(x²+x+1).
(b) y=ln(4+2+1)= ln 7=1.9459…=1.95.
x²+x+1=(x+½)²+¾>0, so the logarithm is defined for every x.
Common error
Taking ln of each term separately: from eʸ=x²+x+1, writing y=ln x²+ln x+ln 1. The logarithm applies to the whole right-hand side: y=ln(x²+x+1).
Whenever g(y)=y (or y+b), the y side integrates to ln|y|. Exponentiate the whole right-hand side: ln|y|=f(x)+c⇒|y|=eᶠ⁽ˣ⁾⁺ᶜ=eᶜ eᶠ⁽ˣ⁾. Replace ±eᶜ by a single constant A: y=Aeᶠ⁽ˣ⁾. Equivalently, write the constant as ln A from the start: ln y=f(x)+ln A. In context questions the quantity is usually positive (a mass, a population), so the modulus can be dropped; say so.
WORKED EXAMPLE · QUESTION
The variables x and y satisfy dy/dx=ysin x, with y>0, and y=3 when x=0. (a) Show that y=3e^(1-cos x). (b) Find the greatest value of y, giving the exact value and the value to 3 significant figures.
SOLUTION
(a) Separate: ∫1/y dy=∫sin x dx, so ln y=-cos x+c (y>0, so no modulus). Exponentiate the whole right-hand side: y=eᶜe^(-cos x)=Ae^(-cos x). Use x=0, y=3 (cos 0=1): 3=Ae⁻¹, so A=3e and y=3e×e^(-cos x)=3e^(1-cos x) ✓.
(b) The power 1-cos x is greatest when cos x=-1: y_(max)=3e²≈22.2 (at x=π,3π,…).
Common error
Writing ln y=-cos x+c as y=e^(-cos x)+c. The constant is part of the power, so it becomes a multiplier: y=Ae^(-cos x). Putting x=0 into the wrong form gives a different, incorrect curve.
After separating, either side may be a rational function such as 2/(x(x+2)) or 1/(P(6-P)). Split it into partial fractions (P4.1) so each piece integrates to a logarithm. Watch the sign: ∫1/(6-P) dP=-ln|6-P|, because the derivative of 6-P is -1. Collect the logarithms into one with the laws of logs, then exponentiate: ln y=ln x/(x+2)+c⇒y=Ax/(x+2). Other separated integrals may need a standard result from the formulae booklet (∫tan x dx=ln|sec x|) or integration by parts (∫xeˣ dx): see P4.5.
WORKED EXAMPLE · QUESTION
(a) Express 2/(x(x+2)) in partial fractions. (b) Given that dy/dx=2y/(x(x+2)) for x>0, y>0, and that y=6 when x=1, find y in terms of x. (c) Describe what happens to y as x becomes large.
SOLUTION
(a) 2≡P(x+2)+Qx; x=0 gives P=1, x=-2 gives Q=-1: 1/x-1/(x+2).
(b) Separate and use (a): ∫1/y dy=∫(1/x-1/(x+2))dx, so ln y=ln x-ln(x+2)+ln A. Combine: ln y=ln Ax/(x+2), so y=Ax/(x+2). With x=1, y=6: 6=A/3, A=18, so y=18x/(x+2).
(c) Divide top and bottom by x: y=18/(1+2/x)→18 as x→∞. y increases towards 18 but never reaches it.
Exam tip
Writing the constant as ln A when every other term is a logarithm lets you combine everything into one log in a single line. Either form of the constant is correct; choose the one that makes the rearranging shortest.
Identify the quantity and the time variable: "the rate of change of the mass m" is dm/dt. "Proportional to" gives a constant: proportional to m is km; inversely proportional to m is k/m; proportional to the square root of h is k√h. If the quantity decreases, write a minus sign and state k>0, so that dm/dt=-km is negative whenever m>0. "Proportional to the difference between θ and 20" gives k(θ-20); "to the product of P and 6-P" gives kP(6-P). When the rate is given for a different quantity (volume, when the question wants depth), connect them with the chain rule first (P4.4).
