← 2018 Paper 1

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))(1+x)^2y''+(1+x)y'+y=4\cos(\log(1+x)).

Technique

Cauchy–Euler substitution t=log(1+x)t=\log(1+x) reduces to (D2+1)y=4cost(D^2+1)y=4\cos t; resonant PI 1D2+1cost=t2sint\frac1{D^2+1}\cos t=\frac t2\sin t.

Solution

This is a Cauchy–Euler (equidimensional) equation in the variable (1+x)(1+x).

Step 1 — Substitution

Let t=log(1+x)t=\log(1+x), so 1+x=et1+x=e^{t} and dtdx=11+x\dfrac{dt}{dx}=\dfrac{1}{1+x}. Writing D=ddtD=\dfrac{d}{dt}, the standard reductions are

(1+x)y=Dy,(1+x)2y=D(D1)y.(1+x)y'=Dy,\qquad (1+x)^2y''=D(D-1)y.

Step 2 — Transform the equation

D(D1)y+Dy+y=4cost    (D2D+D+1)y=4cost    (D2+1)y=4cost.D(D-1)y+Dy+y=4\cos t\;\Longrightarrow\;(D^2-D+D+1)y=4\cos t\;\Longrightarrow\;(D^2+1)y=4\cos t.

A constant-coefficient equation in tt.

Step 3 — Complementary function

m2+1=0m=±im^2+1=0\Rightarrow m=\pm i:

yc=C1cost+C2sint.y_c=C_1\cos t+C_2\sin t.

Step 4 — Particular integral (resonance)

The forcing 4cost4\cos t has frequency matching the roots ±i\pm i, so it resonates. Using 1D2+1cost=t2sint\dfrac{1}{D^2+1}\cos t=\dfrac{t}{2}\sin t (the standard resonant result):

yp=4t2sint=2tsint.y_p=4\cdot\frac{t}{2}\sin t=2t\sin t.

Step 5 — Back-substitute t=log(1+x)t=\log(1+x)

Answer

  y=C1cos(log(1+x))+C2sin(log(1+x))+2log(1+x)sin(log(1+x)).  \boxed{\;y=C_1\cos\big(\log(1+x)\big)+C_2\sin\big(\log(1+x)\big)+2\log(1+x)\,\sin\big(\log(1+x)\big).\;}
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