Skip to content

Edexcel A-level Further Maths revision notes

Differential equations

Section CP-9
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
9 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section CP-9

Checked against Edexcel 9FM0 section CP-9. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

In the exam: Formulae booklet provided · calculator allowed in every paper

Open the printable pack
CP-9.1

Find and use an integrating factor to solve differential equations of form dy/dx + P(x)y = Q(x) and recognise when it is appropriate to do so.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A first-order linear differential equation must first be written as y+P(x)y=Q(x)y'+P(x)y=Q(x). Its integrating factor is I(x)=eP(x)dxI(x)=e^{\int P(x)\,dx}, which may be quoted.
  • Multiplying every term by II makes the left side the product derivative (Iy)(Iy)', so integration gives Iy=IQdx+CIy=\int I Q\,dx+C and division by II gives the general solution.
  • This method is appropriate only when yy and yy' occur linearly and P,QP,Q depend on the independent variable alone.
  • If the coefficient of yy' is not 11, divide through before forming II.
  • Examiners expect the integrating factor, the product-derivative step and the arbitrary constant to be shown.
Worked example

Solve xdydx+2y=x3x\dfrac{dy}{dx}+2y=x^3 for x>0x>0, given that y(1)=2y(1)=2.

  1. 1.Divide by xx: y+2xy=x2y'+\dfrac2x y=x^2, so I=e2/xdx=x2I=e^{\int2/x\,dx}=x^2.
  2. 2.Multiply through: (x2y)=x4(x^2y)'=x^4.
  3. 3.Integrate: x2y=x55+Cx^2y=\dfrac{x^5}{5}+C, hence y=x35+Cx2y=\dfrac{x^3}{5}+Cx^{-2}.
  4. 4.y(1)=2y(1)=2 gives C=95C=\dfrac95.

Answer: y=x35+95x2y=\dfrac{x^3}{5}+\dfrac{9}{5x^2} for x>0x>0.

Common mistakes

  • Don't use e2dxe^{\int2\,dx} before dividing the equation by xx.
  • Don't multiply only the yy-terms by the integrating factor and leave Q(x)Q(x) unchanged.
  • Don't integrate (Iy)(Iy)' but omit the arbitrary constant.

Exam tip

Write the equation in monic linear form before stating the integrating factor; this secures the method logic.

Tier 1 · Easy

ORIGINAL

1.

Find an integrating factor for dydx+3y=x\dfrac{dy}{dx}+3y=x.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

For x>0x>0, solve dydx+2xy=x2\dfrac{dy}{dx}+\dfrac2x y=x^2.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

On 0x<π/20\leq x<\pi/2, solve dydx+2tanxy=sinx\dfrac{dy}{dx}+2\tan x\,y=\sin x subject to y(0)=2y(0)=2. Hence find y(π/3)y(\pi/3).

(7)

(Total for Question 1 is 7 marks)

Your progress and exam materials

This section: Evidence from your answers: 0/9 secureYour confidence: 0 self-rated secureTracker status: 0/9 secure, 0 shaky, 9 unseen

Overall: Evidence from your answers: 0/116 secureYour confidence: 0 self-rated secureTracker status: 0/116 secure, 0 shaky, 116 unseen

Progress is saved on this device for guests and accounts right now; cross-device account sync is not live yet.

CP-9.2

Find both general and particular solutions to differential equations.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A general solution represents a family of solution curves and contains the correct number of independent arbitrary constants: normally one for a first-order equation and two for a second-order equation. A particular solution results when initial or boundary conditions determine all those constants.
  • Conditions involving derivatives must be applied to the derivative of the complete general solution.
  • The phrase particular integral has a different meaning: it is one trial response to a non-homogeneous forcing term, not a solution selected by initial data.
  • The specification also expects sketches of members of a solution family.
  • Examiners therefore look for constants in the general solution, correct use of every condition, and curve features such as common asymptotes or the effect of changing the constant.
Several members of a one-parameter solution family approaching a common oblique asymptote.
Worked example

The general solution is y=Ce2x+x12y=Ce^{-2x}+x-\tfrac12. Find the member through (0,3)(0,3) and state its oblique asymptote.

