Method of variation of parameters

At a Glance

• Frequency: 11 sub-parts across 11 of 13 years (2013, 2014, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (4), 13 (1), 15 (4), 20 (1), 8 (1)
• Average solve time: ~13 min
• Difficulty mix: medium 7, hard 3, easy 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Variation of parameters appears in 11 of the last 13 years — only two years missed, making it the most consistently examined technique in Paper 1 Section B. The method works on any second-order linear ODE regardless of the right-hand side: it handles $\sec$, $\log$, $\tan$, polynomials, and combinations that undetermined coefficients cannot touch. In every question the four-step skeleton is the same; the skill is recognising which variant you need (standard CF, Cauchy-Euler, or reduction-of-order) before the VoP machinery kicks in.

Minimum Theory

The formula. Given $y''+P(x)y'+Q(x)y=g(x)$ with two complementary functions $y_1,y_2$, seek $y_p=u_1y_1+u_2y_2$ where $u_1'=-\frac{y_2\,g}{W},\qquad u_2'=\frac{y_1\,g}{W},\qquad W=y_1y_2'-y_2y_1'.$ This is the standard form — the ODE must have leading coefficient $1$ before you read off $g(x)$. For $ay''+\ldots=R(x)$, divide everything by $a$ first; then $g=R/a$.

Two constraints, two equations. The conditions $u_1'y_1+u_2'y_2=0$ and $u_1'y_1'+u_2'y_2'=g$ uniquely determine $u_1'$ and $u_2'$ via Cramer’s rule — the formulas above.

Getting both CF solutions. The VoP formula needs two independent homogeneous solutions. Three sub-cases arise:

1. Constant coefficients (most common): solve the auxiliary equation; the roots give exponential, trigonometric, or repeated-root CF.
2. Cauchy-Euler ($x^2y''-\ldots$): try $y=x^m$ to get the indicial equation; roots $m_1,m_2$ give $y_1=x^{m_1}$, $y_2=x^{m_2}$.
3. One CF given: use reduction of order ($y_2=v(x)y_1$) or operator factorisation to find the second.

Question Archetypes

Three patterns cover all VoP questions.

variation-of-parameters (5 question(s); 2013, 2014, 2016, 2017, 2018)

Recognition Cues

• The ODE has constant coefficients (or is first-order linear).
• The right-hand side is $\sec(ax)$, $\log x$, $\tan(ax)$, or a polynomial — functions that undetermined coefficients cannot handle, or the problem explicitly says “use variation of parameters.”
• CF is found by the standard auxiliary equation.

Solution Template

1. Write the auxiliary equation and find the two CF solutions $y_1,y_2$.
2. Compute $W=y_1y_2'-y_2y_1'$.
3. Write $u_1'=-y_2g/W$ and $u_2'=y_1g/W$ where $g$ is the RHS of the standard-form ODE.
4. Integrate to find $u_1,u_2$. Use standard integrals: $\int\tan(ax)\,dx=-\tfrac1a\ln|\cos(ax)|$; $\int\sec(2x)\,dx=\tfrac12\ln|\sec2x+\tan2x|$; $\int e^{ax}\sin(bx)\,dx=\tfrac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}$; $\int x\log x\,dx=\tfrac{x^2}{2}\log x-\tfrac{x^2}{4}$.
5. Write $y_p=u_1y_1+u_2y_2$ and simplify using trig identities if needed.
6. General solution: $y=y_c+y_p$.

Worked Example(s)

2013 Paper 1, 2013-P1-Q6b (10 marks)

Solve $y''+a^2y=\sec ax$ by variation of parameters.

CF. Auxiliary equation $r^2+a^2=0\Rightarrow y_1=\cos ax,\,y_2=\sin ax$.

Wronskian. $W=a\cos^2ax+a\sin^2ax=a$.

