Question

Let $y=y\left(x\right)$ be the solution of the differential equation, $\frac{2+\text{sin}x}{y+1}.\frac{dy}{dx}=-\text{cos}x,y>0,y\left(0\right)=1.$
If $y\left(\pi \right)=a,$ and $\frac{dy}{dx}$ at $x=\pi$ is $b$, then the ordered pair $(a,b)$ is equal to:

• A. $(2,\frac{3}{2})$
• B. $(1,1)$
• C. $(2,1)$
• D. $(1,-1)$

Answer 1

The correct option is B $(1,1)$

$\int \frac{dy}{y+1}=\int \frac{-\cos xdx}{2+\sin x}$
$\Rightarrow \ln |y+1|=-\ln |2+\sin x|+k$

at point $(0,1)$
$k=\ln 4$

Now $C:(y+1)(2+\sin x)=4$

$\because y\left(\pi \right)=a\Rightarrow (a+1)(2+0)=4\Rightarrow a=1$

$\because {\begin{array}{c}\frac{dy}{dx}\end{array}|}_{x=\pi }=b\Rightarrow b=-(-1)\left(\frac{2+0}{1+1}\right)$
$\Rightarrow b=1$

$\therefore (a,b)=(1,1)$