  1. 1.Substitute x=0x=0, y=3y=3: 3=C123=C-\dfrac12.
  2. 2.Therefore C=72C=\dfrac72.
  3. 3.As xx\to\infty, Ce2x0Ce^{-2x}\to0, so the curve approaches y=x12y=x-\dfrac12.

Answer: y=72e2x+x12y=\dfrac72e^{-2x}+x-\dfrac12, with oblique asymptote y=x12y=x-\dfrac12.

Common mistakes

  • Don't call a complementary-function-plus-particular-integral expression a particular solution before applying conditions.
  • Don't apply a derivative condition to yy rather than first differentiating the full general solution.
  • Don't sketch every family member through the same point despite the arbitrary constant changing the intercept.

Exam tip

For a family-sketch instruction, identify invariant features and show how the arbitrary constant changes the curves.

Tier 1 · Easy

ORIGINAL

1.

Find the general solution of dydx=4x3\dfrac{dy}{dx}=4x^3, then find the particular solution for which y(1)=3y(1)=3.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Solve yy=0y''-y'=0 subject to y(0)=2y(0)=2 and y(0)=3y'(0)=-3.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Find the particular solution of y+4y=12xy''+4y=12x for which y(0)=1y(0)=1 and y(0)=0y'(0)=0.

(7)

(Total for Question 1 is 7 marks)

CP-9.3

Use differential equations in modelling in kinematics and in other contexts.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A differential-equation model begins by defining the independent variable, dependent variable, units and any positive direction. Translate each verbal rate separately before combining them.
  • In kinematics, v=ds/dtv=ds/dt and a=dv/dta=dv/dt; a resistance force acts opposite to the velocity, so its sign follows the chosen direction.
  • In mixing, the rate of change is input rate minus output rate, with concentration equal to amount divided by volume.
  • After solving, constants, equilibrium values and limiting behaviour must be interpreted in context.
  • Examiners also expect modelling assumptions to be recognised, such as constant coefficients, perfect mixing or a resistance law valid over the stated range, and any non-physical prediction to be identified.
Worked example

A particle falls from rest and satisfies dvdt=120.3v\dfrac{dv}{dt}=12-0.3v, with downward positive. Find v(t)v(t) and the terminal speed.

  1. 1.Rewrite as v+0.3v=12v'+0.3v=12; the equilibrium value satisfies 0.3v=120.3v=12, so it is 4040.
  2. 2.The general solution is v=40+Ce0.3tv=40+Ce^{-0.3t}.
  3. 3.v(0)=0v(0)=0 gives C=40C=-40.
  4. 4.As tt\to\infty, the exponential term tends to zero.

Answer: v=40(1e0.3t)v=40(1-e^{-0.3t}), and the terminal speed is 4040 in the model's speed units.

Common mistakes

  • Don't assign the resistance term the same sign as the velocity under the chosen positive direction.
  • Don't use the incoming concentration for both the input and output rates in a mixing model.
  • Don't report a limiting value without interpreting it as an equilibrium or terminal quantity.

Exam tip

A modelling answer should define signs and units before forming the equation and interpret the limiting value afterwards.

Tier 1 · Easy

ORIGINAL

1.

A body at temperature TT is in a room maintained at 18C18^{\circ}\text{C}. State a differential equation expressing that its cooling rate is proportional to its excess temperature, where k>0k>0.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A particle falls from rest. Taking downward as positive, its speed satisfies dvdt=100.5v\dfrac{dv}{dt}=10-0.5v. Find v(t)v(t) and the terminal speed predicted by the model.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A well-mixed bioreactor contains 120120 litres of liquid and initially 1212 grams of dissolved nutrient. Solution enters and leaves at 33 litres per minute, keeping the volume constant. The incoming concentration is 0.50.5 grams per litre. Form and solve a differential equation for the nutrient mass MM grams, find when M=36M=36, and state one modelling assumption.

