Question

If $\left(\frac{2 + cos x}{3 + y}\right)\frac{d y}{d x}+sin⁡x=0$ 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)

$\frac{d y}{y + 3}=\frac{- sin x}{2 + cos ⁡ x}dx$
$\Rightarrow ln \left(y + 3\right)=ln ⁡ \left(2 + cos ⁡ x\right)+ln ⁡ C$
$\Rightarrow y+3=C\left(\right.2+cos x\left.\right)$
Now, at $x=0,y=1$
$\Rightarrow 4=3C\Rightarrow C=\frac{4}{3}$
$\Rightarrow y=\frac{4}{3}\left(2 + cos x\right)-3$
Hence, $y\left(\frac{\pi }{3}\right)=\frac{4}{3}\left(2 + \frac{1}{2}\right)-3=\frac{2}{3}\cdot 5-3=\frac{1}{3}$