← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q5b — 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 $(D^2+DD'-2D'^2)u=e^{x+y}$, where $D=\partial/\partial x$, $D'=\partial/\partial y$.

Technique

Factor the operator $D^2+DD'-2D'^2=(D-D')(D+2D')$; CF from each factor as $\phi_i(y+a_i x)$; PI with resonance handled by applying the non-resonant factor first ($D+2D'\to 3$), then solving the resonant first-order PDE directly via characteristics.

Solution

Step 1 — Factor the operator

$D^2+DD'-2D'^2=(D-D')(D+2D')$? Check: $(D-D')(D+2D')=D^2+2DD'-D'D-2D'^2=D^2+DD'-2D'^2$ ✓.

So the operator factors as $(D-D')(D+2D')$.

Step 2 — Complementary function

For $(D-aD')u=0$, the general solution is $u=\phi(y+ax)$ (arbitrary function).

• Factor $(D-D')$: solution $\phi_1(y+x)$.
• Factor $(D+2D')$: solution $\phi_2(y-2x)$.

CF: $u_c=\phi_1(y+x)+\phi_2(y-2x)$, where $\phi_1,\phi_2$ are arbitrary functions of one variable.

Step 3 — Particular integral

For RHS $e^{ax+by}$, $D\to a$, $D'\to b$. Here $a=1,b=1$.

$f(D,D')=D^2+DD'-2D'^2$; $f(1,1)=1+1-2=0$ — resonance.

Factor $(D-D')$ kills $e^{x+y}$ (since $D-D'$ applied to $e^{x+y}$ gives $e^{x+y}-e^{x+y}=0$). So we need a $t$-multiplier-style boost.

Standard rule: if $f(D,D')=(D-aD')\cdot g(D,D')$ and $g(a\cdot 1,1\cdot 1)\ne 0$… let me set up properly.

For PI of $e^{x+y}/[(D-D')(D+2D')]$: factor $D+2D'$ at $D=1,D'=1$ gives $1+2=3\ne 0$. So apply $(D+2D')^{-1}$ first (no resonance with this factor), then $(D-D')^{-1}$ (resonance with this factor).

Apply $(D+2D')^{-1}$ to $e^{x+y}$: $1/(D+2D')$ acting on $e^{x+y}$: substitute $D=1,D'=1$, get $1/3$. So $(D+2D')^{-1}e^{x+y}=e^{x+y}/3$.

Apply $(D-D')^{-1}$ to $e^{x+y}/3$: resonance. Use the standard shift rule:

For PDE $(D-D')u=e^{x+y}$: this is a first-order linear PDE. Lagrange: $dx/1=dy/(-1)=du/e^{x+y}$.

$dy/(-1)=dx/1\Rightarrow y=-x+c_1$, i.e. $x+y=c_1$. So $e^{x+y}=e^{c_1}$ constant along characteristics.

$du/e^{x+y}=dx/1\Rightarrow du=e^{x+y}dx=e^{c_1}dx$, so $u=e^{c_1}x+c_2=xe^{x+y}+c_2$.

Hence one PI of $(D-D')u=e^{x+y}$ is $u=xe^{x+y}$.

We had $1/(D-D')$ acting on $e^{x+y}/3$: so $\text{PI}=xe^{x+y}/3$.

Verification (PI for original): Take $u_p=xe^{x+y}/3$.

$D u_p=(1/3)(e^{x+y}+xe^{x+y})=(e^{x+y}/3)(1+x)$. $D^2 u_p=(1/3)[e^{x+y}(1+x)+e^{x+y}]=(e^{x+y}/3)(2+x)$. $D' u_p=(1/3)xe^{x+y}=xe^{x+y}/3$. $DD' u_p=(1/3)[e^{x+y}+xe^{x+y}]=(e^{x+y}/3)(1+x)$. $D'^2 u_p=xe^{x+y}/3$.

$(D^2+DD'-2D'^2)u_p=(e^{x+y}/3)[(2+x)+(1+x)-2x]=(e^{x+y}/3)(3)=e^{x+y}$ ✓.

Step 4 — General solution

Answer