Question

Solve the differential equation: $\sin y\frac{dy}{dx}=\cos y(1-x\cos y)$

• A. $\sec y{e}^{-x}=-e^{-x}(x+1)+c.$
• B. $\sec y{e}^{-x}=e^{-x}(x-1)+c.$
• C. $\sec y{e}^{-x}=-e^{-x}(x-1)+c.$
• D. $\sec y{e}^{-x}=e^{-x}(x+1)+c.$

Answer 1

The correct option is D $\sec y{e}^{-x}=e^{-x}(x+1)+c.$

$\sin y\frac{dy}{dx}=\cos y(1-x\cos y)$

$\Rightarrow \sec y\tan y\frac{dy}{dx}-\sec y=-x$
Put $\sec y=v\Rightarrow \sec y\tan ydy=dx$
$\therefore \frac{dv}{dx}-v.1=-x$ ...(1)
Here $P=-1\Rightarrow -\int dx=-x$
$\therefore I.F.=e^{-x}$
Multiplying (1) by $I.F.$ we get
$e^{-x}\frac{dv}{dx}-e^{-x}v=-x{e}^{-x}$
Integrating both sides we get
$e^{-x}v=-\int x{e}^{-x}dx+c=e^{-x}(x+1)+c$

$\Rightarrow \sec y{e}^{-x}=e^{-x}(x+1)+c$