Home / A-Level Further Maths 9231 / Differential equations / Particular integrals and particular solutions
Cambridge International AS & A Level · Topic 2.6
A-Level Further Maths 9231 / Topic 2.6 / AS & A Level

The part that stays.

The complementary function solves the equation with zero on the right. The particular integral deals with the right-hand side itself. Choose its form from a short table, find the coefficients by substituting, switch to a new form when it clashes with the CF, then fix \(A\) and \(B\) from the initial conditions and read off what happens in the long run.

Topic 2.6All differential equations lessons →
The key idea

For \(a\dfrac{d^{2}y}{dx^{2}} + b\dfrac{dy}{dx} + cy = f(x)\) (or the first order \(b\dfrac{dy}{dx} + cy = f(x)\)), the general solution is CF + PI: the complementary function, with its arbitrary constants, plus any one particular integral. Try a PI of the same type as \(f(x)\): a polynomial of the same degree, \(Ce^{kx}\), or \(C\cos px + D\sin px\) (always both terms). If the trial form is already part of the CF, multiply it by \(x\) (by \(x^{2}\) for a repeated root). Find the constants \(A\) and \(B\) only AFTER adding the PI, using the initial conditions. In a damped model the CF dies away and the PI is the long-term behaviour.

t x 2 −2 2 6 10 √5 solution x(t) PI 2 cos t + sin t CF (transient)
Fig. 1: the solution of x'' + 2x' + 5x = 10 cos t with x = 0 and x' = 0 at t = 0 (worked example 6). The complementary function e^(-t)(-2 cos 2t - 1.5 sin 2t) dies away within about 4 time units, after which the solution is indistinguishable from the particular integral 2 cos t + sin t, an oscillation of amplitude √5 at the driving frequency
Section 01

Choosing the trial form.

A particular integral (PI) is any one function that makes the left side equal to \(f(x)\). Guess its shape from \(f(x)\), with unknown coefficients, substitute it into the equation, and compare coefficients. Differentiating a polynomial, an exponential or a sine-cosine pair gives back the same kind of function, which is why these guesses work.

Right-hand side \(f(x)\)Trial PI
a constant \(k\)a constant \(c\)
a polynomial of degree \(n\), e.g. \(4x^{2}\)a FULL polynomial of degree \(n\), e.g. \(ax^{2} + bx + c\)
\(ke^{bx}\)\(Ce^{bx}\)
\(k\cos px\), \(k\sin px\) or \(k_{1}\cos px + k_{2}\sin px\)\(C\cos px + D\sin px\), always both terms

Two habits matter. Include every lower power in a polynomial trial, even if \(f(x)\) has only \(x^{2}\): the derivatives create \(x\) and constant terms that must be balanced. And use both cosine and sine even when \(f(x)\) has only one of them, because a \(\dfrac{dy}{dx}\) term turns cosines into sines. The same table works for a first order equation such as \(\dfrac{dy}{dx} + 2y = 3e^{x}\).

Worked example A polynomial right-hand side
Find a particular integral of \(\dfrac{d^{2}y}{dx^{2}} - 3\dfrac{dy}{dx} + 2y = 4x^{2}\).
  1. \(f(x)\) is a quadratic, so try \(y = ax^{2} + bx + c\). Then \(y' = 2ax + b\) and \(y'' = 2a\).
  2. Substitute: \[\begin{aligned} &2a - 3(2ax + b) + 2(ax^{2} + bx + c) \\ &= 2ax^{2} + (2b - 6a)x + (2a - 3b + 2c) = 4x^{2}. \end{aligned}\]
  3. \(x^{2}\): \(2a = 4\), so \(a = 2\).   \(x\): \(2b - 6a = 0\), so \(b = 6\).
  4. Constant: \(2a - 3b + 2c = 0\), so \(4 - 18 + 2c = 0\) and \(c = 7\).
  5. Check: \(y'' - 3y' + 2y = 4 - 3(4x + 6) + 2(2x^{2} + 6x + 7) = 4x^{2}\).
Answer: \(y = 2x^{2} + 6x + 7\)
Worked example A sine on the right, both terms in the trial
Find a particular integral of \(\dfrac{d^{2}y}{dx^{2}} + \dfrac{dy}{dx} - 2y = 10\sin 2x\).
  1. Try \(y = C\cos 2x + D\sin 2x\). Then \(y' = -2C\sin 2x + 2D\cos 2x\) and \(y'' = -4C\cos 2x - 4D\sin 2x\).
  2. Collect the cosines: \(-4C + 2D - 2C = -6C + 2D\). Collect the sines: \(-4D - 2C - 2D = -2C - 6D\).
  3. Compare: \(\cos 2x\): \(-6C + 2D = 0\), so \(D = 3C\).   \(\sin 2x\): \(-2C - 6D = 10\).
  4. So \(-2C - 18C = 10\), giving \(C = -\tfrac{1}{2}\) and \(D = -\tfrac{3}{2}\).
Answer: \(y = -\tfrac{1}{2}\cos 2x - \tfrac{3}{2}\sin 2x\)
Your turn
Find a particular integral of the first order equation \(\dfrac{dy}{dx} + 2y = 3e^{x}\).
Examiner trap

Trying only \(C\sin 2x\) because the right side is \(10\sin 2x\) fails as soon as there is a \(\dfrac{dy}{dx}\) term: it produces \(\cos 2x\) terms that nothing can cancel, and the equations for \(C\) are inconsistent. In the same way, trying \(ax^{2}\) alone for \(4x^{2}\) leaves stray \(x\) and constant terms. Use the full form from the table every time.

