← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 13 yrs · Read the full method →

Question

Solve: y′′′−6y′′+12y′−8y=12e2x+27e−xy'''-6y''+12y'-8y=12e^{2x}+27e^{-x}.

Technique

Factor the cubic to the perfect cube (D−2)3(D-2)^3; resonant PI via 1(D−2)3e2x=e2xx3/3!\frac1{(D-2)^3}e^{2x}=e^{2x}x^3/3!; ordinary PI by substituting D=−1D=-1.

Solution

Write as (D3−6D2+12D−8)y=12e2x+27e−x(D^3-6D^2+12D-8)y=12e^{2x}+27e^{-x}.

Step 1 — Complementary function

The auxiliary cubic factors as a perfect cube:

m3−6m2+12m−8=(m−2)3=0⇒m=2,2,2.m^3-6m^2+12m-8=(m-2)^3=0\Rightarrow m=2,2,2.

A triple root gives

yc=(C1+C2x+C3x2) e2x.y_c=(C_1+C_2x+C_3x^2)\,e^{2x}.

Step 2 — Particular integral for 12e2x12e^{2x} (resonant)

Here ϕ(D)=(D−2)3\phi(D)=(D-2)^3 and the forcing e2xe^{2x} has ϕ(2)=0\phi(2)=0 with multiplicity 33, so it is fully resonant. Use 1(D−2)3e2x=e2x1D3(1)=e2x⋅x33!\dfrac1{(D-2)^3}e^{2x}=e^{2x}\dfrac1{D^3}(1)=e^{2x}\cdot\dfrac{x^3}{3!}:

yp1=1(D−2)3 12e2x=12 e2x x36=2x3e2x.y_{p1}=\frac{1}{(D-2)^3}\,12e^{2x}=12\,e^{2x}\,\frac{x^3}{6}=2x^3e^{2x}.

Step 3 — Particular integral for 27e−x27e^{-x} (ordinary)

−1-1 is not a root, so substitute D=−1D=-1: ϕ(−1)=(−1−2)3=−27\phi(-1)=(-1-2)^3=-27:

yp2=1(D−2)3 27e−x=27−27 e−x=−e−x.y_{p2}=\frac{1}{(D-2)^3}\,27e^{-x}=\frac{27}{-27}\,e^{-x}=-e^{-x}.

Step 4 — General solution

Answer

  y=(C1+C2x+C3x2)e2x+2x3e2x−e−x.  \boxed{\;y=(C_1+C_2x+C_3x^2)e^{2x}+2x^3e^{2x}-e^{-x}.\;}
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