Question

If $\left(\frac{2+cosx}{3+y}\right)\frac{dy}{dx}+sinx=0$ and $y\left(0\right)=1,$ then $y\left(\frac{\pi }{3}\right)$ is equal to

• A. $\frac{4}{3}$
• B. $\frac{7}{3}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

Answer: (C)

$\frac{dy}{y+3}=\frac{-sinx}{2+cosx}dx$
$\Rightarrow ln\left(y+3\right)=ln\left(2+cosx\right)+lnC$
$\Rightarrow y+3=C(2+cosx)$
Now, at $x=0,y=1$
$\Rightarrow 4=3C\Rightarrow C=\frac{4}{3}$
$\Rightarrow y=\frac{4}{3}\left(2+cosx\right)-3$
Hence, $y\left(\frac{\pi }{3}\right)=\frac{4}{3}\left(2+\frac{1}{2}\right)-3=\frac{2}{3}\cdot 5-3=\frac{1}{3}$