Second-order linear PDEs with constant coefficients (CF, PI)

At a Glance

• Frequency: 13 sub-parts across 13 of 14 years (2013, 2014, 2015, 2016, 2017, 2018, 2020, 2021, 2022, 2023, 2024, 2025, 2026)
• Priority tier: T1
• Marks (count): 10 (6), 15 (6)
• Average solve time: ~13 min
• Difficulty mix: medium 7, hard 4, easy 1
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom appears in every year except one since 2013, always worth 10 or 15 marks. The method — factor the operator, write the CF, compute a PI term by term — is completely algorithmic once you know the four PI rules. The only genuine difficulty is detecting resonance (where the naive exponential rule gives $0/0$) and applying the $x$-multiplier correction. Mastering that single trap, plus the shift identity for the $x^n\sin/\cos$ RHS, covers every hard case in the corpus.

Minimum Theory

Operator notation and factorisation. Write the PDE as $F(D,D')z=R(x,y)$ where $D=\partial/\partial x$, $D'=\partial/\partial y$. The characteristic auxiliary equation sets $D=m$ and $D'=1$ (or $D/D'=m$): solving $F(m,1)=0$ for $m$ gives the slopes of the characteristic families. Each simple linear factor $(D-mD')$ contributes an arbitrary function $\phi(y+mx)$ to the complementary function (CF). A repeated factor $(D-mD')^k$ contributes $\phi_1(y+mx)+x\phi_2(y+mx)+\cdots+x^{k-1}\phi_k(y+mx)$.

PI rules (four cases).

For a polynomial $R$: expand $1/F(D,D')$ as a formal power series in $D/D'$ (or $D'/D$) and apply to $R$; the series terminates.

Boundary conditions. When boundary curves are given, pin down the arbitrary functions $\phi_i$ by substituting the stated conditions into the general solution.

Question Archetypes

Three patterns cover the corpus.

operator-pde (10 question(s); 2013, 2014, 2015, 2016, 2017, 2018, 2020, 2021, 2022, 2025)

Recognition Cues

• “Solve $(F(D,D'))z=R(x,y)$” with explicit $D,D'$ notation.
• The RHS is a sum of exponentials, polynomials, trig functions, or products thereof.
• No boundary data given — general solution only.

Solution Template

1. Factor. Write $F(D,D')=\prod(D-m_iD')$. Find $m_i$ by solving $F(m,1)=0$ (for second-order, a quadratic; for third-order, a cubic — group and factor).
2. CF. $z_c=\sum\phi_i(y+m_ix)$. For a repeated root $m$: add $x\phi_j(y+mx)$ for each repetition.
3. PI. For each RHS term, apply the appropriate rule. Check for resonance before applying the exponential rule.
4. Sum. $z=z_c+z_p$.

Worked Example(s)

2014 Paper 2, 2014-P2-Q5a (10 marks)

Solve $(2D^2-5DD'+2D'^2)z=24(y-x)$.

Factor. $(2D-D')(D-2D')$. Roots: $m=1/2$ (from $2D-D'=0$) and $m=2$ (from $D-2D'=0$).

CF. $z_c=\phi_1(y+\tfrac12x)+\phi_2(y+2x)$, equivalently $f(2y+x)+g(y+2x)$.

PI. RHS $=24(y-x)$, degree 1. Try cubic ansatz $z_p=Ax^3+Bx^2y$; matching coefficients: $(2D^2-5DD'+2D'^2)(Ax^3+Bx^2y)=12Ax+4By-10Bx$. Set $=(24y-24x)$: $B=6$, $A=3$.

$\boxed{\;z=f(2y+x)+g(y+2x)+3x^3+6x^2y.\;}$

2015 Paper 2, 2015-P2-Q5b (10 marks)

Solve $(D^2+DD'-2D'^2)u=e^{x+y}$.

Factor. $(D-D')(D+2D')$. Roots $m=1$ and $m=-2$.

CF. $u_c=\phi_1(y+x)+\phi_2(y-2x)$.

PI. $F(1,1)=1+1-2=0$: resonance with the factor $(D-D')$.

Apply $(D+2D')^{-1}$ first (non-resonant, value $1+2=3$): gives $e^{x+y}/3$.
Then $(D-D')^{-1}(e^{x+y}/3)$: solve $(D-D')u=e^{x+y}$ via characteristics — $u=xe^{x+y}$ (integration along $x+y=\text{const}$). Divide by 3:

$u_p=\frac{x}{3}e^{x+y}.$

$\boxed{\;u=\phi_1(y+x)+\phi_2(y-2x)+\frac{x}{3}e^{x+y}.\;}$

2017 Paper 2, 2017-P2-Q5a (10 marks)

Solve $(D^2-2DD'+D'^2)z=e^{x+2y}+x^3+\sin2x$.

