← 2013 Paper 1

UPSC 2013 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Variables separable · ODEs · asked 3× in 14 yrs · Read the full method →

Question

yy is a function of xx, such that the differential coefficient dydx\dfrac{dy}{dx} is equal to cos⁡(x+y)+sin⁡(x+y)\cos(x+y)+\sin(x+y). Find out a relation between xx and yy, which is free from any derivative/differential.

Technique

Substitution v=x+yv=x+y converts an x+yx+y-dependent ODE into separable form; half-angle identities collapse 1+cos⁡v+sin⁡v1+\cos v+\sin v; then a clean sec⁡2u/(1+tan⁡u)\sec^{2}u/(1+\tan u) form integrates directly.

Solution

Strategy. The RHS depends only on x+yx+y — use the substitution v=x+yv=x+y to convert to a separable ODE.

Step 1 — Substitute v=x+yv=x+y

v=x+y  ⇒  dvdx=1+dydx,  dydx=dvdx−1v=x+y\;\Rightarrow\;\dfrac{dv}{dx}=1+\dfrac{dy}{dx},\;\dfrac{dy}{dx}=\dfrac{dv}{dx}-1.

The ODE becomes

dvdx−1=cos⁡v+sin⁡v  ⟹  dvdx=1+cos⁡v+sin⁡v.\frac{dv}{dx}-1=\cos v+\sin v\;\Longrightarrow\;\frac{dv}{dx}=1+\cos v+\sin v.

Step 2 — Simplify the RHS using half-angle identities

1+cos⁡v=2cos⁡2(v/2)1+\cos v=2\cos^{2}(v/2) and sin⁡v=2sin⁡(v/2)cos⁡(v/2)\sin v=2\sin(v/2)\cos(v/2), so

dvdx=2cos⁡2 ⁣v2+2sin⁡v2cos⁡v2=2cos⁡v2 ⁣(cos⁡v2+sin⁡v2).\frac{dv}{dx}=2\cos^{2}\!\tfrac{v}{2}+2\sin\tfrac{v}{2}\cos\tfrac{v}{2}=2\cos\tfrac{v}{2}\!\left(\cos\tfrac{v}{2}+\sin\tfrac{v}{2}\right).

Step 3 — Separate and integrate

dv2cos⁡v2 ⁣(cos⁡v2+sin⁡v2)=dx.\frac{dv}{2\cos\tfrac{v}{2}\!\left(\cos\tfrac{v}{2}+\sin\tfrac{v}{2}\right)}=dx.

Let u=v/2,  du=dv/2u=v/2,\;du=dv/2:

ducos⁡u(cos⁡u+sin⁡u)=dx.\frac{du}{\cos u(\cos u+\sin u)}=dx.

Divide numerator and denominator by cos⁡2u\cos^{2}u:

sec⁡2u du1+tan⁡u=dx.\frac{\sec^{2}u\,du}{1+\tan u}=dx.

Substitute w=1+tan⁡u,  dw=sec⁡2u duw=1+\tan u,\;dw=\sec^{2}u\,du:

dww=dx  ⟹  ln⁡∣w∣=x+C  ⟹  ln⁡ ⁣∣1+tan⁡ ⁣x+y2∣=x+C.\frac{dw}{w}=dx\;\Longrightarrow\;\ln|w|=x+C\;\Longrightarrow\;\ln\!\left|1+\tan\!\frac{x+y}{2}\right|=x+C.

Step 4 — Express derivative-free

Exponentiate:

1+tan⁡ ⁣x+y2=A ex,A=±eC  a non-zero constant.1+\tan\!\frac{x+y}{2}=A\,e^{x},\qquad A=\pm e^{C}\;\text{a non-zero constant.}

Answer

  1+tan⁡ ⁣x+y2=A ex.  \boxed{\;1+\tan\!\frac{x+y}{2}=A\,e^{x}.\;}
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