Let y=y(x) be the solution of the differential equation,


Let $y=y(x)$ be the solution of the differential equation,

$\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1$. If $y(\pi)=a$ and

$\frac{d y}{d x}$ at $x=\pi$ is $b$, then the ordered pair $(a, b)$ is equal to :

  1. (1) $\left(2, \frac{3}{2}\right)$

  2. (2) $(1,-1)$

  3. (3) $(1,1)$

  4. (4) $(2,1)$

Correct Option: , 3


The given differential equation is

$\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y>0$

$\Rightarrow \frac{d y}{y+1}=-\frac{\cos x}{2+\sin x} d x$

Integrate both sides,

$\int \frac{d y}{y+1}=\int \frac{(-\cos x) d x}{2+\sin x}$

$\ln |y+1|=-\ln |2+\sin x|+\ln C$

$\Rightarrow \ln |y+1|+\ln |2+\sin x|=\ln C$

$\Rightarrow \ln |(y+1)(2+\sin x)|=\ln C$

$\because y(0)=1 \Rightarrow \ln 4=\ln C \Rightarrow C=4$

$\therefore(y+1)(2+\sin x)=4$

$\Rightarrow y=\frac{4}{2+\sin x}-1$

$\therefore y=\frac{2-\sin x}{2+\sin x} \Rightarrow y(\pi)=\frac{2-\sin \pi}{2+\sin \pi}=1$

$\Rightarrow a=1$

Now, $\frac{d y}{d x}=\frac{(2+\sin x)(-\cos x)-(2-\sin x) \cdot \cos x}{(2+\sin x)^{2}}$

$\left.\frac{d y}{d x}\right|_{x=\pi}=1 \Rightarrow b=1 .$

Ordered pair $(a, b)=(1,1)$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now