| Description | Differential equation | Shape of the solution |
|---|---|---|
| N increases at a rate proportional to N | dN/dt=kN, k>0 | N=N₀eᵏᵗ: exponential growth |
| m decreases at a rate proportional to m | dm/dt=-km, k>0 | m=m₀e⁻ᵏᵗ: decay towards 0 |
| temperature θ falls at a rate proportional to its excess over the surroundings at θₛ | dθ/dt=-k(θ-θₛ) | θ=θₛ+(θ₀-θₛ)e⁻ᵏᵗ: levels off at θₛ |
| Description | Differential equation | Shape of the solution |
|---|---|---|
| depth h falls at a rate proportional to √h | dh/dt=-k√h | √h falls linearly: the tank empties in a finite time |
| P grows at a rate proportional to P(L-P) | dP/dt=kP(L-P) | S-shaped curve levelling off at L |
WORKED EXAMPLE · QUESTION
A drug is injected into a patient's bloodstream. The mass of the drug, m mg, decreases at a rate proportional to the mass present. Initially there is 40 mg; 3 hours later there is 25 mg. (a) Write down a differential equation for m in terms of the time t hours. (b) Show that m=40e⁻ᵏᵗ, where k is a positive constant. (c) Show that k=⅓ln 1.6. (d) Find the time at which the mass has fallen to 5 mg. (e) Find the time taken for the mass to halve.
SOLUTION
(a) Decreasing, so a minus sign with k>0: dm/dt=-km.
(b) Separate (m>0): ∫1/m dm=∫-k dt, so ln m=-kt+c and m=Ae⁻ᵏᵗ. At t=0, m=40, so A=40: m=40e⁻ᵏᵗ ✓.
(c) Use t=3, m=25: e⁻³ᵏ=(25/40), so e³ᵏ=1.6 and k=⅓ln 1.6 ✓ (=0.1567…).
(d) 40e⁻ᵏᵗ=5⇒eᵏᵗ=8⇒t=ln 8/k=3ln 8/ln 1.6=13.273…= 13.3 hours.
(e) eᵏᵗ=2⇒t=3ln 2/ln 1.6= 4.42 hours, the same from any starting mass.
Newton's law of cooling. The rate at which a body's temperature changes is proportional to the difference between its temperature and the temperature of its surroundings. With surroundings at θₛ: dθ/dt=-k(θ-θₛ), k>0.
A hot body (θ>θₛ) cools; a cold body (θ<θₛ) warms up; the same equation covers both. Separate with θ-θₛ as one block: ∫1/(θ-θₛ) dθ=ln|θ-θₛ|. The temperature approaches θₛ but never reaches it in the model, so a "long-term" answer is θₛ.
WORKED EXAMPLE · QUESTION
A bowl of soup at 85 °C is placed in a room at a constant 20 °C. The temperature of the soup, θ °C, t minutes later satisfies dθ/dt=-k(θ-20), where k is a positive constant. After 10 minutes the temperature is 60 °C. (a) Show that θ=20+65e⁻ᵏᵗ. (b) Find the exact value of k. (c) Find the time at which the soup reaches 35 °C. (d) State what happens to θ in the long term.
SOLUTION
(a) Separate (θ>20 throughout): ∫1/(θ-20) dθ=∫-k dt, so ln(θ-20)=-kt+c and θ-20=Ae⁻ᵏᵗ. At t=0, A=85-20=65: θ=20+65e⁻ᵏᵗ ✓.
(b) t=10, θ=60: 65e⁻¹⁰ᵏ=40⇒e¹⁰ᵏ=(13/8)⇒ k=(1/10)ln(13/8) (=0.04855…).
(c) 65e⁻ᵏᵗ=15⇒eᵏᵗ=(13/3)⇒t=(10ln(13/3))/(ln(13/8))=30.20…= 30.2 minutes.
(d) e⁻ᵏᵗ→0 as t→∞, so θ→20: the soup cools towards room temperature but stays above it.
Common error
Using θ=85e⁻ᵏᵗ. That model cools towards 0 °C, not to room temperature. The exponential describes the excess temperature θ-20, which starts at 65, not 85.
Water flowing out through a hole near the base leaves faster when the water is deeper. A common model: the depth falls at a rate proportional to √h, so dh/dt=-k√h. Separate with a power: ∫h^(-1/2) dh=2√h. The solution has √h falling linearly, so, unlike exponential decay, the tank empties in a finite time. The formula for h is a squared bracket, which would start rising again after the tank is empty, so state the interval of validity.
WORKED EXAMPLE · QUESTION
Water leaks from a tank through a small hole in its base. The depth of water is h metres at time t minutes, and the depth decreases at a rate proportional to the square root of the depth. Initially the depth is 1.44 m; after 10 minutes it is 1.00 m. (a) Write down a differential equation for h. (b) Find the general solution. (c) Find the values of the constants, and hence express h in terms of t. (d) Find the time taken for the tank to empty. (e) State, with a reason, the interval of t for which your formula is valid.
