Further Mathematics · Pearson Edexcel 9FM0 · Topic 9

Differential equations

Equations that involve derivatives. You solve first-order equations with an integrating factor and second-order equations with constant coefficients, then use them to model motion, oscillations and populations.

  • 9 sub-topics
  • Taught in Year 13
  • 146 practice questions
  • 9 animated lessons
Practice questions (146) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 9.1Integrating factor methodY13
  2. 9.2General and particular solutionsY13
  3. 9.3Modelling with first-order equationsY13
  4. 9.4Second-order homogeneous equationsY13
  5. 9.5Second-order non-homogeneous equationsY13
  6. 9.6Auxiliary equation casesY13
  7. 9.7Simple harmonic motionY13
  8. 9.8Damped and forced oscillationsY13
  9. 9.9Coupled first-order systemsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 146 questions in this topic, free, no account needed.

Differential equations

Not needed for Year 12Easy

First-order linear equation · Year 13 · 4 marks

  1. Find the general solution of dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y{}+ 2y=ex{}= \mathrm{e}^x.
    (4)
Show worked solutionHide worked solution
  1. The equation is y′+P(x)y=Q(x)y' + P(x)y = Q(x) with P=2P = 2.Integrating factor =e∫2 dx= \mathrm{e}^{\int 2\,\mathrm{d}x}=e2x{}= \mathrm{e}^{2x}
  2. Multiply through by the integrating factor.ddx(ye2x)\dfrac{\mathrm{d}}{\mathrm{d}x}(y\mathrm{e}^{2x})=e3x{}= \mathrm{e}^{3x}
  3. Integrate both sides.ye2xy\mathrm{e}^{2x}=13e3x{}= \tfrac{1}{3}\mathrm{e}^{3x}+c{}+ c
  4. Divide by e2x\mathrm{e}^{2x}.yy=13ex{}= \tfrac{1}{3}\mathrm{e}^x+ce−2x{}+ c\mathrm{e}^{-2x}

Total4

  • M
  • A
  • B
How did it go?

Report a problem with this question

Differential equations

Not needed for Year 12Medium

Integrating factor · Year 13 · 5 marks

  1. Solve dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}+yx{}+ \dfrac{y}{x}=x2{}= x^2, x>0x > 0, given that y=1y = 1 when x=1x = 1.
    (5)
Show worked solutionHide worked solution
  1. The equation is in the form y′+P(x)y=Q(x)y' + P(x)y = Q(x), so the integrating factor is e∫P dx\mathrm{e}^{\int P\,\mathrm{d}x}.Integrating factor =e∫1x dx= \mathrm{e}^{\int \frac{1}{x}\,\mathrm{d}x}=x{}= x
  2. Multiplying by xx makes the left side the derivative of a product.ddx(xy)\dfrac{\mathrm{d}}{\mathrm{d}x}(xy)=x3{}= x^3
  3. Integrate both sides; remember the constant.xy=14x4+cxy = \tfrac{1}{4}x^4 + c
  4. Use the given point to find cc.11=14+c{}= \tfrac{1}{4} + c⇒c{}\Rightarrow c=34{}= \tfrac{3}{4}
  5. Divide by xx to make yy the subject.y=x34+34xy = \dfrac{x^3}{4} + \dfrac{3}{4x}

Total5

  • M
  • A
  • B
How did it go?

Report a problem with this question

Differential equations

Not needed for Year 12Hard

Modelling damped motion · Year 13 · 8 marks

A particle on a spring moves in a straight line. Its displacement xx metres from equilibrium at time tt seconds satisfies d2xdt2+4dxdt+4x=0.\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 0. At t=0t = 0 the particle is at equilibrium, moving with speed 2 m s−12\ \mathrm{m\,s^{-1}}.
  1. (a)
    Find xx in terms of tt.
    (4)
  2. (b)
    Find the greatest displacement of the particle, and state the type of damping.
    (4)
Show worked solutionHide worked solution

Part (a)4 marks

  1. Write the auxiliary equation by replacing each derivative with a power of mm.m2+4m+4m^2 + 4m + 4=0{}= 0⇒m{}\Rightarrow m=−2{}= -2 (repeated)
  2. A repeated root gives this form of general solution.x=(A+Bt)e−2tx = (A + Bt)\mathrm{e}^{-2t}
  3. Use x=0x = 0 at t=0t = 0, then differentiate and use x˙=2\dot{x} = 2 at t=0t = 0.A=0A = 0; x˙\dot{x}=Be−2t{}= B\mathrm{e}^{-2t}−2Bte−2t{}- 2Bt\mathrm{e}^{-2t}, so B=2B = 2
  4. Substitute the constants.x=2te−2tx = 2t\mathrm{e}^{-2t}

Part (b)4 marks

  1. The greatest displacement is where the particle stops, so x˙=0\dot{x} = 0.x˙\dot{x}=2e−2t(1−2t){}= 2\mathrm{e}^{-2t}(1 - 2t)=0{}= 0⇒t{}\Rightarrow t=12{}= \tfrac{1}{2}
  2. Substitute this time into xx.xx=2×12×e−1{}= 2 \times \tfrac{1}{2} \times \mathrm{e}^{-1}=e−1{}= \mathrm{e}^{-1}
  3. Give a sensible number of significant figures.Greatest displacement =0.368= 0.368 m
  4. The auxiliary equation has a repeated root, so the particle returns to equilibrium without oscillating.Critical damping

Total8

  • M
  • A
  • B
How did it go?

Report a problem with this question