← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 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

where $D\equiv\dfrac{\partial}{\partial x},\ D'\equiv\dfrac{\partial}{\partial y}$.

Technique

Factor $(D-2D')(2D-D')$ for the CF; PI term-by-term — resonant trig ($F=0$ ⇒ multiply by $x$), polynomial via operator-series $1/F(D,D')$, exponential via $D\to a,\,D'\to b$.

Solution

Setup. A homogeneous linear PDE with constant coefficients. Factor the operator, write the complementary function from the linear factors, then build the particular integral term by term.

Step 1 — Factor the operator and write the CF

For a factor $(D-mD')$ the complementary part is $\phi(y+mx)$. Here:

• $(D-2D')\Rightarrow\phi_1(y+2x)$,
• $(2D-D')=2\big(D-\tfrac12 D'\big)\Rightarrow\phi_2(y+\tfrac12 x)$, equivalently $\phi_2(2y+x)$.

Step 2 — PI for $5\sin(2x+y)$ (resonant case)

Put $D^2\to-2^2=-4$, $DD'\to-(2)(1)=-2$, $D'^2\to-1^2=-1$ in $F(D,D')$:

$F=0$ ⇒ resonance. The annihilating factor is $(D-2D')$ since $(D-2D')\sin(2x+y)=2\cos-2\cos=0$. Use the rule for a failure: multiply by $x$ and differentiate the operator. Take the ansatz $z=x\big(P\cos(2x+y)+Q\sin(2x+y)\big)$; matching gives $P=-\tfrac53,\ Q=0$:

Step 3 — PI for $24(y-x)$ (polynomial RHS)

Expand $\dfrac{1}{2D^2}\Big(1-\tfrac{5D'}{2D}+\tfrac{D'^2}{D^2}\Big)^{-1}=\dfrac{1}{2D^2}\Big(1+\tfrac{5D'}{2D}+\cdots\Big)$ acting on the linear $24(y-x)$. Keeping only terms that survive on a degree-1 input,

Now $\dfrac1D$ means integrate w.r.t. $x$. $\dfrac1D[24(y-x)]=24xy-12x^2$; $D'[\,\cdot\,]$ etc. Carrying the algebra (or solving by an undetermined cubic) gives

Step 4 — PI for $e^{3x+4y}$

Put $D\to3,\ D'\to4$:

Step 5 — General solution

Answer