(9)

(Total for Question 1 is 9 marks)

CP-9.4

Solve differential equations of form y'' + ay' + by = 0 where a and b are constants by using the auxiliary equation.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For the homogeneous constant-coefficient equation y+ay+by=0y''+ay'+by=0, substitute y=emxy=e^{mx} to form the auxiliary equation m2+am+b=0m^2+am+b=0.
  • Its two roots determine two independent solution terms.
  • Distinct real roots r,sr,s give Aerx+BesxAe^{rx}+Be^{sx}; a repeated root rr gives (A+Bx)erx(A+Bx)e^{rx}; conjugate roots p±iqp\pm iq give epx(Acosqx+Bsinqx)e^{px}(A\cos qx+B\sin qx).
  • Initial conditions are imposed only after the complete real general solution and any needed derivatives have been written.
  • Examiners expect the auxiliary equation, its roots and the matching form of solution, rather than unsupported answers or unresolved complex exponentials.
Worked example

Solve y2y+10y=0y''-2y'+10y=0 subject to y(0)=1y(0)=1 and y(0)=5y'(0)=5.

  1. 1.The auxiliary equation m22m+10=0m^2-2m+10=0 has roots 1±3i1\pm3i.
  2. 2.Thus y=ex(Acos3x+Bsin3x)y=e^x(A\cos3x+B\sin3x).
  3. 3.y(0)=1y(0)=1 gives A=1A=1.
  4. 4.y(0)=A+3B=5y'(0)=A+3B=5 gives B=43B=\dfrac43.

Answer: y=ex(cos3x+43sin3x)y=e^x\left(\cos3x+\dfrac43\sin3x\right).

Common mistakes

  • Don't write the auxiliary coefficient of mm as bb instead of aa.
  • Don't use Aerx+BerxAe^{rx}+Be^{rx} for a repeated root, because this loses independence.
  • Don't convert roots p±iqp\pm iq to trigonometric terms but omit the factor epxe^{px}.

Exam tip

State the root type immediately after solving the auxiliary equation, then write the corresponding real solution form.

Tier 1 · Easy

ORIGINAL

1.

Find the general solution of y+5y+6y=0y''+5y'+6y=0.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the general solution of y6y+9y=0y''-6y'+9y=0.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Solve 2y+4y+10y=02y''+4y'+10y=0 subject to y(0)=2y(0)=2 and y(0)=2y'(0)=2.

(6)

(Total for Question 1 is 6 marks)

CP-9.5

Solve differential equations of form y'' + ay' + by = f(x) by solving the homogeneous case and adding a particular integral to the complementary function (f(x) polynomial, exponential or trigonometric).

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For y+ay+by=f(x)y''+ay'+by=f(x), the general solution is the complementary function plus a particular integral: y=yc+ypy=y_{\mathrm c}+y_{\mathrm p}. The specification uses forcing terms kepxke^{px}, linear or quadratic polynomials, and mcosωx+nsinωxm\cos\omega x+n\sin\omega x.
  • Choose a trial from the same family with enough coefficients to reproduce every term after differentiation.
  • If that trial overlaps the complementary function, multiply the whole trial by xx until it is independent; a repeated overlap may require x2x^2.
  • Substitute to determine the coefficients, combine CF and PI, then apply conditions.
  • Examiners expect resonance to be handled explicitly and conditions to be applied to the complete solution.
Worked example

Find the general solution of y4y+4y=6e2x+3xy''-4y'+4y=6e^{2x}+3x.

  1. 1.The auxiliary equation (m2)2=0(m-2)^2=0 gives yc=(A+Bx)e2xy_{\mathrm c}=(A+Bx)e^{2x}.
  2. 2.Because e2xe^{2x} corresponds to a repeated root, try yp1=Cx2e2xy_{p1}=Cx^2e^{2x}; substitution gives 2Ce2x=6e2x2Ce^{2x}=6e^{2x}, so C=3C=3.
  3. 3.For 3x3x, try yp2=px+qy_{p2}=px+q; substitution gives 4px+(4q4p)=3x4px+(4q-4p)=3x, so p=34p=\dfrac34, q=34q=\dfrac34.

