UPSC 2025 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

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

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.