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Particular integralsAQA A-Level Further Maths: Revision notes

Section 1

General solution = complementary function + particular integral

For y′′+ay′+by=f(x)y''+ay'+by=f(x) the general solution is y=CF+PI.y=\text{CF}+\text{PI}. The complementary function (CF) is the general solution of the homogeneous equation y′′+ay′+by=0y''+ay'+by=0 (found from the auxiliary equation) and contains two arbitrary constants. The particular integral (PI) is any one function that satisfies the full equation and contains no arbitrary constants. Adding the PI shifts the solution so that it also produces f(x)f(x) on the left.

Key termscomplementary functionparticular integral
Common mistake

Putting arbitrary constants in the particular integral. The constants AA and BB belong only to the complementary function.

Section 2

Choosing a trial function

Choose the PI to match the form of f(x)f(x):

  • f(x)f(x) a polynomial of degree nn: a general polynomial of degree nn, e.g. px+qpx+q for 12x12x.
  • f(x)=keλxf(x)=ke^{\lambda x}: peλxpe^{\lambda x}.
  • f(x)=kcos⁡ωxf(x)=k\cos\omega x or ksin⁡ωxk\sin\omega x (or both): pcos⁡ωx+qsin⁡ωxp\cos\omega x+q\sin\omega x, always with both terms, because differentiation swaps sine and cosine. Include every lower power in a polynomial trial: px2+qx+rpx^2+qx+r for a quadratic ff.
Key termstrial function
Common mistake

Using y=pcos⁡xy=p\cos x alone for f(x)=10cos⁡xf(x)=10\cos x. The derivative y′y' brings in sin⁡x\sin x, so both terms are needed.

Section 3

Finding the coefficients

Differentiate the trial function twice, substitute into the equation, then compare coefficients of each type of term. Example: y′′−y′−6y=12xy''-y'-6y=12x with y=px+qy=px+q has y′=py'=p and y′′=0y''=0. So −p−6px−6q=12x-p-6px-6q=12x. Comparing xx: −6p=12-6p=12, so p=−2p=-2. Comparing constants: −p−6q=0-p-6q=0, so q=13q=\frac13. PI: y=−2x+13y=-2x+\frac13. The CF is Ae3x+Be−2xAe^{3x}+Be^{-2x}, so the general solution is y=Ae3x+Be−2x−2x+13y=Ae^{3x}+Be^{-2x}-2x+\frac13.

Key termscomparing coefficients
Exam tip

Check your PI by substituting it back into the original equation.

Section 4

When the trial function is in the complementary function

If f(x)f(x) is a term of the CF, the trial function gives 00 on the left and fails. Multiply it by xx. For y′′−3y′+2y=4exy''-3y'+2y=4e^{x} the CF is Aex+Be2xAe^{x}+Be^{2x}, so try y=pxexy=pxe^{x}. Then y′=p(1+x)exy'=p(1+x)e^{x} and y′′=p(2+x)exy''=p(2+x)e^{x}, and the equation gives −pex=4ex-pe^{x}=4e^{x}, so p=−4p=-4. If a repeated root makes emxe^{mx} and xemxxe^{mx} both part of the CF, multiply by x2x^2 instead. Always find the CF first so that you spot this.

Key termsresonance
Exam tip

Compare the right-hand side with the CF before choosing a trial function.

Section 5

Particular solutions and long-term behaviour

Write the general solution first, then apply two conditions (often yy and dydx\frac{dy}{dx} at x=0x=0) to the general solution, not to the CF alone. For y′′+3y′+2y=10cos⁡xy''+3y'+2y=10\cos x: y=Ae−x+Be−2x+cos⁡x+3sin⁡xy=Ae^{-x}+Be^{-2x}+\cos x+3\sin x. With y(0)=2y(0)=2 and y′(0)=1y'(0)=1 we get A+B=1A+B=1 and A+2B=2A+2B=2, so y=e−2x+cos⁡x+3sin⁡xy=e^{-2x}+\cos x+3\sin x. For large xx the CF terms decay and the PI gives the steady oscillation.

Key termsparticular solution
Common mistake

Using the conditions on the CF before adding the PI. The constants must be found from the full general solution.

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Exam questions on Particular integrals

  1. d2ydx2−dydx−6y=12x\frac{d^2y}{dx^2}-\frac{dy}{dx}-6y=12x.
    Find a particular integral.2 marks
  2. d2ydx2−4dydx+3y=16e−x\frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y=16e^{-x}.
    Find the general solution.2 marks
  3. d2ydx2+3dydx+2y=10cos⁡x\frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=10\cos x.
    Find a particular integral.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).