← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q5d — 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 the partial differential equation (D3−2D2D′−DD′2+2D′3)z=e2x+y+sin⁡(x−2y)(D^3-2D^2D'-DD'^2+2D'^3)z=e^{2x+y}+\sin(x-2y), where D≡∂∂xD\equiv\frac{\partial}{\partial x}, D′≡∂∂yD'\equiv\frac{\partial}{\partial y}.

Technique

Factor the homogeneous operator into (D−mD′)(D-mD') factors for the CF; operator shortcut D→a,D′→bD\to a,D'\to b for the exponential PI (with the x⋅x\cdot correction for the failing factor) and the D2→−a2D^2\to-a^2 rule for the trigonometric PI.

Solution

This is a linear homogeneous (constant-coefficient) PDE: F(D,D′)z=F(D,D')z= RHS.

Step 1 — Complementary function

Factor the operator. With m=D′/Dm=D'/D the auxiliary equation is

m3−2m2−m+2=0 ⇒ (m−1)(m+1)(m−2)=0,m=1,−1,2.m^3-2m^2-m+2=0\ \Rightarrow\ (m-1)(m+1)(m-2)=0,\quad m=1,-1,2.

Equivalently

F(D,D′)=(D−D′)(D+D′)(D−2D′).F(D,D')=(D-D')(D+D')(D-2D').

For each factor (D−mD′)(D-mD') the solution is an arbitrary function ϕ(y+mx)\phi(y+mx). Hence

CF: zc=ϕ1(y+x)+ϕ2(y−x)+ϕ3(y+2x).\boxed{\text{CF}:\ z_c=\phi_1(y+x)+\phi_2(y-x)+\phi_3(y+2x).}

Step 2 — Particular integral for e2x+ye^{2x+y}

Put D→2, D′→1D\to2,\ D'\to1 in FF:

F(2,1)=8−2(4)(1)−(2)(1)+2(1)=8−8−2+2=0.F(2,1)=8-2(4)(1)-(2)(1)+2(1)=8-8-2+2=0.

So (D−2D′)(D-2D') is the failing factor (indeed 2−2(1)=02-2(1)=0). Use the rule for a simple failing linear factor:

1F(D,D′)e2x+y=1(D−D′)(D+D′)⋅1(D−2D′)e2x+y.\frac{1}{F(D,D')}e^{2x+y}=\frac{1}{(D-D')(D+D')}\cdot\frac{1}{(D-2D')}e^{2x+y}.

The non-vanishing factors at (2,1)(2,1) give (2−1)(2+1)=3(2-1)(2+1)=3. For the failing factor, 1D−2D′e2x+y=x e2x+y\dfrac{1}{D-2D'}e^{2x+y}=x\,e^{2x+y}. Therefore

PI1=x e2x+y(2−1)(2+1)=x3 e2x+y.\text{PI}_1=\frac{x\,e^{2x+y}}{(2-1)(2+1)}=\frac{x}{3}\,e^{2x+y}.

Step 3 — Particular integral for sin⁡(x−2y)\sin(x-2y)

For sin⁡(ax+by)\sin(ax+by) with the cubic operator, substitute D2→−a2, D′2→−b2, DD′→−abD^2\to-a^2,\ D'^2\to-b^2,\ DD'\to-ab. Here a=1, b=−2a=1,\ b=-2. Write FF grouping one DD or D′D' out of each cubic term:

F=D⋅D2−2D′⋅D2−D⋅D′2+2D′⋅D′2.F=D\cdot D^2-2D'\cdot D^2-D\cdot D'^2+2D'\cdot D'^2.

Replace D2→−1, D′2→−4D^2\to-1,\ D'^2\to-4:

F→D(−1)−2D′(−1)−D(−4)+2D′(−4)=−D+2D′+4D−8D′=3D−6D′.F\to D(-1)-2D'(-1)-D(-4)+2D'(-4)=-D+2D'+4D-8D'=3D-6D'.

Thus

PI2=13D−6D′sin⁡(x−2y)=13⋅1D−2D′sin⁡(x−2y).\text{PI}_2=\frac{1}{3D-6D'}\sin(x-2y)=\frac{1}{3}\cdot\frac{1}{D-2D'}\sin(x-2y).

Now 1D−2D′sin⁡(x−2y)\dfrac{1}{D-2D'}\sin(x-2y): operate using 1D−2D′sin⁡(ax+by)\dfrac{1}{D-2D'}\sin(ax+by). Multiply numerator and denominator by (D+2D′)(D+2D')? Simpler: note 1D−2D′sin⁡(x−2y)=∫sin⁡(x−2c) \dfrac{1}{D-2D'}\sin(x-2y)=\displaystyle\int\sin(x-2c)\, along characteristics; we instead clear the operator. Apply D−2D′D-2D' to a trial Acos⁡(x−2y)A\cos(x-2y):

(D−2D′)[Acos⁡(x−2y)]=−Asin⁡(x−2y)−2D′[Acos⁡(x−2y)].(D-2D')\big[A\cos(x-2y)\big]=-A\sin(x-2y)-2D'\big[A\cos(x-2y)\big].

Here D′[cos⁡(x−2y)]=2sin⁡(x−2y)D'[\cos(x-2y)]=2\sin(x-2y), so (D−2D′)[Acos⁡]=−Asin⁡(x−2y)−2A⋅(2sin⁡)=−5Asin⁡(x−2y)(D-2D')[A\cos]=-A\sin(x-2y)-2A\cdot(2\sin)=-5A\sin(x-2y). Setting −5A=1-5A=1 gives A=−15A=-\tfrac15, so

1D−2D′sin⁡(x−2y)=−15cos⁡(x−2y),\frac{1}{D-2D'}\sin(x-2y)=-\frac15\cos(x-2y),

and

PI2=13(−15cos⁡(x−2y))=−115cos⁡(x−2y).\text{PI}_2=\frac{1}{3}\left(-\frac15\cos(x-2y)\right)=-\frac{1}{15}\cos(x-2y).

Step 4 — General solution

Answer

  z=ϕ1(y+x)+ϕ2(y−x)+ϕ3(y+2x)+x3e2x+y−115cos⁡(x−2y).  \boxed{\;z=\phi_1(y+x)+\phi_2(y-x)+\phi_3(y+2x)+\frac{x}{3}e^{2x+y}-\frac{1}{15}\cos(x-2y).\;}
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