Integrands ($g=\sec ax$): $u_1'=-\frac{\sin ax\cdot\sec ax}{a}=-\frac{\tan ax}{a}\;\Rightarrow\;u_1=\frac{\ln|\cos ax|}{a^2}.$ $u_2'=\frac{\cos ax\cdot\sec ax}{a}=\frac{1}{a}\;\Rightarrow\;u_2=\frac{x}{a}.$

Particular integral: $y_p=\frac{\cos ax}{a^2}\ln|\cos ax|+\frac{x\sin ax}{a}.$

$\boxed{\;y=C_1\cos ax+C_2\sin ax+\frac{\cos ax}{a^2}\ln|\cos ax|+\frac{x\sin ax}{a}.\;}$

2016 Paper 1, 2016-P1-Q6b (15 marks)

Solve $(D^2+2D+1)y=e^{-x}\log x$ by variation of parameters.

CF. $(D+1)^2=0$ — repeated root $D=-1$: $y_1=e^{-x}$, $y_2=xe^{-x}$.

Wronskian. $W=e^{-2x}$.

Integrands ($g=e^{-x}\log x$): the exponentials cancel: $y_2g/W=x\log x$ and $y_1g/W=\log x$.

$\int x\log x\,dx=\tfrac{x^2}{2}\log x-\tfrac{x^2}{4},\qquad\int\log x\,dx=x\log x-x.$

Assemble: $y_p=-e^{-x}\!\left(\tfrac{x^2}{2}\log x-\tfrac{x^2}{4}\right)+xe^{-x}(x\log x-x)=e^{-x}\!\left(\tfrac{x^2}{2}\log x-\tfrac{3x^2}{4}\right).$

$\boxed{\;y=(c_1+c_2x)e^{-x}+e^{-x}\!\left(\frac{x^2}{2}\log x-\frac{3x^2}{4}\right).\;}$

2018 Paper 1, 2018-P1-Q6c (13 marks)

Solve $y''+16y=32\sec2x$.

CF. $r^2+16=0\Rightarrow y_1=\cos4x,\,y_2=\sin4x$; $W=4$.

Compute $u_1,u_2$ ($g=32\sec2x$): $u_1'=-\frac{\sin4x\cdot32\sec2x}{4}=-8\cdot\frac{\sin4x}{\cos2x}.$ Use $\sin4x=2\sin2x\cos2x$: $u_1'=-16\sin2x\Rightarrow u_1=8\cos2x$.

$u_2'=\frac{\cos4x\cdot32\sec2x}{4}=8\cdot\frac{\cos4x}{\cos2x}.$ Use $\cos4x=2\cos^22x-1$: $u_2'=16\cos2x-8\sec2x\Rightarrow u_2=8\sin2x-4\ln|\sec2x+\tan2x|$.

Combine: $u_1y_1+u_2y_2$ has elementary part $8(\cos2x\cos4x+\sin2x\sin4x)=8\cos2x$: $y_p=8\cos2x-4\sin4x\ln|\sec2x+\tan2x|.$

$\boxed{\;y=C_1\cos4x+C_2\sin4x+8\cos2x-4\sin4x\ln|\sec2x+\tan2x|.\;}$

2014 Paper 1, 2014-P1-Q6a (10 marks)

Solve $y'-5y=\sin x$ by variation of parameters.

First-order case. CF: $y_1=e^{5x}$. Try $y_p=u(x)e^{5x}$; substituting: $u'e^{5x}=\sin x\Rightarrow u'=e^{-5x}\sin x$.

Using $\int e^{ax}\sin bx=\tfrac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}$ with $a=-5,b=1$: $u=-\frac{e^{-5x}(5\sin x+\cos x)}{26}\;\Rightarrow\;y_p=-\frac{5\sin x+\cos x}{26}.$

$\boxed{\;y=Ce^{5x}-\frac{5\sin x+\cos x}{26}.\;}$

2017 Paper 1, 2017-P1-Q7b-ii (8 marks)

Solve $y''-y'-2y=44-76x-48x^2$ by variation of parameters.

