← 2022 Paper 2

UPSC 2022 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 $(D^2+DD'-6D'^2)z=x^2\sin(x+y)$ where $D=\partial/\partial x$, $D'=\partial/\partial y$.

Technique

Factor operator $f(D,D')=(D+3D')(D-2D')$; CF $\phi_1(y+2x)+\phi_2(y-3x)$; PI for $x^2 e^{i(x+y)}$ via shift theorem; extract imaginary part.

Solution

Step 1 — Factor operator

$D^2+DD'-6D'^2=(D+3D')(D-2D')$? Check: $(D+3D')(D-2D')=D^2-2DD'+3D'D-6D'^2=D^2+DD'-6D'^2$ ✓.

Step 2 — Complementary function

CF from each factor:

• $(D-2D')$ gives $\phi_1(y+2x)$.
• $(D+3D')$ gives $\phi_2(y-3x)$.

$z_c=\phi_1(y+2x)+\phi_2(y-3x)$.

Step 3 — Particular integral

For RHS $=x^2\sin(x+y)$, write $\sin(x+y)=\Im e^{i(x+y)}$.

Particular for $x^2 e^{i(x+y)}$: this has the form $x^2\cdot e^{ax+by}$ with $a=b=i$ but multiplied by polynomial.

Standard trick: Use the operator inverse $\dfrac{1}{f(D,D')}$ acting on polynomial times exponential.

$\dfrac{1}{f(D,D')}[x^2 e^{i(x+y)}]=e^{i(x+y)}\cdot\dfrac{1}{f(D+i,D'+i)}[x^2]$.

(Standard shift: $f(D,D')[e^{ax+by}P(x,y)]=e^{ax+by}f(D+a,D'+b)[P]$.)

Compute $f(D+i,D'+i)$: $(D+i)^2+(D+i)(D'+i)-6(D'+i)^2$ $=D^2+2iD-1+DD'+iD+iD'-1-6(D'^2+2iD'-1)$ $=D^2+DD'-6D'^2+2iD+iD+iD'-12iD'+(-1-1+6)$ $=D^2+DD'-6D'^2+3iD-11iD'+4$.

Hmm careful: $-1+(-1)+(-6)(-1)=-1-1+6=4$. ✓

$2iD+iD-12iD'+iD'$: let me redo coefficient of $D$ and $D'$:

• From $(D+i)^2=D^2+2iD-1$: coefficient of $D$ is $2i$.
• From $(D+i)(D'+i)=DD'+iD+iD'-1$: coefficient of $D$ is $i$, coefficient of $D'$ is $i$.
• From $-6(D'+i)^2=-6D'^2-12iD'+6$: coefficient of $D'$ is $-12i$.

Coefficient of $D$: $2i+i=3i$. Coefficient of $D'$: $i-12i=-11i$.

So $f(D+i,D'+i)=D^2+DD'-6D'^2+3iD-11iD'+4$.

Acting on $x^2$: only $D$-terms produce nonzero results.

$D[x^2]=2x$, $D^2[x^2]=2$, $D'[x^2]=0$, $D'^2[x^2]=0$, $DD'[x^2]=0$.

$f(D+i,D'+i)[x^2]=2+0-0+3i(2x)-11i(0)+4(x^2)=2+6ix+4x^2$.

We need $1/f(D+i,D'+i)[x^2]$, i.e., solve $f(D+i,D'+i)[P]=x^2$ for polynomial $P$.

Let $P=Ax^2+Bx+C$ (no $y$ since RHS is $y$-free).

$f(D+i,D'+i)[P]=2A+3i(2Ax+B)-11i\cdot 0+4(Ax^2+Bx+C)$ $=2A+6iAx+3iB+4Ax^2+4Bx+4C$ $=4Ax^2+(6iA+4B)x+(2A+3iB+4C)$.

Match to $x^2$:

• $x^2$: $4A=1\Rightarrow A=1/4$.
• $x$: $6iA+4B=0\Rightarrow 4B=-6iA=-6i/4=-3i/2\Rightarrow B=-3i/8$.
• constant: $2A+3iB+4C=0\Rightarrow 1/2+3i\cdot(-3i/8)+4C=0\Rightarrow 1/2+9/8+4C=0\Rightarrow 4C=-13/8\Rightarrow C=-13/32$.

So $P=\dfrac{x^2}{4}-\dfrac{3ix}{8}-\dfrac{13}{32}$.

Step 4 — Particular integral (complex form)

$z_p^{(\mathbb C)}=e^{i(x+y)}\!\left[\dfrac{x^2}{4}-\dfrac{3ix}{8}-\dfrac{13}{32}\right]$.

Step 5 — Take imaginary part

$e^{i(x+y)}=\cos(x+y)+i\sin(x+y)$.

$z_p^{(\mathbb C)}=(\cos(x+y)+i\sin(x+y))\!\left[\dfrac{x^2}{4}-\dfrac{13}{32}-\dfrac{3ix}{8}\right]$.

Real part = $\cos(x+y)\cdot(x^2/4-13/32)+\sin(x+y)\cdot(3x/8)$.

Imaginary part = $\sin(x+y)\cdot(x^2/4-13/32)-\cos(x+y)\cdot(3x/8)$.

We want $\Im z_p^{(\mathbb C)}$ (since RHS is $\sin$):

Step 6 — General solution

Answer