← 2021 Paper 2

UPSC 2021 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

Find general solution of $(D^2-D'^2-3D+3D')z=xy+e^{x+2y}$.

Technique

Factor operator; CF from each factor; PI for $e^{x+2y}$ requires resonance handling; PI for $xy$ via change of variables $u=x+y,v=x-y$.

Solution

Step 1 — Factor operator

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

Step 2 — Complementary function

From $(D-D')$: $z=\phi_1(y+x)$.

From $(D+D'-3)$: $z=e^{3x}\phi_2(y-x)$ (since $D+D'-3$ acting on $e^{ax}f(y-x)$ gives $(a-3)e^{ax}f$ plus the $f$-derivative parts; for $a=3$, the $f$-derivative parts cancel because $D+D'$ applied to $f(y-x)$ gives $-f'+f'=0$).

Verify: $z=e^{3x}\phi_2(y-x)$, $Dz=3e^{3x}\phi_2-e^{3x}\phi_2'$, $D'z=e^{3x}\phi_2'$. $(D+D'-3)z=3e^{3x}\phi_2-e^{3x}\phi_2'+e^{3x}\phi_2'-3e^{3x}\phi_2=0$ ✓.

So CF $=\phi_1(y+x)+e^{3x}\phi_2(y-x)$.

Step 3 — Particular integral for $e^{x+2y}$

$f(D,D')=(D-D')(D+D'-3)$.

At $D=1,D'=2$: $f(1,2)=(1-2)(1+2-3)=(-1)(0)=0$. Resonance!

The factor $(D+D'-3)$ vanishes at $(1,2)$. Standard resonance handling:

$\dfrac{1}{(D-D')(D+D'-3)}e^{x+2y}$: apply $(D-D')^{-1}$ first (where it’s not resonant — $(D-D')$ at $(1,2)$ gives $-1$, nonzero):

$\dfrac{1}{D-D'}e^{x+2y}=\dfrac{e^{x+2y}}{-1}=-e^{x+2y}$.

Now apply $(D+D'-3)^{-1}$ to $-e^{x+2y}$ — resonant.

For resonance with $(D+D'-3)$: use $1/(D+D'-3)\cdot e^{x+2y}=x\cdot e^{x+2y}/[\partial(D+D'-3)/\partial D]=x\cdot e^{x+2y}/1=x e^{x+2y}$ (taking $x$-derivative; or $y$-derivative gives same).

Actually for the resonant case $f(D,D')[e^{ax+by}]=0$ at $(a,b)$, the PI is $x\cdot e^{ax+by}/f'_D(a,b)$ if $f'_D\ne 0$, or $y\cdot e^{ax+by}/f'_{D'}(a,b)$.

$f=(D+D'-3)$, $f'_D=1$.

$\dfrac{1}{D+D'-3}e^{x+2y}=x e^{x+2y}$.

So PI for $e^{x+2y}$: $\dfrac{1}{(D-D')(D+D'-3)}e^{x+2y}=(-1)\cdot x e^{x+2y}=-x e^{x+2y}$.

Step 4 — Particular integral for $xy$

$\dfrac{1}{(D-D')(D+D'-3)}xy$.

Expand $1/(D+D'-3)=-\dfrac{1}{3}\cdot\dfrac{1}{1-(D+D')/3}=-\dfrac{1}{3}\sum_{n=0}^\infty(D+D')^n/3^n$.

Acting on $xy$ (degree 2 polynomial), only $n=0,1,2$ contribute.

$n=0$: $1\cdot xy=xy$. $n=1$: $(D+D')xy=y+x$. $n=2$: $(D+D')^2 xy=(D+D')(y+x)=(0+1)+(1+0)=2$. (Both $D(y+x)=1$ and $D'(y+x)=1$, so $(D+D')(y+x)=2$.) $n\ge 3$: 0.

So $\dfrac{1}{D+D'-3}xy=-\dfrac{1}{3}\!\left[xy+\dfrac{y+x}{3}+\dfrac{2}{9}+0\right]=-\dfrac{xy}{3}-\dfrac{x+y}{9}-\dfrac{2}{27}$.

Now apply $1/(D-D')$ to this. Let $g=-xy/3-(x+y)/9-2/27$.

