Question

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

• A. $\frac{1}{3}$
• B. $-\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{1}{3}$

$(2+sinx)\frac{dy}{dx}+(y+1)cosx=0,y\left(0\right)=1,y\left(\frac{\pi }{2}\right)=?$

$\frac{dy}{dx}+\frac{cosx}{2+sinx}(y+1)=0$

$\frac{dy}{dx}=-\frac{cosx}{2+sinx}(y+1)$

$\int \frac{dy}{y+1}=-\int \frac{cosx}{2+sinx}dx$

$ln(y+1)=-ln(2+sinx)+c$

$y\left(0\right)=1$

$ln\left(2\right)=-ln\left(2\right)+c$

$c=ln\left(4\right)$

$ln(y+1)=-ln(2+sinx)+ln\left(4\right)$

$ln(y+1)=-ln(2+1)+ln4$

$ln(y+1)=ln\left(\frac{4}{3}\right)$

$y=\frac{4}{3}-1=\frac{1}{3}$