← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Second-order linear PDEs with constant coefficients (CF, PI) · PDEs · asked 12× in 13 yrs · Read the full method →

Question

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

Technique

Factor into $(D-D')^2$ (repeated linear factor → CF $\phi_1+x\phi_2$); term-by-term PI: exponential rule $D\to a,D'\to b$; inverse-operator expansion for the polynomial; $D^2\to-a^2$ for the trig term.

Solution

Setup. The operator factors as a perfect square:

The general solution is $z=\text{CF}+\text{PI}$, and since the RHS is a sum of three terms we compute a particular integral for each separately.

Step 1 — Complementary function

The auxiliary equation comes from $D-mD'=0$ with $(D-D')^2$, i.e. a repeated factor $m=1$ (treating $D=mD'$, the symbolic roots of $m^2-2m+1=0$ are $m=1,1$). For a repeated factor $(D-mD')^2$ the CF is

with $\phi_1,\phi_2$ arbitrary functions.

Step 2 — PI for $e^{x+2y}$

For an exponential $e^{ax+by}$ put $D\to a,\ D'\to b$. Here $a=1,\ b=2$:

So

Step 3 — PI for $x^3$

Since $x^3$ is independent of $y$, every $D'$-term annihilates it ($D'x^3=0$). Hence only the leading term survives:

Step 4 — PI for $\sin 2x$

For $\sin 2x$ (a function of $x$ only), use $D^2\to-2^2=-4$, $DD'\to0$, $D'^2\to0$:

Step 5 — General solution

Answer