← 2013 Paper 2
UPSC 2013 Maths Optional Paper 2 Q6a — 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
Solve
( D 2 + D D ′ − 6 D ′ 2 ) z = x 2 sin ( x + y ) , (D^{2}+DD'-6D'^{2})z=x^{2}\sin(x+y), ( D 2 + D D ′ − 6 D ′ 2 ) z = x 2 sin ( x + y ) , where D D D D ′ D' D ′ ∂ ∂ x \dfrac{\partial}{\partial x} ∂ x ∂ ∂ ∂ y \dfrac{\partial}{\partial y} ∂ y ∂
Technique
Operator factorisation for CF; complex-exponential shift identity + polynomial PI for the particular solution.
Solution
Strategy. Factor the operator → complementary function. Use complex exponential method (sin ( x + y ) = Im e i ( x + y ) \sin(x+y)=\operatorname{Im}\,e^{i(x+y)} sin ( x + y ) = Im e i ( x + y ) f ( D , D ′ ) [ e a x + b y ϕ ] = e a x + b y f ( D + a , D ′ + b ) ϕ f(D,D')[e^{ax+by}\phi]=e^{ax+by}f(D+a,D'+b)\phi f ( D , D ′ ) [ e a x + b y ϕ ] = e a x + b y f ( D + a , D ′ + b ) ϕ
Step 1 — Factor the operator and find CF
D 2 + D D ′ − 6 D ′ 2 = ( D + 3 D ′ ) ( D − 2 D ′ ) . D^{2}+DD'-6D'^{2}=(D+3D')(D-2D'). D 2 + D D ′ − 6 D ′ 2 = ( D + 3 D ′ ) ( D − 2 D ′ ) . Two characteristic roots in D / D ′ D/D' D / D ′ r = − 3 r=-3 r = − 3 D + 3 D ′ = 0 D+3D'=0 D + 3 D ′ = 0 r = 2 r=2 r = 2 D − 2 D ′ = 0 D-2D'=0 D − 2 D ′ = 0
For each root r r r f ( y + r x ) f(y+rx) f ( y + r x )
z c = f ( y − 3 x ) + g ( y + 2 x ) , f , g arbitrary . z_c=f(y-3x)+g(y+2x),\qquad f,g\text{ arbitrary}. z c = f ( y − 3 x ) + g ( y + 2 x ) , f , g arbitrary . Step 2 — Particular integral via complex shift
Write sin ( x + y ) = Im e i ( x + y ) \sin(x+y)=\operatorname{Im}\,e^{i(x+y)} sin ( x + y ) = Im e i ( x + y )
( D 2 + D D ′ − 6 D ′ 2 ) z = x 2 e i ( x + y ) (D^{2}+DD'-6D'^{2})z=x^{2}e^{i(x+y)} ( D 2 + D D ′ − 6 D ′ 2 ) z = x 2 e i ( x + y ) then take imaginary part.
Shift: z p C = e i ( x + y ) Φ ( x , y ) z_p^{\mathbb C}=e^{i(x+y)}\,\Phi(x,y) z p C = e i ( x + y ) Φ ( x , y ) Φ = [ f ( D + i , D ′ + i ) ] − 1 x 2 \Phi=[f(D+i,D'+i)]^{-1}\,x^{2} Φ = [ f ( D + i , D ′ + i ) ] − 1 x 2 f ( D , D ′ ) = D 2 + D D ′ − 6 D ′ 2 f(D,D')=D^{2}+DD'-6D'^{2} f ( D , D ′ ) = D 2 + D D ′ − 6 D ′ 2
Compute f ( D + i , D ′ + i ) f(D+i,D'+i) f ( D + i , D ′ + i )
( D + i ) 2 + ( D + i ) ( D ′ + i ) − 6 ( D ′ + i ) 2 = D 2 + 2 i D − 1 + D D ′ + i D + i D ′ − 1 − 6 D ′ 2 − 12 i D ′ + 6 (D+i)^{2}+(D+i)(D'+i)-6(D'+i)^{2}=D^{2}+2iD-1+DD'+iD+iD'-1-6D'^{2}-12iD'+6 ( D + i ) 2 + ( D + i ) ( D ′ + i ) − 6 ( D ′ + i ) 2 = D 2 + 2 i D − 1 + D D ′ + i D + i D ′ − 1 − 6 D ′ 2 − 12 i D ′ + 6 = D 2 + D D ′ − 6 D ′ 2 + 3 i D − 11 i D ′ + 4. =D^{2}+DD'-6D'^{2}+3iD-11iD'+4. = D 2 + D D ′ − 6 D ′ 2 + 3 i D − 11 i D ′ + 4. So we need [ 4 + L ] − 1 x 2 [4+L]^{-1}x^{2} [ 4 + L ] − 1 x 2 L = 3 i D − 11 i D ′ + D 2 + D D ′ − 6 D ′ 2 L=3iD-11iD'+D^{2}+DD'-6D'^{2} L = 3 i D − 11 i D ′ + D 2 + D D ′ − 6 D ′ 2
