UPSC 2016 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

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

Question

Solve the partial differential equation

\frac{\partial^3 z}{\partial x^3}-2\frac{\partial^3 z}{\partial x^2\partial y}-\frac{\partial^3 z}{\partial x\partial y^2}+2\frac{\partial^3 z}{\partial y^3}=e^{x+y}.

Technique

Factor the constant-coefficient operator via the auxiliary cubic $m^3-2m^2-m+2=(m-2)(m-1)(m+1)$ ; simple factors give CF $\sum\phi_i(y+m_ix)$ ; for the PI, substitution $D\to1,D'\to1$ kills the factor $(D-D')$ (resonance), handled by $\tfrac{1}{D-D'}e^{x+y}=xe^{x+y}$ with the remaining factors evaluated numerically.

Solution

Setup. With $D=\partial/\partial x,\ D'=\partial/\partial y$ , the equation is

\big(D^3-2D^2D'-DD'^2+2D'^3\big)z=e^{x+y}.

This is a linear homogeneous (constant-coefficient) operator, solvable by factoring into linear factors $D-mD'$ .

Step 1 — Factor the operator

Put $D=mD'$ (auxiliary): $m^3-2m^2-m+2=0$ . Group:

m^2(m-2)-(m-2)=(m-2)(m^2-1)=(m-2)(m-1)(m+1).

Roots $m=2,\,1,\,-1$ , so

D^3-2D^2D'-DD'^2+2D'^3=(D-2D')(D-D')(D+D').

Step 2 — Complementary function

Each simple factor $(D-mD')$ contributes an arbitrary function $\phi(y+mx)$ :

\boxed{\text{CF}=\phi_1(y+2x)+\phi_2(y+x)+\phi_3(y-x),}

with $\phi_1,\phi_2,\phi_3$ arbitrary.

Step 3 — Particular integral (resonance)

For $e^{ax+by}$ substitute $D\to a,\ D'\to b$ ; here $a=b=1$ :

\phi(1,1)=(1-2)(1-1)(1+1)=(-1)(0)(2)=0.

The factor $(D-D')$ vanishes at $(1,1)$ — resonance. Treat the vanishing factor by the rule $\dfrac{1}{D-D'}e^{x+y}=x\,e^{x+y}$ (the multiplier $x$ arises because $b-a\cdot1=0$ for this factor), while the other two factors are evaluated at $(1,1)$ :

\text{PI}=\frac{e^{x+y}}{(D-2D')(D+D')}\Big|_{D=1,D'=1}\cdot\frac{1}{(D-D')}e^{x+y} =\frac{1}{(1-2)(1+1)}\cdot x\,e^{x+y}=\frac{x\,e^{x+y}}{-2}.

\boxed{\text{PI}=-\tfrac12\,x\,e^{x+y}.}

Step 4 — General solution

Answer

\boxed{\;z=\phi_1(y+2x)+\phi_2(y+x)+\phi_3(y-x)-\tfrac12\,x\,e^{x+y}.\;}