Factor. $(D-D')^2$: repeated root $m=1$.

CF. $z_c=\phi_1(y+x)+x\phi_2(y+x)$.

PI for $e^{x+2y}$. $F(1,2)=(1-2)^2=1\ne0$: PI $=e^{x+2y}/1=e^{x+2y}$.

PI for $x^3$. Only $D,D^2$ act (since $D'x^3=0$): $1/D^2\cdot x^3=x^5/20$.

PI for $\sin2x$. $D^2\to-4$, $DD'\to0$, $D'^2\to0$: $F\to-4$, PI $=-\sin2x/4$.

$\boxed{\;z=\phi_1(y+x)+x\phi_2(y+x)+e^{x+2y}+\frac{x^5}{20}-\frac{\sin2x}{4}.\;}$

2016 Paper 2, 2016-P2-Q7a (15 marks)

Solve $(D^3-2D^2D'-DD'^2+2D'^3)z=e^{x+y}$.

Factor. Auxiliary $m^3-2m^2-m+2=(m-2)(m-1)(m+1)$: roots $m=2,1,-1$.

CF. $z_c=\phi_1(y+2x)+\phi_2(y+x)+\phi_3(y-x)$.

PI. $F(1,1)=(1-2)(1-1)(1+1)=0$: resonance on $(D-D')$. Non-vanishing factors at $(1,1)$: $(1-2)(1+1)=-2$.

$z_p=\frac{xe^{x+y}}{-2}=-\frac{x}{2}e^{x+y}.$

$\boxed{\;z=\phi_1(y+2x)+\phi_2(y+x)+\phi_3(y-x)-\frac{x}{2}e^{x+y}.\;}$

2018 Paper 2, 2018-P2-Q7a (15 marks)

Solve $(2D^2-5DD'+2D'^2)z=5\sin(2x+y)+24(y-x)+e^{3x+4y}$.

Factor. $(D-2D')(2D-D')$; CF $=\phi_1(y+2x)+\phi_2(2y+x)$.

PI for $5\sin(2x+y)$. $D^2\to-4$, $DD'\to-2$, $D'^2\to-1$: $F=2(-4)-5(-2)+2(-1)=0$. Resonance! The factor $(D-2D')$ annihilates $\sin(2x+y)$ (since $2x+y$ matches the slope $m=2$). Trial $z=x(P\cos(2x+y)+Q\sin(2x+y))$; matching gives PI $=-\tfrac53x\cos(2x+y)$.

PI for $24(y-x)$. Polynomial ansatz (cubic): PI $=\tfrac{12}{7}x^2y-\tfrac{12}{7}xy^2=\tfrac{12}{7}xy(x-y)$.

PI for $e^{3x+4y}$. $F(3,4)=-10\ne0$: PI $=-\tfrac{1}{10}e^{3x+4y}$.

$\boxed{\;z=\phi_1(y+2x)+\phi_2(2y+x)-\frac{5}{3}x\cos(2x+y)+\frac{12}{7}xy(x-y)-\frac{1}{10}e^{3x+4y}.\;}$

2013/2022 Paper 2, 2013-P2-Q6a and 2022-P2-Q7a (15 marks each)

Solve $(D^2+DD'-6D'^2)z=x^2\sin(x+y)$.

Sources:,

Factor. $(D+3D')(D-2D')$; CF $=f(y-3x)+g(y+2x)$.

PI. RHS is $x^2\cdot\sin(x+y)$ — use the shift identity. Replace $\sin(x+y)=\operatorname{Im}(e^{i(x+y)})$ and find PI of $x^2e^{i(x+y)}$. The shifted operator $F(D+i,D'+i)$ has non-zero constant term $F(i,i)=i^2+i^2-6i^2=4\ne0$, so no resonance. Expand $(4+L)^{-1}$ with $L=3iD-11iD'+\ldots$:

$\Phi=[4+L]^{-1}x^2=\tfrac{1}{4}\bigl[x^2-\tfrac{6ix+2}{4}+\tfrac{-18}{16}\bigr]=\tfrac{x^2}{4}-\tfrac{3ix}{8}-\tfrac{13}{32}$.

