← 2025 Paper 2

UPSC Maths 2025 Paper 2 Q5a — Solution

10 marks · Section B

Question

Find the solution of the equation $(D^2 + DD' - 2D'^2)z = y\sin x$, where $D \equiv \dfrac{\partial}{\partial x}$ and $D' \equiv \dfrac{\partial}{\partial y}$.

Technique

This is a linear PDE with constant coefficients, homogeneous in $D, D'$. Get the complementary function by factorising the operator, and the particular integral by undetermined coefficients (with a polynomial-in-$y$ factor in the trial, since the operator annihilates pure functions of $x$).

Solution

Factorise the operator. $D^2 + DD' - 2D'^2 = (D - D')(D + 2D').$ Check: $(D-D')(D+2D') = D^2 + 2DD' - DD' - 2D'^2 = D^2 + DD' - 2D'^2.$ ✓

Complementary function. For a factor $(D - mD')$, the solution is an arbitrary function $\phi(y + mx)$.

• Factor $(D - D')$: here $m = 1$, giving $f(y + x)$.
• Factor $(D + 2D')$: here $m = -2$, giving $g(y - 2x)$.

$z_c = f(y + x) + g(y - 2x),$ with $f, g$ arbitrary twice-differentiable functions.

Particular integral. We want $z_p$ with $(D^2 + DD' - 2D'^2)z_p = y\sin x$. To account for both the trigonometric factor and the linear factor $y$, take the trial $z_p = y(a\sin x + b\cos x) + (c\sin x + d\cos x).$

Compute the derivatives: $D z_p = y(a\cos x - b\sin x) + (c\cos x - d\sin x),$ $D^2 z_p = y(-a\sin x - b\cos x) + (-c\sin x - d\cos x),$ $D' z_p = a\sin x + b\cos x,\qquad D'^2 z_p = 0,$ $DD' z_p = a\cos x - b\sin x.$

Therefore $(D^2 + DD' - 2D'^2)z_p = y(-a\sin x - b\cos x) + (-c\sin x - d\cos x) + (a\cos x - b\sin x).$

Match against $y\sin x + 0$:

• Coefficient of $y\sin x$: $-a = 1 \Rightarrow a = -1.$
• Coefficient of $y\cos x$: $-b = 0 \Rightarrow b = 0.$
• Coefficient of $\sin x$ (non-$y$): $-c - b = 0 \Rightarrow c = 0.$
• Coefficient of $\cos x$ (non-$y$): $-d + a = 0 \Rightarrow d = a = -1.$

Hence $z_p = -y\sin x - \cos x.$

General solution. $z = f(y + x) + g(y - 2x) - y\sin x - \cos x.$

Answer

$\boxed{\,z = f(y + x) + g(y - 2x) - y\sin x - \cos x\,}$ where $f, g$ are arbitrary functions.