Section 02

When the trial form is in the CF.

If the trial form is already part of the complementary function, substituting it gives zero on the left, however you choose \(C\). The fix is to multiply the trial form by \(x\), and by \(x^{2}\) when the clash is with a repeated root. So always find the CF FIRST, then choose the PI.

EquationCFUsual trialUse instead
\(y'' - y' - 6y = 10e^{3x}\)\(Ae^{3x} + Be^{-2x}\)\(Ce^{3x}\) (in the CF)\(Cxe^{3x}\)
\(y'' - 4y' + 4y = 6e^{2x}\)\((A + Bx)e^{2x}\)\(Ce^{2x}\), \(Cxe^{2x}\) (both in the CF)\(Cx^{2}e^{2x}\)
\(y'' + 9y = 12\cos 3x\)\(A\cos 3x + B\sin 3x\)\(C\cos 3x + D\sin 3x\) (in the CF)\(x(C\cos 3x + D\sin 3x)\)
\(y'' + 3y' = 6\)\(A + Be^{-3x}\)a constant (in the CF)\(ax\)

In the last row \(y = ax\) gives \(0 + 3a = 6\), so the PI is \(2x\). Exam questions often hand you the form, for example 'given that \(kx\sin 3x\) is a particular integral, find \(k\)'. Then you only need to differentiate carefully, with the product rule, and compare coefficients.

Worked example An exponential that clashes
Find a particular integral of \(\dfrac{d^{2}y}{dx^{2}} - \dfrac{dy}{dx} - 6y = 10e^{3x}\).
  1. Auxiliary equation \(m^{2} - m - 6 = 0\), so \(m = 3\) or \(m = -2\) and the CF is \(Ae^{3x} + Be^{-2x}\).
  2. \(e^{3x}\) is in the CF, so try \(y = Cxe^{3x}\). Product rule: \(y' = C(1 + 3x)e^{3x}\) and \(y'' = C(6 + 9x)e^{3x}\).
  3. Substitute: \[\begin{aligned} &C\big[(6 + 9x) - (1 + 3x) - 6x\big]e^{3x} \\ &= 5Ce^{3x} = 10e^{3x}. \end{aligned}\]
  4. The \(x\) terms cancel, as they must, and \(C = 2\).
Answer: \(y = 2xe^{3x}\)
Worked example A given form: find k
Given that \(y = kx\sin 3x\) is a particular integral of \(\dfrac{d^{2}y}{dx^{2}} + 9y = 12\cos 3x\), find the value of \(k\).
  1. \(y' = k\sin 3x + 3kx\cos 3x\) (product rule).
  2. \(y'' = 3k\cos 3x + 3k\cos 3x - 9kx\sin 3x = 6k\cos 3x - 9kx\sin 3x\).
  3. \(y'' + 9y = 6k\cos 3x - 9kx\sin 3x + 9kx\sin 3x = 6k\cos 3x\).
  4. Compare with \(12\cos 3x\): \(6k = 12\).
Answer: \(k = 2\), so the PI is \(y = 2x\sin 3x\)
Your turn
Find a particular integral of \(\dfrac{d^{2}y}{dx^{2}} - 4\dfrac{dy}{dx} + 4y = 6e^{2x}\).
Examiner trap

If your comparison gives \(0 = 10\), your trial form was in the CF: you skipped finding the CF first. And differentiate \(xe^{3x}\) or \(x\sin 3x\) with the product rule: writing \(y' = 3Cxe^{3x}\) loses the term \(Ce^{3x}\) that carries the whole answer.

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Which is the correct form to try for a particular integral of \(\dfrac{d^{2}y}{dx^{2}} + 3\dfrac{dy}{dx} + 2y = 3\sin 2x\)?

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Which is the simplest suitable trial form for a particular integral of \(\dfrac{d^{2}y}{dx^{2}} - 5\dfrac{dy}{dx} + 6y = 2x + 1\)?

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The complementary function of \(\dfrac{d^{2}y}{dx^{2}} - 4y = 5e^{2x}\) is \(Ae^{2x} + Be^{-2x}\). Which trial form should be used for the particular integral?

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What is the general solution of \(\dfrac{d^{2}y}{dx^{2}} + 4y = 8\)?

Section 03

Particular solutions and the long run.

The general solution is \(y = \text{CF} + \text{PI}\), with two arbitrary constants for a second order equation (one for first order). Two initial conditions, usually \(y\) and \(\dfrac{dy}{dx}\) at \(x = 0\), fix \(A\) and \(B\) and give the particular solution. Always add the PI before using the conditions: the PI contributes to \(y(0)\) and \(y'(0)\) too.

