UPSC 2016 Maths Optional Paper 1 Q6a — Step-by-Step Solution

10 marks · Section B

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

Question

Solve: $\{y(1-x\tan x)+x^2\cos x\}\,dx-x\,dy=0$ .

Recognise linearity in $y$ ; integrating factor $\mu=\dfrac{1}{x\cos x}$ (from $\int(-1/x+\tan x)\,dx$ ); the RHS collapses to $1$ .

Divide by $dx$ and isolate $\dfrac{dy}{dx}$ :

x\frac{dy}{dx}=y(1-x\tan x)+x^2\cos x,

\frac{dy}{dx}-\frac{1-x\tan x}{x}\,y=x\cos x.

This is linear with $P(x)=-\dfrac{1-x\tan x}{x}=-\dfrac1x+\tan x$ .

\int P\,dx=\int\!\left(-\frac1x+\tan x\right)dx=-\ln x-\ln\cos x=-\ln(x\cos x).

\mu=e^{-\ln(x\cos x)}=\frac{1}{x\cos x}.

\frac{d}{dx}\!\left(\frac{y}{x\cos x}\right)=\mu\cdot x\cos x=\frac{1}{x\cos x}\cdot x\cos x=1.

Integrating,

\frac{y}{x\cos x}=x+C.

\boxed{\;y=x\cos x\,(x+C).\;}