Question

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

• A. $\sin y=\sin x-1-c{e}^{-\sin x}\sin x\cos xdx$
• B. $\sin y=\sin x-1+c{e}^{\sin x}\sin x\cos xdx$
• C. $-\sin y=\sin x-1+c{e}^{-\sin x}\sin x\cos xdx$
• D. $\sin y=\sin x-1+c{e}^{-\sin x}\sin x\cos xdx$

Answer 1

Answer: (D)

$\sin y=\sin x-1+c{e}^{-\sin x}\sin x\cos xdx$

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

Put $\sin y=v\Rightarrow \cos ydy=dv$

$\therefore \frac{dv}{dx}+v\cos x=\sin x$ ...(1)

Here $P=\cos x\Rightarrow \int PdP=\int \cos xdx=\sin x$

$\therefore I.F.=e^{\sin x}$

Multiplying (1) by $I.F.$ we get

$e^{\sin x}\frac{dv}{dx}+e^{\sin x}.v\cos x=e^{\sin x}.\sin x$

Integrating both sides, we get

$v.e{.}^{\sin x}=\int e^{\sin x}\sin x\cos xdx$

$\therefore \sin y=\sin x-1+c{e}^{-\sin x}\sin x\cos xdx$