In models, \(t\) is time. When every root of the auxiliary equation has a negative real part, each CF term contains \(e^{-kt}\) with \(k \gt 0\), so the CF tends to zero: it is the transient. What remains for large \(t\) is the PI, the steady response to the forcing term. For a damped spring driven by \(10\cos t\), the motion settles to an oscillation at the driving frequency whose amplitude is fixed by the PI, whatever the starting position. Fig. 1 shows this for worked example 6.

An amplitude \(R\) comes from writing \(C\cos t + D\sin t = R\cos(t - \alpha)\) with \(R = \sqrt{C^{2} + D^{2}}\).

Worked example Fixing A and B
Solve \(\dfrac{d^{2}y}{dx^{2}} + 3\dfrac{dy}{dx} + 2y = 12e^{x}\), given that \(y = 0\) and \(\dfrac{dy}{dx} = 0\) when \(x = 0\).
  1. CF: \(m^{2} + 3m + 2 = 0\), \(m = -1\) or \(-2\), so CF \(= Ae^{-x} + Be^{-2x}\).
  2. PI: try \(Ce^{x}\) (no clash): \(C + 3C + 2C = 6C = 12\), so \(C = 2\).
  3. General solution: \(y = Ae^{-x} + Be^{-2x} + 2e^{x}\), so \(y' = -Ae^{-x} - 2Be^{-2x} + 2e^{x}\).
  4. \(x = 0\): \(A + B + 2 = 0\) and \(-A - 2B + 2 = 0\).
  5. Adding: \(-B + 4 = 0\), so \(B = 4\) and \(A = -6\).
Answer: \(y = -6e^{-x} + 4e^{-2x} + 2e^{x}\)
Worked example A forced damped oscillation
The displacement \(x\) of a particle at time \(t\) satisfies \(\dfrac{d^{2}x}{dt^{2}} + 2\dfrac{dx}{dt} + 5x = 10\cos t\), and \(x = 0\), \(\dfrac{dx}{dt} = 0\) when \(t = 0\). Find \(x\) in terms of \(t\) and describe the motion for large \(t\).
  1. CF: \(m^{2} + 2m + 5 = 0\), \(m = -1 \pm 2i\), so CF \(= e^{-t}(A\cos 2t + B\sin 2t)\).
  2. PI: try \(C\cos t + D\sin t\). Cosines: \(-C + 2D + 5C = 4C + 2D = 10\). Sines: \(-D - 2C + 5D = 4D - 2C = 0\).
  3. So \(C = 2D\), \(10D = 10\): \(D = 1\), \(C = 2\). PI \(= 2\cos t + \sin t\).
  4. \(x(0) = A + 2 = 0\), so \(A = -2\). Differentiating, \(x'(0) = -A + 2B + 1 = 0\), so \(2 + 2B + 1 = 0\) and \(B = -\tfrac{3}{2}\).
  5. So \[\begin{aligned} x = {}&e^{-t}\big(-2\cos 2t - \tfrac{3}{2}\sin 2t\big) \\ &+ 2\cos t + \sin t. \end{aligned}\]
  6. As \(t \to \infty\), \(e^{-t} \to 0\), so \(x \approx 2\cos t + \sin t = \sqrt{5}\cos(t - \alpha)\), \(\tan\alpha = \tfrac{1}{2}\).
Answer: \(x = e^{-t}\big(-2\cos 2t - \tfrac{3}{2}\sin 2t\big) + 2\cos t + \sin t\); for large \(t\) it oscillates with period \(2\pi\) and amplitude \(\sqrt{5}\)
Your turn
Solve \(\dfrac{dy}{dx} + 3y = 6x\), given that \(y = 0\) when \(x = 0\).
Examiner trap

Do not find \(A\) and \(B\) from the CF alone and then add the PI on: the conditions belong to the WHOLE solution, and the PI changes both \(y(0)\) and \(y'(0)\). In a long-term question, 'the motion dies away' is wrong when there is a forcing term: only the CF dies away, and the PI remains.

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Find a particular integral of \(\dfrac{d^{2}y}{dx^{2}} + 2\dfrac{dy}{dx} - 3y = 10e^{2x}\).

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Given that \(y = kxe^{-x}\) is a particular integral of \(\dfrac{d^{2}y}{dx^{2}} + 3\dfrac{dy}{dx} + 2y = 5e^{-x}\), find \(k\).

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Find the solution of \(\dfrac{dy}{dx} + 2y = 4\) for which \(y = 0\) when \(x = 0\).

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A damped oscillator satisfies \(\dfrac{d^{2}x}{dt^{2}} + 4\dfrac{dx}{dt} + 5x = 8\sin t\). Which statement describes \(x\) for large \(t\), whatever the initial conditions?

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Trial forms chosen, clashes fixed, constants found, long run read off.

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Six original Cambridge-style questions
Trial forms with and without a clash, polynomial and trigonometric PIs, a given form \(kxe^{2x}\) used to reach a particular solution, and a forced damped oscillation with its transient and long-term amplitude.