CF. $m^2-m-2=0\Rightarrow m=2,-1$; $y_1=e^{-x}$, $y_2=e^{2x}$; $W=3e^x$.

Using $\int e^x P\,dx=e^x(P-P'+P''-\cdots)$ with $P=44-76x-48x^2$: $u_1y_1+u_2y_2=24x^2+14x-5\qquad(\text{exponential factors cancel}).$

$\boxed{\;y=C_1e^{-x}+C_2e^{2x}+24x^2+14x-5.\;}$

Common Traps

• The Wronskian for $\{\cos ax,\sin ax\}$ is $a$, not $1$. A dropped factor of $a$ changes every subsequent result.
• For $\sec2x$: the key identities are $\sin4x/\cos2x=2\sin2x$ and $\cos4x/\cos2x=2\cos2x-\sec2x$. Without them the integrals look intractable.
• Repeated root gives $y_2=xe^{-x}$ — using $e^{-x}$ twice gives $W=0$, and the formula collapses.
• In the 2016 problem, the exponentials cancel in both integrands, leaving $\int x\log x$ and $\int\log x$ — keep $\log x$, do not differentiate it away.
• The formula has a minus sign on the first piece: $y_p=-y_1\int\tfrac{y_2g}{W}+y_2\int\tfrac{y_1g}{W}$.

vop-cauchy-euler (2 question(s); 2019, 2024)

Recognition Cues

• The homogeneous ODE has the form $x^2y''-2xy'+2y=0$ or similar (powers of $x$ matching the order of differentiation — Euler-Cauchy type).
• The full equation is $x^2y''-\ldots=x^n\sin x$ or similar polynomial-trig forcing.
• You must find $y_1,y_2$ first by the indicial method, then apply VoP to the normalised equation.

Solution Template

1. Solve the homogeneous Cauchy-Euler ODE by trying $y=x^m$; get the indicial equation.
2. Write down $y_1=x^{m_1}$ and $y_2=x^{m_2}$.
3. Divide the full ODE by $x^2$ to get standard form; read off $g(x)$ from the normalised RHS.
4. Compute $W=y_1y_2'-y_2y_1'$.
5. Apply VoP: $u_1'=-y_2g/W$, $u_2'=y_1g/W$; integrate; assemble $y_p$.

Worked Example(s)

2019 Paper 1, 2019-P1-Q7a (15 marks) and 2024 Paper 1, 2024-P1-Q6c-ii (10 marks)

Find the independent CF solutions of $x^2y''-2xy'+2y=x^3\sin x$ and find the general solution by variation of parameters.

Sources:,

Indicial equation. Try $y=x^m$ in $x^2y''-2xy'+2y=0$: $m(m-1)-2m+2=m^2-3m+2=(m-1)(m-2)=0$, giving $y_1=x,\qquad y_2=x^2.$

Standard form. Divide by $x^2$: $y''-(2/x)y'+(2/x^2)y=x\sin x$, so $g(x)=x\sin x$.

Wronskian. $W=x\cdot2x-x^2\cdot1=x^2$.

VoP integrals: $u_1'=-\frac{x^2\cdot x\sin x}{x^2}=-x\sin x\;\Rightarrow\;u_1=x\cos x-\sin x.$ $u_2'=\frac{x\cdot x\sin x}{x^2}=\sin x\;\Rightarrow\;u_2=-\cos x.$

Particular integral: $y_p=(x\cos x-\sin x)\cdot x+(-\cos x)\cdot x^2=x^2\cos x-x\sin x-x^2\cos x=-x\sin x.$

$\boxed{\;y=C_1x+C_2x^2-x\sin x.\;}$

Common Traps

• The forcing in the VoP formulas is $g=x\sin x$ (the normalised RHS after dividing by $x^2$), not the raw $x^3\sin x$. Using the raw RHS gives $y_p$ wrong by a factor of $x^2$.
• The $x^2\cos x$ terms cancel in $y_p$ — a nice simplification; do not leave them unsimplified.
• The Wronskian is $W=x^2$ (from $y_1=x,y_2=x^2$); it appears in every denominator.

vop-after-reduction (4 question(s); 2020, 2021, 2022, 2025)

Recognition Cues

• One CF solution $y_1$ is given (e.g., $y_1=e^{-x}$, $y_1=x$).
• Or the equation has variable coefficients and you must find $y_2$ yourself.
• The second step is always: find $y_2$ by reduction of order (try $y_2=v(x)y_1$) or operator factorisation.