Take imaginary part after multiplying by $e^{i(x+y)}$:

$z_p=\!\left(\frac{x^2}{4}-\frac{13}{32}\right)\!\sin(x+y)-\frac{3x}{8}\cos(x+y).$

$\boxed{\;z=f(y-3x)+g(y+2x)+\!\left(\frac{x^2}{4}-\frac{13}{32}\right)\!\sin(x+y)-\frac{3x}{8}\cos(x+y).\;}$

2020 Paper 2, 2020-P2-Q5d (10 marks)

Solve $(D^3-2D^2D'-DD'^2+2D'^3)z=e^{2x+y}+\sin(x-2y)$.

Factor. $(D-D')(D+D')(D-2D')$; CF $=\phi_1(y+x)+\phi_2(y-x)+\phi_3(y+2x)$.

PI for $e^{2x+y}$. $F(2,1)=0$ (since $D-2D'$ vanishes at $m=2$, $b/a=1/2$). Non-vanishing factors: $(2-1)(2+1)=3$. PI $=xe^{2x+y}/3$.

PI for $\sin(x-2y)$. Replace $D^2\to-1$, $D'^2\to-4$, $DD'\to2$ (since $a=1,b=-2$, $ab=-2$). Cubic operator reduces to $3D-6D'$: $\dfrac{1}{3D-6D'}\sin(x-2y)=-\dfrac{1}{15}\cos(x-2y)$.

$\boxed{\;z=\phi_1(y+x)+\phi_2(y-x)+\phi_3(y+2x)+\frac{x}{3}e^{2x+y}-\frac{\cos(x-2y)}{15}.\;}$

2025 Paper 2, 2025-P2-Q5a (10 marks)

Solve $(D^2+DD'-2D'^2)z=y\sin x$.

Factor. $(D-D')(D+2D')$; CF $=f(y+x)+g(y-2x)$.

PI. RHS $=y\sin x$. Trial: $z_p=y(a\sin x+b\cos x)+(c\sin x+d\cos x)$. Apply the operator: only $D^2$ and $DD'$ act non-trivially. Matching coefficients of $y\sin x$, $y\cos x$, $\sin x$, $\cos x$:

$a=-1$, $b=0$, $c=0$, $d=-1$.

$\boxed{\;z=f(y+x)+g(y-2x)-y\sin x-\cos x.\;}$

2021 Paper 2, 2021-P2-Q7a (15 marks)

Solve $(D^2-D'^2-3D+3D')z=xy+e^{x+2y}$.

Factor. $D^2-D'^2-3D+3D'=(D-D')(D+D')-3(D-D')=(D-D')(D+D'-3)$.

CF. From $(D-D')$: $\phi_1(y+x)$. From $(D+D'-3)$: $e^{3x}\phi_2(y-x)$ (verify: $D+D'-3$ kills $e^{3x}f(y-x)$ since $D+D'$ kills functions of $y-x$ and $3e^{3x}$ cancels).

PI for $e^{x+2y}$. $F(1,2)=(1-2)(1+2-3)=(-1)(0)=0$: resonance on $(D+D'-3)$. Apply $(D-D')^{-1}$ first (value $-1$): gives $-e^{x+2y}$. Then $(D+D'-3)^{-1}e^{x+2y}=xe^{x+2y}$. So PI $=-xe^{x+2y}$.

PI for $xy$. Use $(u,v)=(x+y,x-y)$ substitution to make the operator separable; result is a degree-3 polynomial in $(x,y)$ (full form in the source).

$\boxed{\;z=\phi_1(y+x)+e^{3x}\phi_2(y-x)+(\text{poly PI})-xe^{x+2y}.\;}$

Common Traps

• Resonance detection is step zero. Before applying $e^{ax+by}/F(a,b)$, always check whether $F(a,b)=0$. If it does, identify which factor vanishes and apply the $x$-multiplier rule only to that factor, dividing by the remaining factors evaluated at $(a,b)$.
• Repeated-root CF requires $x\phi_2$. For $(D-D')^2$, the CF is $\phi_1(y+x)+x\phi_2(y+x)$ — a common error is to write only $\phi_1$.
• Trig PI with odd-degree operators (2020-Q5d): first reduce all $D^2\to-a^2$, $D'^2\to-b^2$, $DD'\to-ab$, which simplifies the cubic operator to a first-degree expression; then invert that.
• Shift identity for $x^n\sin/\cos$ (2013, 2022): write $\sin(ax+by)=\operatorname{Im}(e^{i(ax+by)})$; shift the operator by $(a,b)$ using $F(D+a,D'+b)$; the constant term is $F(ia,ib)$, which must be non-zero (no resonance after the shift); the polynomial expansion terminates because $D^3 x^2=0$.
• $D'$ kills functions of $x$ only: when the RHS is $x^n$ (no $y$), every $D'$ term in the operator annihilates it, and only powers of $D$ act.

operator-pde-boundary (1 question(s); 2023)

Recognition Cues

• “Find the surface satisfying [PDE] and passing through [two lines or curves].”
• General solution has arbitrary functions $\phi_i$; boundary conditions pin them down.

