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UPSC 2026 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

Second-order linear PDEs with constant coefficients (CF, PI) · PDEs · asked 13× in 14 yrs · Read the full method →

Question

Solve the partial differential equation

where $D \equiv \dfrac{\partial}{\partial x}$, $D' \equiv \dfrac{\partial}{\partial y}$, $z = z(x, y)$.

Technique

A homogeneous linear PDE with constant coefficients (every term is of order $2$), so the complementary function comes from the auxiliary equation $2m^2 - 5m + 2 = 0$ with $z=\phi(y+mx)$. The trigger for the particular integral is that both pieces of the right-hand side are functions of the single combination $u = y - x$, i.e. of $y+mx$ with $m=-1$. For such a right-hand side a homogeneous operator of degree $n$ collapses to the constant $F(m,1)$ times $d^n/du^n$, so the PI is obtained by $n$ plain integrations — provided $F(-1,1)\neq 0$, which must be checked because $F(-1,1)=0$ is exactly the resonant case where the shortcut fails.

Solution

Step 1 — Factorise the operator

Treat $F(D,D') = 2D^2 - 5DD' + 2D'^2$ as a quadratic form in $D, D'$:

Check by expansion: $(2D-D')(D-2D') = 2D^2 - 4DD' - DD' + 2D'^2 = 2D^2 - 5DD' + 2D'^2$. ✓

The operator is homogeneous of degree $2$, so the equation is a homogeneous linear PDE with constant coefficients and the standard theory applies: general solution $=$ complementary function $+$ particular integral, with two arbitrary functions.

Step 2 — The complementary function

The convention, stated (this is a mark). For a homogeneous operator, put $z = \phi(y + mx)$ with $\phi$ arbitrary and twice differentiable. Then

so

Hence $\phi(y+mx)$ solves $F(D,D')z=0$ iff $m$ is a root of the auxiliary equation

Solving: $2m^2-5m+2=(2m-1)(m-2)=0$, so

The roots are real and distinct, so no repeated-root modification (no factor $x$) is needed, and

with $\phi_1,\phi_2$ arbitrary twice-differentiable functions. Since $\phi_2\!\left(y+\tfrac x2\right) = \psi\!\left(2y+x\right)$ on writing $\psi(\tau)=\phi_2(\tau/2)$ — and $\psi$ is arbitrary exactly when $\phi_2$ is — the CF may equally be written

Two arbitrary functions, matching the order of the equation. ✓

Step 3 — Why the right-hand side is special, and the lemma that exploits it

The right-hand side is

which is a function of the single variable $u = y - x = y + mx$ with $m = -1$. This is the whole point of the question.

Lemma. If $F(D,D')$ is homogeneous of degree $n$ with constant coefficients and $G$ is $n$-times differentiable, then

Proof. Each term of $F$ is $c_k D^{\,n-k}D'^k$, and $D^{\,n-k}D'^{k}G(y+mx) = m^{\,n-k}\,G^{(n)}(y+mx)$, because each $D$ contributes a factor $m$ and each $D'$ a factor $1$. Summing over $k$ gives $\left(\sum_k c_k m^{\,n-k}\right)G^{(n)} = F(m,1)\,G^{(n)}$. $\square$

(Equation $(\ast)$ of Step 2 is this lemma with $n=2$, $G=\phi$.)

Corollary (the shortcut, with its hypothesis). If the right-hand side is $g(y+mx)$ and $F(m,1)\neq 0$, then a particular integral is

the constants of integration being discardable because they generate polynomials in $u$ of degree $<n$, which already lie in the complementary function.

The hypothesis must be checked here, not assumed. With $m=-1$,

Equivalently: $m=-1$ is not a root of $2m^2-5m+2=0$ (the roots are $2$ and $\tfrac12$), so the direction $y-x$ is not a characteristic of the operator and the case is non-resonant. Had $-1$ been a root, the two-fold integration would have produced $0=R$ and the method would have collapsed, requiring the usual $x\cdot(\,\cdot\,)$ modification instead.

Discardability of the integration constants, verified explicitly. A general linear function $\alpha(y-x)+\beta$ does lie in the CF: taking $\phi(\tau)=A\tau$, $\psi(\tau)=B\tau$ gives $A(y+2x)+B(2y+x) = (A+2B)y+(2A+B)x$, and $A=-\alpha,\;B=\alpha$ reproduces $\alpha(y-x)$; the constant $\beta$ is absorbed by a constant $\phi$. So dropping them costs nothing.

Step 4 — The particular integral

By the Corollary with $n=2$, $m=-1$, $F(-1,1)=9$ and $g(u) = 24u + \sin u$:

Hence

Direct verification by substitution (do this in the hall — it is two lines and it insures the whole PI). Write $u=y-x$, so $\partial u/\partial x=-1$, $\partial u/\partial y=+1$. For any twice-differentiable $f$,

so

With $f(u) = \tfrac49 u^3 - \tfrac19\sin u$ we get $f''(u) = \tfrac{8}{3}u + \tfrac19\sin u$, and

(Cross-check of the trigonometric part by the standard operator rule. For $\sin(ax+by)$ one replaces $D^2\to -a^2$, $DD'\to -ab$, $D'^2\to -b^2$. Here $a=-1$, $b=1$, giving $2(-1)-5(1)+2(-1) = -9 \neq 0$, so

in agreement.)

Step 5 — General solution

$\phi,\psi$ arbitrary twice-differentiable functions. $\qquad\blacksquare$

Step 6 — Remark: a second, equally correct PI for the polynomial part

Expanding $F^{-1}$ in powers of $D'/D$ (the routine textbook route) gives, for the part $24(y-x)$,

and indeed $2D^2(6x^2y+3x^3) - 5DD'(6x^2y+3x^3) = 2(12y+18x) - 5(12x) = 24(y-x)$. ✓

This looks different from $\tfrac49(y-x)^3$, but the two differ by a complementary function — as they must:

Both are therefore correct particular integrals and both give the same general solution. Writing this identity out (one line) pre-empts the examiner who has only the other form on the marking scheme.

Answer

with $\phi,\psi$ arbitrary twice-differentiable functions (auxiliary roots $m=2,\tfrac12$; PI legitimate because $F(-1,1)=9\neq0$).