Further Mathematics · AQA 7367 · Topic I

Differential equations

Integrating factors, second-order equations, simple harmonic motion, damping and coupled equations such as predator–prey models.

  • 11 sub-topics
  • Taught in Year 13
  • 163 practice questions
  • 11 animated lessons
Practice questions (163) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA 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. I1Integrating factor for dy/dx + P(x)y = Q(x)Y13
  2. I2General and particular solutionsY13
  3. I3Differential equations in kinematics and other modelling contextsY13
  4. I4Second order homogeneous equations using the auxiliary equationY13
  5. I5Second-order non-homogeneous differential equationsY13
  6. I6Sign of the discriminant and the form of the solutionY13
  7. I7Simple harmonic motion, ẍ = −ω²xY13
  8. I8Damped oscillations: light, critical and heavy dampingY13
  9. I9Coupled first-order simultaneous differential equationsY13
  10. I10Hooke's law to form the equation of simple harmonic motionY13
  11. I11Damped motion with damping force proportional to velocityY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 163 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)
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  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}

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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)
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  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}

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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)
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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

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