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UPSC 2018 Maths Optional Paper 1 Q7a — Step-by-Step Solution
13 marks · Section B
Euler-Cauchy equation · ODEs · asked 8× in 13 yrs · Read the full method →
Question
Solve: (1+x)2y′′+(1+x)y′+y=4cos(log(1+x)).
Technique
Cauchy–Euler substitution t=log(1+x) reduces to (D2+1)y=4cost; resonant PI D2+11cost=2tsint.
Solution
This is a Cauchy–Euler (equidimensional) equation in the variable (1+x).
Step 1 — Substitution
Let t=log(1+x), so 1+x=et and dxdt=1+x1. Writing D=dtd, the standard reductions are
(1+x)y′=Dy,(1+x)2y′′=D(D−1)y.
D(D−1)y+Dy+y=4cost⟹(D2−D+D+1)y=4cost⟹(D2+1)y=4cost.
A constant-coefficient equation in t.
Step 3 — Complementary function
m2+1=0⇒m=±i:
yc=C1cost+C2sint.
Step 4 — Particular integral (resonance)
The forcing 4cost has frequency matching the roots ±i, so it resonates. Using D2+11cost=2tsint (the standard resonant result):
yp=4⋅2tsint=2tsint.
Step 5 — Back-substitute t=log(1+x)
Answer
y=C1cos(log(1+x))+C2sin(log(1+x))+2log(1+x)sin(log(1+x)).