← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Linear first-order · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Solve the differential equation:

Technique

Linear first-order ODE; integrating factor $\mu=\exp(\int P\,dx)=x\sec x$; after multiplying, the RHS simplifies to $\sec^2 x$, integrable by inspection.

Solution

Strategy. Divide by $x\cos x$ to get standard linear form $y'+P(x)y=Q(x)$, find integrating factor. Watch for a clever rearrangement — the coefficient $x\sin x+\cos x$ is exactly $\dfrac{d}{dx}(x\sin x)$… let’s check.

Step 1 — Recognise exact form

Note $\dfrac{d}{dx}(x\sin x)=\sin x+x\cos x$. That’s not quite $x\sin x+\cos x$. But $\dfrac{d}{dx}(\text{something involving }x\sec x)$? Let’s compute $\dfrac{d}{dx}(y\cdot x\sec x)=x\sec x\cdot y'+y(\sec x+x\sec x\tan x)$. Multiply through by $\cos x$: not matching directly.

Try the standard approach.

Step 2 — Standard form

Divide by $x\cos x$ (assuming $x\cos x\ne 0$):

So $P(x)=\tan x+1/x$, $Q(x)=\sec x/x$.

Step 3 — Integrating factor

Drop absolute values: $\mu=\dfrac{x}{\cos x}=x\sec x$.

Step 4 — Multiply and integrate

Therefore

Step 5 — Solve for $y$

Answer