← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q5a — Solution

10 marks · Section B

Question

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

Technique

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

Solution

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

Step 1 — Substitute $v=x+y$

$v=x+y\;\Rightarrow\;\dfrac{dv}{dx}=1+\dfrac{dy}{dx},\;\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$.

The ODE becomes

Step 2 — Simplify the RHS using half-angle identities

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

Step 3 — Separate and integrate

Let $u=v/2,\;du=dv/2$:

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

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

Step 4 — Express derivative-free

Exponentiate:

Answer