Solution Template

1. Accept the given $y_1$ (or find it by inspection/trial).
2. Reduction of order: substitute $y_2=v(x)y_1$ into the homogeneous ODE; reduce to a first-order ODE in $u=v'$; integrate twice to get $v$ and hence $y_2$.
3. Divide the full ODE by its leading coefficient to get standard form; read off $g(x)$.
4. Compute $W=y_1y_2'-y_2y_1'$.
5. Apply VoP: $u_1'=-y_2g/W$, $u_2'=y_1g/W$; integrate; assemble $y_p$.

Worked Example(s)

2021 Paper 1, 2021-P1-Q8a-ii (10 marks)

Solve $x^2y''-2x(1+x)y'+2(1+x)y=x^3$ by variation of parameters, given $y_1=x$ satisfies the homogeneous equation.

Find $y_2$. Try $y_2=v(x)\cdot x$. Substituting into the homogeneous ODE and simplifying: $x^3v''-2x^3v'=0\;\Rightarrow\;v''=2v'\;\Rightarrow\;u=v'=e^{2x}\;\Rightarrow\;v=e^{2x}/2.$ Take $y_2=xe^{2x}$.

Standard form. Divide by $x^2$: $g(x)=x^3/x^2=x$.

Wronskian. $W=x\cdot(1+2x)e^{2x}-xe^{2x}\cdot1=2x^2e^{2x}$.

VoP: $u_2'=\tfrac{x\cdot x}{2x^2e^{2x}}=\tfrac12e^{-2x}\Rightarrow u_2=-\tfrac14e^{-2x}$. Then $u_1'=-\tfrac12\Rightarrow u_1=-x/2$.

$y_p=(-x/2)(x)+(-e^{-2x}/4)(xe^{2x})=-x^2/2-x/4.$

The $-x/4$ is absorbed into $C_1x$:

$\boxed{\;y=C_1x+C_2xe^{2x}-\frac{x^2}{2}.\;}$

2022 Paper 1, 2022-P1-Q6b (15 marks)

Solve $(x^2-1)y''-2xy'+2y=(x^2-1)^2$ by variation of parameters, given $y_1=x$.

Find $y_2$. Try $y_2=vx$; substituting into the homogeneous equation: $x(x^2-1)v''-2v'=0\;\Rightarrow\;\frac{dv'}{v'}=\frac{2\,dx}{x(x^2-1)}.$ Partial fractions: $\tfrac{2}{x(x-1)(x+1)}=-\tfrac2x+\tfrac1{x-1}+\tfrac1{x+1}$. Integrating: $v'=(x^2-1)/x^2=1-1/x^2$, so $v=x+1/x$ and $y_2=x^2+1$.

Standard form. Divide by $(x^2-1)$: $g=x^2-1$.

Wronskian. $W=x\cdot2x-(x^2+1)\cdot1=x^2-1$.

VoP: $u_2'=x\Rightarrow u_2=x^2/2$; $u_1'=-(x^2+1)\Rightarrow u_1=-x^3/3-x$.

$y_p=(-x^3/3-x)\cdot x+(x^2/2)\cdot(x^2+1)=\frac{x^4}{6}-\frac{x^2}{2}.$

$\boxed{\;y=C_1x+C_2(x^2+1)+\frac{x^4}{6}-\frac{x^2}{2}.\;}$

2020 Paper 1, 2020-P1-Q6a (20 marks)

Solve $y''+(1-\cot x)y'-y\cot x=\sin^2 x$ by variation of parameters, given $y_1=e^{-x}$.