Answer: y=(A+Bx)e2x+3x2e2x+34x+34y=(A+Bx)e^{2x}+3x^2e^{2x}+\dfrac34x+\dfrac34.

Common mistakes

  • Don't use Ce2xCe^{2x} as the trial when e2xe^{2x} is already a repeated complementary-function term.
  • Don't use a constant trial for a linear forcing term and cannot match the coefficient of xx.
  • Don't apply initial conditions before adding the particular integral to the complementary function.

Exam tip

Before choosing a PI trial, compare every forcing term with the auxiliary roots and mark any overlap.

Tier 1 · Easy

ORIGINAL

1.

Find the general solution of yy2y=6y''-y'-2y=6.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the general solution of y4y=8e2xy''-4y=8e^{2x}.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Solve y+2y+5y=10cosxy''+2y'+5y=10\cos x subject to y(0)=1y(0)=1 and y(0)=0y'(0)=0.

(9)

(Total for Question 1 is 9 marks)

CP-9.6

Understand and use the relationship between the cases when the discriminant of the auxiliary equation is positive, zero and negative and the form of solution of the differential equation.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For the auxiliary equation m2+am+b=0m^2+am+b=0, the discriminant Δ=a24b\Delta=a^2-4b determines the form of the differential-equation solution. If Δ>0\Delta>0, two distinct real roots give two exponential terms.
  • If Δ=0\Delta=0, the repeated root rr gives (A+Bx)erx(A+Bx)e^{rx}. If Δ<0\Delta<0, roots p±iqp\pm iq give epx(Acosqx+Bsinqx)e^{px}(A\cos qx+B\sin qx).
  • The imaginary part qq produces oscillation; the real part pp controls exponential growth or decay.
  • Examiners expect parameter ranges to follow from a correctly simplified discriminant and the transition case to use the repeated-root form.
  • A negative real root alone does not imply oscillation.
Worked example

Classify the solutions of y+2ky+9y=0y''+2ky'+9y=0 for k>0k>0 and write the solution when the damping is critical.

  1. 1.The auxiliary discriminant is (2k)236=4(k29)(2k)^2-36=4(k^2-9).
  2. 2.It is negative for 0<k<30<k<3, zero for k=3k=3, and positive for k>3k>3.
  3. 3.At k=3k=3, the repeated auxiliary root is 3-3.

Answer: The roots are complex for 0<k<30<k<3, repeated for k=3k=3, and distinct real for k>3k>3; at k=3k=3, y=(A+Bx)e3xy=(A+Bx)e^{-3x}.

Common mistakes

  • Don't use a2+4ba^2+4b instead of a24ba^2-4b for the auxiliary discriminant.
  • Don't write two identical exponential terms when the discriminant is zero.
  • Don't say that any negative auxiliary root makes the solution oscillatory.

Exam tip

For a parameter classification, solve the discriminant inequalities and state the equality case separately.

Tier 1 · Easy

ORIGINAL

1.

For y+4y+8y=0y''+4y'+8y=0, state the sign of the auxiliary discriminant and hence write the general solution.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Given k>0k>0, find the value of kk for which y+ky+16y=0y''+ky'+16y=0 has a repeated auxiliary root. Write the corresponding general solution.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Classify the auxiliary roots of y+2(p+1)y+(p2+4p+8)y=0y''+2(p+1)y'+(p^2+4p+8)y=0 for all real pp. Also give the general solution at the transition value.