Solution Template

1. Solve the homogeneous PDE to get the general solution $z=f(\xi_1)+xg(\xi_1)$ (or sum of $f_i(\xi_i)$).
2. Apply first BC (simpler curve, often $x=0$ or $z=0$): determines one arbitrary function.
3. Apply second BC (substitute into the remaining general solution): get an equation for the second arbitrary function; solve by substitution.
4. Write the explicit surface.

Worked Example(s)

2023 Paper 2, 2023-P2-Q6a (15 marks)

Find the surface through $z=x=0$ and $z-1=x-y=0$ satisfying $z_{xx}-4z_{xy}+4z_{yy}=0$.

Factor. $(D-2D')^2$: repeated root $m=2$. General solution: $z=f(y+2x)+xg(y+2x)$.

BC 1 ($x=0$, $z=0$): $f(y)=0$ for all $y$, so $f\equiv0$. Now $z=xg(y+2x)$.

BC 2 ($x=y$, $z=1$): $x\,g(3x)=1$, so $g(3x)=1/x$. Set $u=3x$: $g(u)=3/u$.

$z=x\cdot\frac{3}{y+2x}=\boxed{\;\frac{3x}{2x+y}.\;}$

Verify: $z_{xx}-4z_{xy}+4z_{yy}=0$ ✓; $z|_{x=0}=0$ ✓; $z|_{y=x}=3x/3x=1$ ✓.

Common Traps

• For repeated root $(D-mD')^2$: the second piece uses $x\cdot g(y+mx)$, not $y\cdot g$. Using $y$ would give a different (incorrect) surface.
• When inverting $g(3x)=1/x$: write $u=3x$ and express $g$ as a function of a single argument $u$; $g(u)=3/u$ (not $3/x$, which would still depend on the original variable).

verify-pde-solution (1 question(s); 2024)

Recognition Cues

• “Show that $u=f(\xi)+g(\eta)$ satisfies [wave/Laplace/heat equation] when [condition on parameters].”
• No solving required — only differentiation via the chain rule.

Solution Template

1. Compute $u_{xx}$, $u_{yy}$, $u_{tt}$ (or whatever derivatives the PDE involves) using $\xi_x,\xi_y,\xi_t,\eta_x,\eta_y,\eta_t$.
2. Substitute into the PDE; collect terms in $f''(\xi)+g''(\eta)$.
3. The PDE holds iff the coefficient of $f''+g''$ satisfies the stated algebraic condition.

Worked Example(s)

2024 Paper 2, 2024-P2-Q5a (10 marks)

Show $u=f(x-kt+i\alpha y)+g(x-kt-i\alpha y)$ satisfies $u_{xx}+u_{yy}=u_{tt}/c^2$ when $\alpha^2=1-k^2/c^2$.

With $\xi=x-kt+i\alpha y$ and $\eta=x-kt-i\alpha y$:

$u_{xx}=f''+g''$; $\quad u_{yy}=(i\alpha)^2f''+(-i\alpha)^2g''=-\alpha^2(f''+g'')$; $\quad u_{tt}=k^2(f''+g'')$.

$u_{xx}+u_{yy}=(1-\alpha^2)(f''+g''),\qquad \frac{u_{tt}}{c^2}=\frac{k^2}{c^2}(f''+g'').$

Equating: $1-\alpha^2=k^2/c^2$, i.e. $\alpha^2=1-k^2/c^2$. $\blacksquare$

Common Traps

• $(i\alpha)^2=-\alpha^2$, not $+\alpha^2$ or $-i\alpha^2$. Track $i^2=-1$ carefully.

Marks-Aware Writing

10-mark questions (compulsory Q5 slot): Show factorisation (one line), write CF (one line), compute PI with the specific rule named, write the general solution. For resonance: one sentence (”$F(a,b)=0$, so multiply by $x$”), then the corrected PI.

15-mark questions (Q6–Q8 slot): Show factorisation with the auxiliary equation, verify the factorisation. Write CF with correct argument form. For multi-term RHS: handle each PI term in a labelled sub-step. Verify the PI by substituting back (one line, or quote sympy if time is short). Box the general solution.

For the shift-identity questions (2013, 2022): explicitly write $F(D+i,D'+i)$, simplify, identify the constant term $F(i,i)\ne0$, expand the geometric series, and extract the real/imaginary part.

Practice Set