SOLUTION
(a) dh/dt=-k√h, k>0.
(b) Separate: ∫h^(-1/2) dh=∫-k dt, so 2√h=-kt+c.
(c) t=0, h=1.44: c=2×1.2=2.4. t=10, h=1: 2=-10k+2.4, so k=0.04. Then √h=1.2-0.02t and h=(1.2-0.02t)².
(d) h=0: 1.2-0.02t=0⇒ t=60 minutes.
(e) 0≤t≤60: after 60 minutes the tank is empty, but (1.2-0.02t)² would increase again.
Exam tip
With two unknown constants (c and k), use the initial condition first: at t=0 the kt term vanishes and c comes out alone. Then the second data point gives k.
Unlimited growth dP/dt=kP is unrealistic for a population with limited food or space. Multiplying by (L-P) makes the growth rate fall to zero as P approaches the limit L. Separating gives ∫1/(P(L-P)) dP: split into partial fractions, integrate to logs (watch the minus sign on ln|L-P|), combine and exponentiate. Make P the subject by collecting the P terms on one side and factorising. The answer usually comes out in the form L/(1+Be⁻ᶜᵗ), which shows P→L at once.
WORKED EXAMPLE · QUESTION
The population of fish in a lake, P hundred, t years after the lake is stocked, is modelled by dP/dt=(P(6-P))/15, 0<P<6. Initially there are 100 fish. (a) Express 6/(P(6-P)) in partial fractions. (b) Show that ln(5P/(6-P))=0.4t. (c) Hence show that P=6/(1+5e^(-0.4t)). (d) Find the time taken for the population to reach 400 fish. (e) State the long-term size of the population according to the model.
SOLUTION
(a) 6≡a(6-P)+bP; P=0 gives a=1, P=6 gives b=1: 1/P+1/(6-P).
(b) Separate: 15/(P(6-P))=(5/2)×6/(P(6-P)), so ∫(5/2)(1/P+1/(6-P))dP=∫1 dt. The second term gives -ln(6-P): (5/2)[ln P-ln(6-P)]=t+c, so ln P/(6-P)=0.4t+c′. At t=0, P=1 (100 fish): c′=ln⅕. So ln P/(6-P)-ln⅕=0.4t, i.e. ln(5P/(6-P))=0.4t ✓.
(c) Remove the log: 5P=(6-P)e^(0.4t), so P(5+e^(0.4t))=6e^(0.4t) and P=6e^(0.4t)/(5+e^(0.4t))=6/(1+5e^(-0.4t)) ✓ (divide top and bottom by e^(0.4t)).
(d) 400 fish means P=4: 1+5e^(-0.4t)=1.5⇒e^(-0.4t)=0.1⇒t=2.5ln 10= 5.76 years.
(e) e^(-0.4t)→0, so P→6: 600 fish.
Common error
Integrating 1/(6-P) as +ln(6-P). The derivative of 6-P is -1, so the integral is -ln(6-P). With the wrong sign the solution has no limiting value and part (e) cannot come out as 6.
Exam tip 1 — separate fully before you integrate
Write a line of the form ∫(only y) dy=∫(only x) dx with both integral signs showing. Factorise the right-hand side first if needed: xy-2x=x(y-2).
Exam tip 2 — the constant goes in as soon as you integrate
Add +c on the line where the integral signs disappear. Adding it after rearranging (for example after exponentiating) puts it in the wrong place, and the particular solution that follows is wrong.
Exam tip 3 — keep constants exact
Leave k as ⅓ln 1.6 or (1/10)ln(13/8) and use the exact value (or the full calculator value) in later parts. Rounding k to 2 significant figures early can move a final answer by a whole unit in the third figure.
Exam tip 4 — check the sign and the units
For a decreasing quantity the equation must make the derivative negative. After solving, check: does the formula give the initial value at t=0? Does it move in the right direction? State answers with units and in the words of the question ("600 fish", "30.2 minutes").