(6)

(Total for Question 1 is 6 marks)

CP-9.7

Solve the equation for simple harmonic motion x'' = -omega^2 x and relate the solution to the motion.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Simple harmonic motion about equilibrium satisfies x=ω2xx''=-\omega^2x: acceleration is proportional to displacement and directed towards equilibrium.
  • Its general solution is x=Acosωt+Bsinωtx=A\cos\omega t+B\sin\omega t, equivalently x=Rcos(ωt+ϕ)x=R\cos(\omega t+\phi).
  • The amplitude is R=A2+B2R=\sqrt{A^2+B^2}, the period is T=2π/ωT=2\pi/\omega, maximum speed is ωR\omega R, and maximum acceleration magnitude is ω2R\omega^2R.
  • At equilibrium the speed magnitude is greatest and acceleration is zero; at an extreme position the speed is zero and acceleration magnitude is greatest.
  • Examiners expect initial conditions to determine phase and amplitude and signs to be interpreted using the stated positive direction.
Displacement in simple harmonic motion, showing amplitude, equilibrium crossings and one period.
Worked example

A particle satisfies x=25xx''=-25x, with x(0)=3x(0)=3 and x(0)=20x'(0)=-20. Find its displacement, amplitude, period and maximum speed.

  1. 1.ω=5\omega=5, so x=Acos5t+Bsin5tx=A\cos5t+B\sin5t.
  2. 2.x(0)=3x(0)=3 gives A=3A=3, while x(0)=5B=20x'(0)=5B=-20 gives B=4B=-4.
  3. 3.R=32+(4)2=5R=\sqrt{3^2+(-4)^2}=5, T=2π/5T=2\pi/5, and vmax=ωR=25v_{\max}=\omega R=25.

Answer: x=3cos5t4sin5tx=3\cos5t-4\sin5t, amplitude 55, period 2π/52\pi/5, maximum speed 2525.

Common mistakes

  • Don't use ω=25\omega=25 after comparing x=25xx''=-25x with x=ω2xx''=-\omega^2x.
  • Don't calculate amplitude as A+BA+B instead of A2+B2\sqrt{A^2+B^2}.
  • Don't place maximum acceleration at equilibrium rather than at the extreme positions.

Exam tip

When asked to relate the solution to motion, state where speed and acceleration attain their extreme values.

Tier 1 · Easy

ORIGINAL

1.

A particle satisfies x=9xx''=-9x. Write its general displacement and state its period.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

An SHM particle satisfies x=4xx''=-4x, with x(0)=4x(0)=4 and x(0)=6x'(0)=6. Find its displacement, amplitude, period and maximum speed.

(6)

(Total for Question 1 is 6 marks)

Tier 3 · Hard

ORIGINAL

1.

A particle in SHM satisfies x=16xx''=-16x, x(0)=3/2x(0)=\sqrt3/2 and x(0)=2x'(0)=-2. Express xx in the form Rcos(4t+ϕ)R\cos(4t+\phi), find the first positive time at which it crosses equilibrium moving in the negative direction, and find its acceleration when x=1/2x=-1/2.

(8)

(Total for Question 1 is 8 marks)

CP-9.8

Model damped oscillations using second order differential equations and interpret their solutions.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A free damped oscillator can be modelled by x+2λx+ω02x=0x''+2\lambda x'+\omega_0^2x=0 with λ>0\lambda>0, where resistance is proportional to the derivative of displacement. Complex roots give underdamped motion: oscillations within a decaying exponential envelope.
  • A repeated negative root gives critical damping, the fastest non-oscillatory return to equilibrium.
  • Distinct negative real roots give overdamping and a slower non-oscillatory return.
  • A periodic driving force on the right produces forced vibration; the complementary function is the transient response and the particular integral is the persistent steady response.
  • Examiners expect classification from the auxiliary roots and an interpretation of decay, oscillation, frequency and long-term motion.
An underdamped displacement oscillating inside a decaying exponential envelope.
Worked example

Classify x+8x+25x=0x''+8x'+25x=0 and write the form of its general solution.

  1. 1.The auxiliary equation is m2+8m+25=0m^2+8m+25=0.
  2. 2.Its roots are m=4±3im=-4\pm3i, so the motion is underdamped.
  3. 3.The negative real part supplies the decaying envelope and the imaginary part supplies angular frequency 33.

Answer: x=e4t(Acos3t+Bsin3t)x=e^{-4t}(A\cos3t+B\sin3t); the motion oscillates with exponentially decreasing amplitude.

