Edexcel A-level Further Maths coverage

Differential equations

Section CP-9
9 spec leafs

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

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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.

  • Put a first-order linear equation into the form y+P(x)y=Q(x)y'+P(x)y=Q(x) before choosing the integrating factor I(x)=eP(x)dxI(x)=e^{\int P(x)\,dx}.
  • After multiplication, the left side is the product derivative (Iy)(Iy)', so integrate Iy=IQdx+CIy=\int IQ\,dx+C and then divide by II.
  • Use an integrating factor when yy and yy' occur linearly with coefficients depending only on the independent variable.
  • A common error is to use the coefficient of yy' in the exponent; first divide through so that the coefficient of yy' is 11.

Tier 1 · Easy

2 marks
ORIGINAL

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

Tier 2 · Standard

5 marks
ORIGINAL

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

Tier 3 · Hard

7 marks
ORIGINAL

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).

CP-9.2

Find both general and particular solutions to differential equations.

  • A general solution contains the full number of arbitrary constants: one for a first-order equation and two for a second-order equation. Varying the constant generates a family of solution curves, and you are expected to sketch representative members of that family (e.g. y=Cex+x1y=Ce^{-x}+x-1 for several values of CC).
  • A particular solution is obtained when initial or boundary conditions determine all arbitrary constants.
  • Apply conditions only after differentiating the complete general solution as many times as necessary.
  • Do not call a particular integral a particular solution: a particular integral handles the forcing term, while a particular solution also uses the supplied conditions.

Tier 1 · Easy

3 marks
ORIGINAL

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

Tier 2 · Standard

5 marks
ORIGINAL

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

Tier 3 · Hard

7 marks
ORIGINAL

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

CP-9.3

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

  • Define the dependent variable, independent variable, units and positive direction before translating rates into a differential equation.
  • In kinematics, a=dv/dta=dv/dt and v=ds/dtv=ds/dt; resistance acts opposite to motion, so its sign depends on the chosen direction.
  • After solving, interpret equilibrium values, limiting behaviour and constants in the original context rather than leaving a bare formula.
  • A mathematically correct solution can still expose a weak model; check assumptions such as constant coefficients, perfect mixing and whether predicted quantities remain physical.

Tier 1 · Easy

2 marks
ORIGINAL

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.

Tier 2 · Standard

5 marks
ORIGINAL

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.

Tier 3 · Hard

9 marks
ORIGINAL

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.

CP-9.4

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

  • For y+ay+by=0y''+ay'+by=0, try y=emxy=e^{mx} to obtain the auxiliary equation m2+am+b=0m^2+am+b=0.
  • Convert the roots into the matching real general solution, retaining two independent arbitrary constants.
  • Initial conditions determine the constants only after the full complementary solution and its derivative have been written.
  • Do not leave complex exponentials in a requested real solution; combine conjugate roots into sine and cosine terms.

Tier 1 · Easy

3 marks
ORIGINAL

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

Tier 2 · Standard

3 marks
ORIGINAL

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

Tier 3 · Hard

6 marks
ORIGINAL

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

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).

  • Write the general solution as complementary function plus particular integral: y=yc+ypy=y_{\mathrm c}+y_{\mathrm p}.
  • Choose a trial particular integral of the same family as f(x)f(x), with enough undetermined coefficients for every term produced by differentiation.
  • If the proposed trial duplicates a term in the complementary function, multiply it by xx repeatedly until it is independent.
  • A common error is to apply initial conditions to the complementary function alone; combine CF and PI first.

Tier 1 · Easy

4 marks
ORIGINAL

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

Tier 2 · Standard

5 marks
ORIGINAL

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

Tier 3 · Hard

9 marks
ORIGINAL

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

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.

  • For m2+am+b=0m^2+am+b=0, a positive discriminant gives distinct real roots and two exponential terms.
  • A zero discriminant gives a repeated real root rr and the solution (A+Bx)erx(A+Bx)e^{rx}.
  • A negative discriminant gives roots p±iqp\pm iq and the real solution epx(Acosqx+Bsinqx)e^{px}(A\cos qx+B\sin qx).
  • Do not infer oscillation from a negative real root alone: oscillation comes from a non-zero imaginary part.

Tier 1 · Easy

3 marks
ORIGINAL

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

Tier 2 · Standard

4 marks
ORIGINAL

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.

Tier 3 · Hard

6 marks
ORIGINAL

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.

CP-9.7

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

  • Simple harmonic motion about x=0x=0 satisfies x=ω2xx''=-\omega^2x, so acceleration is proportional to displacement and directed towards equilibrium.
  • The 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 2π/ω2\pi/\omega, and the maximum speed is ωR\omega R.
  • Keep displacement, velocity and acceleration signs distinct; maximum acceleration magnitude occurs at the extreme positions, not at equilibrium.

Tier 1 · Easy

3 marks
ORIGINAL

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

Tier 2 · Standard

6 marks
ORIGINAL

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.

Tier 3 · Hard

8 marks
ORIGINAL

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.

CP-9.8

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

  • A free damped oscillator is modelled by x+2λx+ω02x=0x''+2\lambda x'+\omega_0^2x=0 with λ>0\lambda>0; the velocity term removes energy.
  • Complex auxiliary roots give underdamped oscillation with a decaying exponential envelope; a repeated root gives critical damping.
  • Two distinct negative real roots give overdamping: the system returns without oscillating, more slowly than at critical damping.
  • A periodic driving force appears on the right side and produces a persistent steady response; do not mistake the decaying complementary function for the complete motion.

Tier 1 · Easy

4 marks
ORIGINAL

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

Tier 2 · Standard

7 marks
ORIGINAL

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.

Tier 3 · Hard

10 marks
ORIGINAL

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.

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.

  • For a coupled first-order system, eliminate one dependent variable to obtain a second-order equation for the other, then recover the eliminated variable.
  • Translate each interaction term carefully: a transfer leaving one population or compartment may enter the other with the opposite sign.
  • Use both initial conditions, including any derivative value inferred from an original equation, and verify the final pair in both equations.
  • Interpret signs, equilibrium points and limiting behaviour in context; an algebraic solution that predicts negative populations may invalidate the model after that time.

Tier 1 · Easy

3 marks
ORIGINAL

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

Tier 2 · Standard

7 marks
ORIGINAL

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

Tier 3 · Hard

10 marks
ORIGINAL

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.