← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 14 yrs · Read the full method →

Question

Find a particular integral of d2ydx2+y=ex/2sin⁡x32\dfrac{d^2y}{dx^2}+y=e^{x/2}\sin\dfrac{x\sqrt3}{2}.

Technique

Complex-exponential operator method, yp=Im⁡1D2+1e(12+i32)xy_p=\operatorname{Im}\dfrac{1}{D^2+1}e^{(\frac12+i\frac{\sqrt3}{2})x}; substitute D→D\to the complex exponent.

Solution

Write D≡ddxD\equiv\dfrac{d}{dx}. The equation is (D2+1)y=ex/2sin⁡32x(D^2+1)y=e^{x/2}\sin\dfrac{\sqrt3}{2}x.

Step 1 — Check for resonance

The complementary function comes from D2+1=0⇒D=±iD^2+1=0\Rightarrow D=\pm i, i.e. cos⁡x,sin⁡x\cos x,\sin x. The forcing term ex/2sin⁡32xe^{x/2}\sin\frac{\sqrt3}{2}x is the imaginary part of e(12+i32)xe^{(\frac12+i\frac{\sqrt3}{2})x}, whose exponent 12+i32\frac12+i\frac{\sqrt3}{2} is not a root of D2+1D^2+1. Hence there is no resonance and the operator method applies directly.

Step 2 — Complex-exponential method

Take yp=Im⁡ zpy_p=\operatorname{Im}\,z_p where

zp=1D2+1 e(12+i32)x.z_p=\frac{1}{D^2+1}\,e^{\left(\frac12+i\frac{\sqrt3}{2}\right)x}.

Substitute D→12+i32D\to\frac12+i\frac{\sqrt3}{2}:

(12+i32)2+1=(14−34+i32)+1=12+i32.\left(\tfrac12+i\tfrac{\sqrt3}{2}\right)^2+1=\left(\tfrac14-\tfrac34+i\tfrac{\sqrt3}{2}\right)+1=\tfrac12+i\tfrac{\sqrt3}{2}.

So

zp=e(12+i32)x12+i32.z_p=\frac{e^{\left(\frac12+i\frac{\sqrt3}{2}\right)x}}{\tfrac12+i\tfrac{\sqrt3}{2}}.

Step 3 — Rationalise the denominator

The denominator 12+i32=eiπ/3\tfrac12+i\tfrac{\sqrt3}{2}=e^{i\pi/3} has modulus 11. Multiply by its conjugate 12−i32\tfrac12-i\tfrac{\sqrt3}{2} (denominator 14+34=1\tfrac14+\tfrac34=1):

zp=(12−i32)ex/2(cos⁡32x+isin⁡32x).z_p=\left(\tfrac12-i\tfrac{\sqrt3}{2}\right)e^{x/2}\Big(\cos\tfrac{\sqrt3}{2}x+i\sin\tfrac{\sqrt3}{2}x\Big).

Step 4 — Take the imaginary part

Im⁡ zp=ex/2[12sin⁡32x−32cos⁡32x].\operatorname{Im}\,z_p=e^{x/2}\Big[\tfrac12\sin\tfrac{\sqrt3}{2}x-\tfrac{\sqrt3}{2}\cos\tfrac{\sqrt3}{2}x\Big].

Answer

  yp=12ex/2sin⁡32x−32ex/2cos⁡32x.  \boxed{\;y_p=\frac{1}{2}e^{x/2}\sin\frac{\sqrt3}{2}x-\frac{\sqrt3}{2}e^{x/2}\cos\frac{\sqrt3}{2}x.\;}
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