Question

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

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

Answer 1

The correct option is B

$(1,1)$

Explanation for the correct option

Finding the ordered pair $(\mathrm{a},\mathrm{b})$

Given,

$$\begin{gathered} \frac{2+\mathrm{sinx}}{(\mathrm{y}+1)}\frac{\mathrm{dy}}{\mathrm{dx}}=-\cos \mathrm{x},\mathrm{y}>0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}+1}=-\frac{\mathrm{cosx}}{2+\mathrm{sinx}}\mathrm{dx} \end{gathered}$$

On integrating both sides, we get:

$$\begin{gathered} \int \frac{\mathrm{dy}}{\mathrm{y}+1}=-\int \frac{\mathrm{cosx}}{2+\mathrm{sinx}}\mathrm{dx} \\ \Rightarrow \ln \left|\mathrm{y}+1\right|=-\ln \left|2+\mathrm{sinx}\right|+\mathrm{lnk} \\ \Rightarrow \mathrm{y}\left(\mathrm{x}\right)+1=\frac{\mathrm{K}}{2+\mathrm{sinx}}(\mathrm{y}+1)>0 \\ \Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\frac{\mathrm{K}}{2+\mathrm{sinx}}-1 \end{gathered}$$

Given,

$\begin{array}{rcl}\mathrm{y}\left(0\right)& =& 1\\ & \Rightarrow & 1=\frac{K}{2+0}-1\\ & \Rightarrow & \mathrm{K}=4\end{array}$

Therefore,

$\begin{array}{rcl}\mathrm{y}\left(\mathrm{x}\right)& =& \frac{4}{2+\mathrm{sinx}}-1\\ & \Rightarrow & \mathrm{y}\left(\mathrm{\pi }\right)=2-1\\ & \Rightarrow & \mathrm{a}=1\end{array}$

And,

$\begin{array}{rcl}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=\mathrm{\pi }}& =& \frac{-\mathrm{cosx}}{2+\mathrm{sinx}}\left(\mathrm{y}\right(\mathrm{x})+1)\\ & \Rightarrow & \mathrm{b}=1\end{array}$

So $(a,b)=(1,1)$.

Hence, the correct answer is option (B).