Question

Solve $\sin x\frac{dy}{dx}+2y+\sin x(1+\cos x)=0.$

• A. $-y=-x+\sin x+c$
• B. $y=-x-\sin x+c$
• C. $y=-x+\sin x+c$
• D. None of these.

Answer 1

Answer: (C)

$y=-x+\sin x+c$

$\sin x\frac{dy}{dx}+2y+\sin x(1+\cos x)=0.$
$\frac{dy}{dx}+2yco\sec x=-(1+\cos x).$
I.F. $=e^{\int 2co\sec dx}=e^{2\ln \tan \frac{x}{2}}={\tan }^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}.$
Multiplying both sides by I.F. and integrating.
$y.{\tan }^2\frac{x}{2}=-\int (1-\cos x)dx=-x+\sin x+c$