Particular integral via operator method

At a Glance

• Frequency: 7 sub-parts across 5 of 14 years (2016, 2017, 2018, 2019, 2021)
• Priority tier: T2
• Marks (count): 10 (5), 8 (1), 9 (1)
• Average solve time: ~9 min
• Difficulty mix: medium 7
• Section: B | Dominant type: computation

Why This Chapter Matters

The operator method for particular integrals is the standard UPSC technique for Q5 and Q6 ODE questions. It appears 7 times across 5 years; every occurrence is medium difficulty, so reliable execution yields guaranteed marks. The method has four sub-routines — exponential shift for polynomial forcings, complex-exponential for trig, resonance multiplier, and variable-coefficient substitutions — all of which recur predictably. Mastering the decision tree below unlocks not just this atom but half the ODE problems in Section B.

Minimum Theory

Operator notation. Write a constant-coefficient linear ODE as $f(D)y = g(x)$ where $D = d/dx$ and $f(D) = a_n D^n + \cdots + a_1 D + a_0$. The complementary function (CF) is the general solution of $f(D)y=0$, found by solving $f(m)=0$. The particular integral (PI) is any one solution of $f(D)y=g(x)$, denoted formally by $y_p = \frac{1}{f(D)}g(x)$.

Core rules for the PI operator:

Exponential shift. To find $\frac{1}{f(D)}e^{ax}P(x)$ where $P(x)$ is a polynomial: shift the operator by $a$ (replace $D$ by $D+a$), then apply $\frac{1}{f(D+a)}$ to $P(x)$ as a binomial series in $D$ (since $D^k P = 0$ for $k > \deg P$).

Complex-exponential method. For trig forcing $e^{ax}\sin bx$ or $e^{ax}\cos bx$, write the forcing as $\operatorname{Im}[e^{(a+ib)x}]$ or $\operatorname{Re}[e^{(a+ib)x}]$. Compute $\frac{1}{f(D)}e^{(a+ib)x} = \frac{e^{(a+ib)x}}{f(a+ib)}$ (non-resonant case) and take the appropriate part. Rationalise the complex denominator by multiplying by the conjugate.

Question Archetypes

operator-pi (5 question(s); 2016, 2018, 2019, 2021)

Find a particular integral via inverse-operator / exponential-shift methods

Recognition Cues

The ODE has constant coefficients. Forcing terms of the form $e^{ax}$, $x^n e^{ax}$, $e^{ax}\cos bx$, $e^{ax}\sin bx$, or sums thereof. The question asks to “find a particular integral” or “solve.”

Solution Template

1. Find the CF: solve $f(m)=0$; write $y_c$.
2. For each forcing term, check if it is resonant (exponent $a$ is a root of $f(m)=0$).
3. Non-resonant $e^{ax}P(x)$: shift $D \to D+a$, expand $\frac{1}{f(D+a)}$ as series in $D$, apply to $P(x)$.
4. Resonant $e^{ax}$ of multiplicity $n$: PI is $e^{ax}\cdot x^n/n!$ times a constant; or, after shifting, use $\frac{1}{D^n}\cdot P(x) = \frac{x^n}{n!}$ (integrate $P$ $n$ times, ignoring constants absorbed into CF).
5. Trig: use complex exponential method; take Re or Im part.
6. Sum the PIs; write the general solution $y = y_c + y_p$.

Worked Example 1 — exponential-shift, polynomial

2018 Paper 1, 2018-P1-Q5a (10 marks)

Solve $y'' - y = x^2 e^{2x}$.

CF. $m^2-1=0 \Rightarrow m=\pm1$. So $y_c = C_1 e^x + C_2 e^{-x}$.

