Question

Solve the differential equation: $\frac{dy}{dx}+y\cos x=\sin x\cos x$

• A. $y=(\sin x+1)+c{e}^{-\sin x}.$
• B. $y=(\sin x-1)+c{e}^{\sin x}.$
• C. $y=(\sin x-1)+c{e}^{-\sin x}.$
• D. None of these.

Answer 1

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

$\frac{dy}{dx}+y\cos x=\sin x\cos x$ ...(1)
Here $P=\cos x\Rightarrow \int Pdx=\int \cos xdx=\sin xdx$
$\therefore I.F=e^{\sin x}$
Multiplying (1) by $I.F,$ we get
$e^{\sin x}\frac{dy}{dx}+y\cos x{e}^{\sin x}=\sin x\cos x{e}^{\sin x}$
Integrating both sides w.r.t. x, we get
$e^{\sin x}y=\int \sin x\cos x{e}^{\sin x}+c$
$\Rightarrow y{e}^{\sin x}=e^{\sin x}\left(\sin x-1\right)+c$
$\Rightarrow y=\left(\sin x-1\right)+c{e}^{-\sin x}$