← 2014 Paper 2

UPSC 2014 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 the partial differential equation $(2D^{2}-5DD'+2D'^{2})z=24(y-x)$.

Technique

Operator factorisation for CF; polynomial-ansatz PI.

Solution

Strategy. Factor the operator → CF as sum of $f(y+rx)$ for each root $r$. PI for polynomial RHS by direct ansatz or operator inversion.

Step 1 — Factor the operator

$2D^{2}-5DD'+2D'^{2}=(2D-D')(D-2D')$.

Verify: $(2D-D')(D-2D')=2D^{2}-4DD'-DD'+2D'^{2}=2D^{2}-5DD'+2D'^{2}$ ✓.

Step 2 — Complementary function

Family 1: $(2D-D')z=0$, i.e., $2z_x-z_y=0$. Characteristics: $dy/dx=-2$, so $z=f(x+2y)$ for arbitrary $f$ (the curve $x+2y=$ const is the characteristic).

Family 2: $(D-2D')z=0$, i.e., $z_x-2z_y=0$. Characteristics: $dy/dx=-1/2$, so $z=g(2x+y)$.

Step 3 — Particular integral

For polynomial RHS $24(y-x)$ of degree 1, try a polynomial PI of degree 3 (operator is degree 2, raising the degree by 2 when inverted).

Ansatz: $z_p=\alpha x^{3}+\beta x^{2}y+\gamma xy^{2}+\delta y^{3}$.

Compute $(2D^{2}-5DD'+2D'^{2})z_p$:

• $D^{2}z_p=6\alpha x+2\beta y$.
• $DD'z_p=2\beta x+2\gamma y$.
• $D'^{2}z_p=2\gamma x+6\delta y$.

$2D^{2}-5DD'+2D'^{2}$ acting:

Set $=24(y-x)$:

• Coefficient of $x$: $12\alpha-10\beta+4\gamma=-24$.
• Coefficient of $y$: $4\beta-10\gamma+12\delta=24$.

Try $\gamma=\delta=0$. Then $12\alpha-10\beta=-24$ and $4\beta=24$, so $\beta=6$, $12\alpha-60=-24$, $12\alpha=36$, $\alpha=3$.

So $z_p=3x^{3}+6x^{2}y$.

Step 4 — Verify

$D z_p=9x^{2}+12xy$, $D'z_p=6x^{2}$, $D^{2}z_p=18x+12y$, $DD'z_p=12x$, $D'^{2}z_p=0$.

$(2D^{2}-5DD'+2D'^{2})z_p=2(18x+12y)-5(12x)+0=36x+24y-60x=24y-24x=24(y-x)$ ✓.

Step 5 — General solution

Answer