PI. $f(D)=D^2-1$, forcing $x^2 e^{2x}$, $a=2$: $f(2)=3\ne0$ (non-resonant). Apply shift: $y_p = e^{2x}\frac{1}{(D+2)^2-1}x^2 = e^{2x}\frac{1}{D^2+4D+3}x^2.$ Expand $\frac{1}{3+4D+D^2} = \frac{1}{3}\left(1-\frac{4D+D^2}{3}+\frac{(4D)^2}{9}-\cdots\right)$:

$D^2$-coefficient: $\frac{16}{9}-\frac{1}{3}=\frac{13}{9}$. Apply to $x^2$ (where $D(x^2)=2x$, $D^2(x^2)=2$): $\frac{1}{D^2+4D+3}x^2 = \frac{1}{3}\!\left[x^2-\frac{4}{3}(2x)+\frac{13}{9}(2)\right] = \frac{9x^2-24x+26}{27}.$ $\boxed{y = C_1 e^x + C_2 e^{-x} + \frac{(9x^2-24x+26)e^{2x}}{27}.}$

Worked Example 2 — resonant forcing

2018 Paper 1, 2018-P1-Q5c (10 marks)

Solve $y''' - 6y'' + 12y' - 8y = 12e^{2x} + 27e^{-x}$.

CF. $m^3-6m^2+12m-8=(m-2)^3=0$; triple root $m=2$. So $y_c=(C_1+C_2x+C_3x^2)e^{2x}$.

PI for $12e^{2x}$ (resonant, multiplicity 3). $y_{p1} = \frac{12}{(D-2)^3}e^{2x} = 12\,e^{2x}\cdot\frac{1}{D^3}(1) = 12\,e^{2x}\cdot\frac{x^3}{6} = 2x^3e^{2x}.$

PI for $27e^{-x}$ (non-resonant; $f(-1)=(-3)^3=-27$). $y_{p2} = \frac{27}{(D-2)^3}e^{-x} = \frac{27}{-27}e^{-x} = -e^{-x}.$ $\boxed{y = (C_1+C_2x+C_3x^2)e^{2x} + 2x^3e^{2x} - e^{-x}.}$

Worked Example 3 — complex-exponential method

2016 Paper 1, 2016-P1-Q5a (10 marks)

Find a PI of $y'' + y = e^{x/2}\sin\!\frac{\sqrt3}{2}x$.

CF roots: $\pm i$. The forcing is $\operatorname{Im}[e^{(1/2+i\sqrt3/2)x}]$; the exponent $\frac{1}{2}+i\frac{\sqrt3}{2}$ is not $\pm i$ (no resonance).

Compute $f\!\left(\tfrac12+i\tfrac{\sqrt3}{2}\right) = \left(\tfrac12+i\tfrac{\sqrt3}{2}\right)^2+1 = \left(\tfrac14-\tfrac34+i\tfrac{\sqrt3}{2}\right)+1 = \tfrac12+i\tfrac{\sqrt3}{2}$.

So $z_p = \dfrac{e^{(1/2+i\sqrt3/2)x}}{\frac12+i\frac{\sqrt3}{2}}$. Rationalise by multiplying by conjugate ($|\cdot|^2=1$):

$z_p = \left(\tfrac12-i\tfrac{\sqrt3}{2}\right)e^{x/2}\!\left(\cos\tfrac{\sqrt3}{2}x + i\sin\tfrac{\sqrt3}{2}x\right).$

Take imaginary part: $\boxed{y_p = e^{x/2}\!\left(\frac12\sin\frac{\sqrt3}{2}x - \frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}x\right).}$

Worked Example 4 — repeated root with double integration

2019 Paper 1, 2019-P1-Q5b (10 marks)

Solve $y'' - 4y' + 4y = 3x^2 e^{2x}\sin 2x$.

CF. $(D-2)^2=0$, repeated root $m=2$: $y_c=(C_1+C_2x)e^{2x}$.

PI. The forcing carries $e^{2x}$, which is resonant (root of multiplicity 2). Substitute $y = e^{2x}u$: since $(D-2)(e^{2x}u)=e^{2x}Du$, the equation reduces to $u'' = 3x^2\sin 2x$.