Find $y_2$. Factor the operator: $(D-\cot x)(D+1)y=0$. Letting $v=(D+1)y$: $(D-\cot x)v=0\Rightarrow v=\sin x$. Solve $(D+1)y=\sin x$: integrating factor $e^x$: $y_2=\sin x-\cos x$.

Wronskian. $W=2e^{-x}\sin x$.

VoP: $u_2'=\frac{e^{-x}\sin^2 x}{2e^{-x}\sin x}=\frac{\sin x}{2}\;\Rightarrow\;u_2=-\frac{\cos x}{2}.$ $u_1'=-\frac{(\sin x-\cos x)\sin^2 x}{2e^{-x}\sin x}=-\frac{e^x}{4}(1-\cos2x-\sin2x)\;\Rightarrow\;u_1=\frac{e^x}{20}(3\sin2x-\cos2x-5).$

Assembling and simplifying (constant terms cancel): $y_p=-\frac{1}{10}\sin2x+\frac{1}{5}\cos2x.$

$\boxed{\;y=C_1e^{-x}+C_2(\sin x-\cos x)-\frac{\sin2x}{10}+\frac{\cos2x}{5}.\;}$

2025 Paper 1, 2025-P1-Q8a (15 marks)

Solve $(x+2)y''-(2x+5)y'+2y=(1+x)e^x$ by variation of parameters.

CF by inspection. Try $y=e^{mx}$: matching coefficients of $x$ gives $m=2$, yielding $y_1=e^{2x}$. Try $y=\alpha x+\beta$: gives $y_2=2x+5$.

Standard form. Divide by $(x+2)$: $g=(1+x)e^x/(x+2)$.

Wronskian. $W=-4(x+2)e^{2x}$.

VoP: $u_2'=-\tfrac{(x+1)e^x}{4(x+2)^2}$; use $\tfrac{d}{dx}\!\tfrac{e^x}{x+2}=\tfrac{(x+1)e^x}{(x+2)^2}$, so $u_2=-\tfrac{e^x}{4(x+2)}$.

After assembling, both terms combine to give $y_p=-e^x$.

$\boxed{\;y=C_1e^{2x}+C_2(2x+5)-e^x.\;}$

Common Traps

• Variable-coefficient ODEs require standard form first. Divide by the leading coefficient before reading off $g(x)$ for the VoP integrands.
• The reduction-of-order substitution $y_2=v y_1$ always reduces to a first-order ODE in $u=v'$, separable or easily solvable — keep substituting until you see it.
• For the 2022 problem, the partial-fraction calculation $\tfrac{2}{x(x^2-1)}=-\tfrac{2}{x}+\tfrac{1}{x-1}+\tfrac{1}{x+1}$ is the key step; skip it and the whole integral breaks.
• The constant term $-x/4$ from $y_p$ in the 2021 problem merges into $C_1x$ in the general solution — this is correct and saves rewriting.
• In the 2020 operator-factorisation: the factor $(D-\cot x)$ is the left factor; the second CF is found by solving a first-order equation for $v=(D+1)y$.

Marks-Aware Writing

8-mark questions (2017): CF + Wronskian (one line each), two VoP integrals (show clearly), assemble and simplify. Since forcing is polynomial, the exponentials in $u_1y_1+u_2y_2$ cancel — state this explicitly.

10-mark questions (2013, 2014, 2024): Write CF and Wronskian; show both integrands; integrate; write $y_p$ neatly; state the general solution. Four to five working lines suffice.

13-15-mark questions (2016, 2018, 2019, 2022, 2025): Full layout — CF with justification, Wronskian calculation (show the $2\times2$ determinant), both integrands and their integrals step by step, assembled $y_p$, boxed final answer.

20-mark questions (2020): All of the above, plus the derivation of $y_2$ (operator factorisation or reduction of order written out in full), Wronskian computation, all four integrals shown. Double-check the constant cancellations.

Practice Set