UPSC 2016 Maths Optional Paper 1 Q5a — Step-by-Step Solution
10 marks · Section B
Particular integral via operator method · ODEs · asked 7× in 13 yrs · Read the full method →
Question
Find a particular integral of dx2d2y+y=ex/2sin2x3.
Technique
Complex-exponential operator method, yp=ImD2+11e(21+i23)x; substitute D→ the complex exponent.
Solution
Write D≡dxd. The equation is (D2+1)y=ex/2sin23x.
Step 1 — Check for resonance
The complementary function comes from D2+1=0⇒D=±i, i.e. cosx,sinx. The forcing term ex/2sin23x is the imaginary part of e(21+i23)x, whose exponent 21+i23 is not a root of D2+1. Hence there is no resonance and the operator method applies directly.
Step 2 — Complex-exponential method
Take yp=Imzp where
zp=D2+11e(21+i23)x.
Substitute D→21+i23:
(21+i23)2+1=(41−43+i23)+1=21+i23.
So
zp=21+i23e(21+i23)x.
Step 3 — Rationalise the denominator
The denominator 21+i23=eiπ/3 has modulus 1. Multiply by its conjugate 21−i23 (denominator 41+43=1):