Double-integrate $3x^2\sin 2x$ (using IBP twice): $u' = 3\!\left(-\frac{x^2\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}\right) = -\frac32 x^2\cos 2x+\frac32 x\sin 2x+\frac34\cos 2x.$ $u = -\frac34 x^2\sin 2x - \frac32 x\cos 2x + \frac98\sin 2x.$ $\boxed{y = (C_1+C_2x)e^{2x} + e^{2x}\!\left(-\frac34 x^2\sin 2x - \frac32 x\cos 2x + \frac98\sin 2x\right).}$

Worked Example 5 — two PIs combined

2021 Paper 1, 2021-P1-Q5a (10 marks)

Solve $y'' + 2y = x^2 e^{3x} + e^x\cos 2x$.

CF. $D^2+2=0 \Rightarrow D=\pm i\sqrt2$: $y_c=C_1\cos\sqrt2\,x+C_2\sin\sqrt2\,x$.

PI for $x^2e^{3x}$. Shift: $e^{3x}\frac{1}{D^2+6D+11}x^2 = \frac{e^{3x}}{11}\!\left[x^2-\frac{12x+2}{11}+\frac{72}{121}\right] = e^{3x}\cdot\frac{121x^2-132x+50}{1331}$.

PI for $e^x\cos 2x$. Shift $D\to D+1$: $e^x\frac{1}{D^2+2D+3}\cos 2x$. On $\cos 2x$: $D^2\to-4$, so denominator $\to 2D-1$. Rationalise via $2D+1$: $\frac{(2D+1)\cos 2x}{4D^2-1}\Big|_{D^2\to-4} = \frac{-4\sin 2x+\cos 2x}{-17} = \frac{4\sin 2x-\cos 2x}{17}.$ So $y_{p2} = \frac{e^x(4\sin 2x-\cos 2x)}{17}$. $\boxed{y = C_1\cos\sqrt2\,x+C_2\sin\sqrt2\,x + \frac{e^{3x}(121x^2-132x+50)}{1331} + \frac{e^x(4\sin 2x-\cos 2x)}{17}.}$

Common Traps

• Check resonance before computing. If $a$ is a root of $f(m)=0$ of multiplicity $k$, the exponential shift gives $\frac{1}{D^k}P(x)$ (integrate $k$ times), not a simple substitution. Applying $D \to a$ to a resonant term gives $0/0$ — a dead giveaway.
• Multiplicity for repeated roots. For $(m-2)^3$, the forcing $e^{2x}$ resonates with multiplicity 3. The PI carries $x^3/3!$, not $x$ or $x^2$.
• Keep the $D^2$ term in the series. In $\frac{1}{f(D+a)}P(x)$, expanding to order $D^2$ is needed for a degree-2 polynomial $P$. Stopping at $D^1$ gives an incorrect constant term.
• Complex method: correct part. For $\sin bx$ forcing take $\operatorname{Im}[\cdot]$; for $\cos bx$ take $\operatorname{Re}[\cdot]$. Swapping them gives the wrong sign on the trig component.

reducible-pi (1 question(s); 2017)

Reduce by substitution to constant coefficients, then find PI

Recognition Cues

The ODE has variable coefficients but shows a pattern that a substitution (typically $t = x^2$, $t = \ln x$ for Euler-Cauchy) can remove. After substitution, the constant-coefficient operator PI applies.

Solution Template

1. Identify the substitution (e.g., $t=x^2$, so $\frac{d}{dx}=2x\frac{d}{dt}$ and $\frac{d^2}{dx^2}=2\frac{d}{dt}+4x^2\frac{d^2}{dt^2}$).
2. Rewrite the ODE entirely in terms of $t$.
3. Solve the constant-coefficient ODE in $t$ (CF + PI via operator).
4. Back-substitute $t \to$ original variable.

Worked Example

2017 Paper 1, 2017-P1-Q7b-i (9 marks)

Solve $xy'' - y' - 4x^3y = 8x^3\sin(x^2)$.