Common mistakes

  • Don't classify complex roots as overdamping because their real part is negative.
  • Don't interpret the complementary function as the complete long-term motion in a forced-vibration equation.
  • Don't use displacement rather than velocity in the linear resistance term.

Exam tip

For an interpret instruction, connect the real root part to decay and the imaginary part to oscillation.

Tier 1 · Easy

ORIGINAL

1.

Classify the motion governed by x+6x+25x=0x''+6x'+25x=0 and write its general solution.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the positive value of cc for which x+cx+9x=0x''+cx'+9x=0 is critically damped. For this value, solve the equation when x(0)=2x(0)=2 and x(0)=4x'(0)=-4, and interpret the long-term motion.

(7)

(Total for Question 1 is 7 marks)

Tier 3 · Hard

ORIGINAL

1.

A forced damped oscillator satisfies x+4x+20x=10cos(2t)x''+4x'+20x=10\cos(2t), with x(0)=0x(0)=0 and x(0)=0x'(0)=0. Find x(t)x(t) and describe the long-term motion.

(10)

(Total for Question 1 is 10 marks)

CP-9.9

Analyse and interpret models with one independent variable and two dependent variables as a pair of coupled first order simultaneous equations and solve them, e.g. predator-prey models.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The specified coupled models are linear first-order systems with one independent variable and two dependent variables, of the form x=ax+by+f(t)x'=ax+by+f(t) and y=cx+dy+g(t)y'=cx+dy+g(t). One method differentiates an equation, eliminates the other dependent variable and solves the resulting second-order equation; the eliminated variable is then recovered from an original equation.
  • Initial data may also determine a derivative value through the system.
  • Interaction terms must carry consistent signs: a transfer leaving one compartment may enter another.
  • Examiners expect both final functions to satisfy both original equations and the model to be interpreted through equilibria, signs and limiting behaviour.
  • Predictions such as negative populations mark a limit of the model's validity.
Worked example

Solve x=3x+yx'=3x+y, y=x+3yy'=x+3y with x(0)=2x(0)=2 and y(0)=0y(0)=0.

  1. 1.From y=x3xy=x'-3x, differentiate and substitute into y=x+3yy'=x+3y to get x6x+8x=0x''-6x'+8x=0.
  2. 2.The auxiliary roots are 22 and 44, so x=Ae2t+Be4tx=Ae^{2t}+Be^{4t}.
  3. 3.y=x3x=Ae2t+Be4ty=x'-3x=-Ae^{2t}+Be^{4t}.
  4. 4.A+B=2A+B=2 and A+B=0-A+B=0, giving A=B=1A=B=1.

Answer: x=e2t+e4tx=e^{2t}+e^{4t} and y=e2t+e4ty=-e^{2t}+e^{4t}.

Common mistakes

  • Don't eliminate a variable but forget to recover it after solving the second-order equation.
  • Don't use only one initial condition even though the coupled system supplies two.
  • Don't change the sign of a transfer term in one equation without making the corresponding contextual interpretation.

Exam tip

Substitute the final pair into both first-order equations; this catches elimination and sign errors quickly.

Tier 1 · Easy

ORIGINAL

1.

Given u=2u+vu'=2u+v and v=3uv'=3u, eliminate vv to obtain a second-order differential equation for uu.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Solve x=x+2yx'=x+2y, y=2x+yy'=2x+y subject to x(0)=3x(0)=3 and y(0)=1y(0)=1.

(7)

(Total for Question 1 is 7 marks)

Tier 3 · Hard

ORIGINAL

1.

Amounts A(t)A(t) and B(t)B(t) in two connected compartments are modelled by A=2A+BA'=-2A+B and B=2A3BB'=2A-3B, with A(0)=4A(0)=4 and B(0)=1B(0)=1. Find A(t)A(t) and B(t)B(t), determine exactly when BB is greatest, and state the long-term prediction.

(10)

(Total for Question 1 is 10 marks)

Want help turning these notes into marks?

Bring a tricky specification point or a recent answer, and we can work through the method and exam wording together.