Step 3 — Apply L L L x 2 x^{2} x 2
Note D ′ x 2 = 0 D'x^{2}=0 D ′ x 2 = 0 D ′ 2 x 2 = 0 D'^{2}x^{2}=0 D ′ 2 x 2 = 0 D D ′ x 2 = 0 DD'\,x^{2}=0 D D ′ x 2 = 0 D D D D 2 D^{2} D 2 x x x
L x 2 = 3 i D x 2 + D 2 x 2 = 3 i ( 2 x ) + 2 = 6 i x + 2. Lx^{2}=3iD\,x^{2}+D^{2}x^{2}=3i(2x)+2=6ix+2. L x 2 = 3 i D x 2 + D 2 x 2 = 3 i ( 2 x ) + 2 = 6 i x + 2. L 2 x 2 = L ( 6 i x + 2 ) = 3 i D ( 6 i x + 2 ) + D 2 ( 6 i x + 2 ) = 3 i ( 6 i ) + 0 = − 18. L^{2}x^{2}=L(6ix+2)=3iD(6ix+2)+D^{2}(6ix+2)=3i(6i)+0=-18. L 2 x 2 = L ( 6 i x + 2 ) = 3 i D ( 6 i x + 2 ) + D 2 ( 6 i x + 2 ) = 3 i ( 6 i ) + 0 = − 18. L 3 x 2 = L ( − 18 ) = 0 ( constants annihilated by D , D ′ ) . L^{3}x^{2}=L(-18)=0\;(\text{constants annihilated by }D,D'). L 3 x 2 = L ( − 18 ) = 0 ( constants annihilated by D , D ′ ) . So the formal expansion terminates:
Φ = [ 4 + L ] − 1 x 2 = 1 4 [ 1 − L 4 + L 2 16 − ⋯ ] x 2 = 1 4 [ x 2 − 6 i x + 2 4 + − 18 16 ] . \Phi=[4+L]^{-1}x^{2}=\frac{1}{4}\!\left[1-\frac{L}{4}+\frac{L^{2}}{16}-\cdots\right]\!x^{2}=\frac{1}{4}\!\left[x^{2}-\frac{6ix+2}{4}+\frac{-18}{16}\right]. Φ = [ 4 + L ] − 1 x 2 = 4 1 [ 1 − 4 L + 16 L 2 − ⋯ ] x 2 = 4 1 [ x 2 − 4 6 i x + 2 + 16 − 18 ] . Simplify:
Φ = x 2 4 − 6 i x + 2 16 − 18 64 = x 2 4 − 3 i x 8 − 1 8 − 9 32 = x 2 4 − 3 i x 8 − 13 32 . \Phi=\frac{x^{2}}{4}-\frac{6ix+2}{16}-\frac{18}{64}=\frac{x^{2}}{4}-\frac{3ix}{8}-\frac{1}{8}-\frac{9}{32}=\frac{x^{2}}{4}-\frac{3ix}{8}-\frac{13}{32}. Φ = 4 x 2 − 16 6 i x + 2 − 64 18 = 4 x 2 − 8 3 i x − 8 1 − 32 9 = 4 x 2 − 8 3 i x − 32 13 . Step 4 — Real PI
z p C = e i ( x + y ) ( x 2 4 − 3 i x 8 − 13 32 ) z_p^{\mathbb C}=e^{i(x+y)}\!\left(\frac{x^{2}}{4}-\frac{3ix}{8}-\frac{13}{32}\right) z p C = e i ( x + y ) ( 4 x 2 − 8 3 i x − 32 13 )
Take imaginary part. With a = x 2 4 − 13 32 a=\tfrac{x^{2}}{4}-\tfrac{13}{32} a = 4 x 2 − 32 13 b = − 3 x 8 b=-\tfrac{3x}{8} b = − 8 3 x
( a + i b ) ( cos θ + i sin θ ) → Im a sin θ + b cos θ , (a+ib)(\cos\theta+i\sin\theta)\;\xrightarrow{\operatorname{Im}}\;a\sin\theta+b\cos\theta, ( a + ib ) ( cos θ + i sin θ ) Im a sin θ + b cos θ , with θ = x + y \theta=x+y θ = x + y
z p = ( x 2 4 − 13 32 ) sin ( x + y ) − 3 x 8 cos ( x + y ) . z_p=\left(\frac{x^{2}}{4}-\frac{13}{32}\right)\sin(x+y)-\frac{3x}{8}\cos(x+y). z p = ( 4 x 2 − 32 13 ) sin ( x + y ) − 8 3 x cos ( x + y ) . Step 5 — General solution
Answer
z = f ( y − 3 x ) + g ( y + 2 x ) + ( x 2 4 − 13 32 ) sin ( x + y ) − 3 x 8 cos ( x + y ) . \boxed{\;z=f(y-3x)+g(y+2x)+\left(\frac{x^{2}}{4}-\frac{13}{32}\right)\sin(x+y)-\frac{3x}{8}\cos(x+y).\;} z = f ( y − 3 x ) + g ( y + 2 x ) + ( 4 x 2 − 32 13 ) sin ( x + y ) − 8 3 x cos ( x + y ) .