Let $t=x^2$. Then $\frac{dy}{dx}=2x\frac{dy}{dt}$ and $\frac{d^2y}{dx^2}=2\frac{dy}{dt}+4x^2\frac{d^2y}{dt^2}$.

Substitute into the ODE (the $2x\,dy/dt$ and $-y'=-2x\,dy/dt$ terms cancel): $4x^3\frac{d^2y}{dt^2} - 4x^3y = 8x^3\sin t \;\Rightarrow\; y'' - y = 2\sin t.$

CF of $y''-y=0$: $m=\pm1$, so $y_c = C_1e^t+C_2e^{-t}$.

PI. $D^2\to-1$ for $\sin t$: $\frac{1}{D^2-1}\sin t = \frac{2\sin t}{-1-1} = -\sin t$.

Back-substitute $t=x^2$: $\boxed{y = C_1 e^{x^2} + C_2 e^{-x^2} - \sin(x^2).}$

Common Traps

• The transform formula $d^2y/dx^2 = 2\,dy/dt + 4x^2\,d^2y/dt^2$ must be memorised for $t=x^2$; the $2\,dy/dt$ term is exactly what cancels $-y'$ in this ODE — it is not a coincidence but the design of the substitution.
• PI: $D^2 \to -1$ (not $-2$) for $\sin t$; $f(-1)=-1-1=-2$, giving $y_p = 2\sin t/(-2) = -\sin t$.

simultaneous-odes (1 question(s); 2017)

Solve a system of simultaneous linear ODEs by decoupling

Recognition Cues

Two (or more) ODEs in two unknown functions $y(x)$ and $z(x)$ (or $u,v$) linked by the same operator $D$. Symmetric structure (same $D+a$ on both) suggests adding/subtracting; general structure suggests elimination of one variable.

Solution Template

1. Add and subtract the ODEs (for symmetric systems) to form two decoupled equations in $y+z$ and $y-z$.
2. Solve each decoupled equation as a standard single ODE.
3. Recover $y$ and $z$ from the sums/differences.

Worked Example

2017 Paper 1, 2017-P1-Q6a-i (8 marks)

Solve $(D+1)y = z+e^x$ and $(D+1)z = y+e^x$.

Decouple. Let $u=y+z$, $w=y-z$. Adding: $(D+1)u = u+2e^x \Rightarrow Du=2e^x$. Subtracting: $(D+1)w = -w \Rightarrow (D+2)w=0$.

Solve. $u = 2e^x+2C_1$ and $w=2C_2e^{-2x}$.

Recover: $\boxed{y = C_1+C_2e^{-2x}+e^x, \qquad z = C_1-C_2e^{-2x}+e^x.}$

Common Traps

• Two constants, not four. A system of two first-order ODEs is effectively second order — only two arbitrary constants appear in the general solution.
• The constant in $u$. The ODE $Du=2e^x$ integrates to $u=2e^x+\text{const}$; this constant appears in both $y$ and $z$, so it counts once (call it $2C_1$).

Marks-Aware Writing

For a 10-mark “find the PI” question: State the CF (even if brief — it shows you know the structure), write the operator expression for the PI, show the shift or series expansion explicitly, and box the final answer. Skipping the CF and going straight to the PI loses 2–3 marks if the PI is wrong (no recovery path for the examiner).

For a resonant forcing: State explicitly that $a$ is a root of multiplicity $k$ and therefore the standard formula would give $0/0$; then apply the resonance rule. This explanation earns method marks even if arithmetic slips occur.

Practice Set

• 2023-P1-Q6a (15 m) — — third-order ODE with two forcing terms
• 2022-P1-Q8a-ii (10 m) — — reducible ODE via Euler substitution
• 2021-P1-Q6b (15 m) — — simultaneous ODEs (non-symmetric)
• 2018-P1-Q7a (13 m) — — two coupled oscillator equations
• 2019-P1-Q6c-i (10 m) — — operator PI with trig polynomial