Question

For the differential equation $\tan y\frac{dy}{dx}+\tan x=\cos y{\cos }^{3}x.$ Its solution is $\sec x\sec y=\frac{x}{A}+\frac{1}{B}\sin 2x+c.$ Find B/A?

Answer 1

$\tan y\frac{dy}{dx}+\tan x=\cos y{\cos }^{3}x\Rightarrow \tan y\sec y\frac{dy}{dx}+\tan x\sec y={\cos }^{3}x$
Put $\sec y=v\Rightarrow \tan y\sec ydy=dv$
$\therefore \frac{dv}{dx}+v\tan x={\cos }^{3}x$ ...(1)
Here $P=\tan x\Rightarrow \int Pdx=\int \tan xdx=\log \sec x$
$\therefore I.F.=e^{\log \sec x}=\sec x$
Multiplying (1) by $I.F.$ we get
$\sec x\frac{dv}{dx}+v\tan x\sec x={\cos }^{2}x$
Integrating both sides w.r.t $x$ we get
$v\sec x=\int {\cos }^{2}xdx+c=\frac{x}{2}+\frac{1}{4}\sin 2x+c$
$\Rightarrow \sec x\sec y=\frac{x}{2}+\frac{1}{4}\sin 2x+c$