Exam tip 5 — "long term" and "comment on the model"
For long-term behaviour let t→∞ and say which term tends to zero. For a comment on the model, link it to the context: a population cannot be a fraction of a fish, a tank cannot have negative depth, room temperature might not stay constant.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| ln y=2x+c gives y=e²ˣ+c. | e²ˣ⁺ᶜ=eᶜe²ˣ: the constant multiplies. | y=Ae²ˣ. |
| Each side needs its own constant. | Two constants combine into one, c₂-c₁. | One constant, on one side, is enough for a first-order equation. |
| ∫1/(6-P) dP=ln(6-P). | The inner derivative of 6-P is -1. | -ln|6-P|+c. |
| "Decreases at a rate proportional to m" is dm/dt=km. | With k>0 this makes m increase. | dm/dt=-km, k>0. |
| Newton's law of cooling gives θ=θ₀e⁻ᵏᵗ. | That cools towards 0 °C. | θ-θₛ decays: θ=θₛ+(θ₀-θₛ)e⁻ᵏᵗ. |
| A solution formula holds for all t≥0. | The context can stop the model, e.g. an empty tank. | State the interval of validity (0≤t≤60 in worked example 7). |
P4.1, P4.4 and P4.5
The separated integrals use partial fractions (P4.1) and substitution or integration by parts (P4.5); connected rates of change (P4.4) turn a rate for volume into a rate for depth before you separate.
P3.5 Exponential and logarithmic functions
eˣ, ln x, ∫1/x dx=ln|x| and the exponential growth and decay models of P3 are exactly the solutions of dN/dt=±kN.
Mechanics and IAL Physics
Variable acceleration links displacement, velocity and acceleration by differentiation and integration with respect to time, the same "rate of change" language. Radioactive decay (dN/dt=-λN), capacitor discharge and half-life are proportional-decay differential equations with exponential solutions.
| Formula | Use | In the formulae list? |
|---|---|---|
| dy/dx=f(x)g(y)⇒∫1/g(y) dy=∫f(x) dx | separating the variables | No — learn it |
| ∫1/x dx=ln|x|+c; ∫f′(x)/f(x) dx=ln|f(x)|+c | the y side of most model equations | No — learn it |
| ln|y|=f(x)+c⇒y=Aeᶠ⁽ˣ⁾ | removing the logarithm | No — derive it |
| dN/dt=kN⇒N=N₀eᵏᵗ | proportional growth (k>0) or decay (k<0) | No — derive it |
| dθ/dt=-k(θ-θₛ)⇒θ=θₛ+(θ₀-θₛ)e⁻ᵏᵗ | Newton's law of cooling | No — derive it |
| dh/dt=-k√h⇒2√h=-kt+c | draining tank | No — derive it |
| ∫tan x dx=ln|sec x|+c; ∫udv/dx dx=uv-∫vdu/dx dx | harder separated integrals | Yes |
1. How many arbitrary constants does the general solution of a first-order differential equation contain, and how is it removed?
2. Solve dy/dx=2y, given that y=5 when x=0.
3. The volume V of a melting block of ice decreases at a rate proportional to V^(2/3). Write down a differential equation.
4. A model gives θ=15+50e^(-0.1t). What are the initial and long-term temperatures?
5. Is dy/dx=xy-3x separable?
1. One. A given pair of values (an initial or boundary condition) fixes it and gives the particular solution.
2. ∫1/y dy=∫2 dx, so ln|y|=2x+c and y=Ae²ˣ. With y=5 at x=0, A=5: y=5e²ˣ.
3. dV/dt=-kV^(2/3), where k is a positive constant (minus because V decreases).
4. At t=0, θ=15+50=65; as t→∞, e^(-0.1t)→0, so θ→15, the temperature of the surroundings.
5. Yes: xy-3x=x(y-3), so ∫1/(y-3) dy=∫x dx, giving y=3+Ae^(x²/2).
Check understanding
The answer and each step appear one click at a time.
The variables x and y, where y > 0, satisfy the differential equation dy/dx = 4x³ ÷ y.
(a) Find the general solution, giving your answer in the form y² = f(x). [3]
(b) Find the particular solution for which y = 3 when x = 1. [2]
(c) Hence find the value of y when x = 2, giving your answer to 3 significant figures. [1]
Mark scheme click to reveal each step
∫ y dy = ∫ 4x³ dx
½y² = x⁴ + c
y² = 2x⁴ + C
9 = 2 + C
y² = 2x⁴ + 7
y = √39 = 6.24 (6.2449…)
A curve satisfies the differential equation dy/dx = 2xy.
(a) Show that y = Aeˣ², where A is a constant, satisfies the differential equation. [2]
(b) Given that the curve passes through (0, 5), find the value of A. [1]
(c) Find the positive value of x for which y = 20, giving your answer to 3 significant figures. [2]
Mark scheme click to reveal each step
dy/dx = A × 2xeˣ² (chain rule)
= 2x(Aeˣ²) = 2xy as required
5 = Ae⁰, so A = 5
eˣ² = 4 ⇒ x² = ln 4
x = √(ln 4) = 1.18 (1.1774…)
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