$1/(D-D')[g]$: this involves integrating along characteristics $x-y=$ const. For polynomial $g$:

$1/(D-D')[xy]$: we need $w$ such that $(D-D')w=xy$, i.e., $w_x-w_y=xy$. Try $w=ax^2 y+bx y^2+\cdots$. $w_x=2axy+by^2$, $w_y=ax^2+2bxy$. $w_x-w_y=-ax^2+2(a-b)xy+by^2$. Match to $xy$: $a=0,b=0,2(a-b)=1$. Contradiction.

Try $w=$ polynomial of higher degree. Let $w=A x^2 y+B xy^2$. $w_x=2Axy+By^2$, $w_y=Ax^2+2Bxy$. $w_x-w_y=-Ax^2+2(A-B)xy+By^2$. Match $xy$: $-A=0, 2(A-B)=1, B=0\Rightarrow A=B=0$, contradiction.

Try $w=A x^2 y+B x y^2$ + lower. Doesn’t work.

Try $w=(x+y)g(x,y)$ for some $g$. Actually polynomial solution requires careful matching.

Alternative: use $1/(D-D')[xy]=$ (operator series). $1/(D-D')=-1/(D'-D)=-1/D'\cdot 1/(1-D/D')$… not converging on polynomials.

Hmm. The structure $\dfrac{1}{D-D'}$ on polynomials is not as simple as the constant-coefficient case.

Alternative method: substitute $u=x+y,v=x-y$. Then $D=\partial_x=\partial_u+\partial_v$, $D'=\partial_y=\partial_u-\partial_v$. $D-D'=2\partial_v$, $D+D'=2\partial_u$.

$D+D'-3=2\partial_u-3$.

The original operator: $(D-D')(D+D'-3)=2\partial_v(2\partial_u-3)$.

$xy=(u^2-v^2)/4$.

$\dfrac{1}{(D-D')(D+D'-3)}\!\left[\dfrac{u^2-v^2}{4}\right]=\dfrac{1}{4\cdot 2\partial_v(2\partial_u-3)}(u^2-v^2)$.

$1/(2\partial_u-3)$: standard. For polynomial $f(u)$: $1/(2\partial_u-3)\cdot u^2$: solve $(2\partial_u-3)w=u^2$. Try $w=au^2+bu+c$: $2(2au+b)-3(au^2+bu+c)=u^2\Rightarrow -3a u^2+(4a-3b)u+(2b-3c)=u^2\Rightarrow a=-1/3, b=-4/9, c=-8/27$. So $1/(2\partial_u-3)\cdot u^2=-u^2/3-4u/9-8/27$.

$1/(2\partial_u-3)\cdot (-v^2)=-v^2$ multiplied by $1/(2\partial_u-3)$ applied to a constant in $u$: $1/(2\partial_u-3)\cdot 1=-1/3$ (since $(2\partial_u-3)(-1/3)=1$). Wait that’s wrong: $(2\partial_u-3)\cdot(-1/3)=0-3(-1/3)=1$ ✓.

So $1/(2\partial_u-3)\cdot(-v^2)=-v^2\cdot(-1/3)=v^2/3$.

Sum: $1/(2\partial_u-3)\cdot(u^2-v^2)=-u^2/3-4u/9-8/27+v^2/3$.

Now apply $1/(2\partial_v)$ (=$(1/2)\int dv$):

$\dfrac{1}{2\partial_v}\!\left[-\dfrac{u^2}{3}-\dfrac{4u}{9}-\dfrac{8}{27}+\dfrac{v^2}{3}\right]=\dfrac{1}{2}\!\left[-\dfrac{u^2 v}{3}-\dfrac{4uv}{9}-\dfrac{8v}{27}+\dfrac{v^3}{9}\right]$.

Multiply by $1/4$ (from $xy=(u^2-v^2)/4$):

$1/(4\cdot 2\partial_v(2\partial_u-3))[u^2-v^2]=\dfrac{1}{8}\!\left[-\dfrac{u^2 v}{3}-\dfrac{4uv}{9}-\dfrac{8v}{27}+\dfrac{v^3}{9}\right]$.

Convert back $u=x+y,v=x-y$:

$y_{p,1}=\dfrac{1}{8}\!\left[-\dfrac{(x+y)^2(x-y)}{3}-\dfrac{4(x+y)(x-y)}{9}-\dfrac{8(x-y)}{27}+\dfrac{(x-y)^3}{9}\right]$.

This is messy but acceptable.

Step 5 — General solution

Answer