Let y = y(x) be the solution of the differential equation, 2 + sinx / y + 1 . dy/dx = − cosx, y > 0, y(0) = 1.

Question

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

(1) $\left(2, {3\over2}\right)$

(2) (1, –1)

(3) (2, 1)

(4) (1, 1)

Answer 1

Ans. (4) (1, 1)

Sol. ${dy\over1+y}={-\cos x\over2+\sin x}dx$

$\ln(1+y)=-\ln(2+\sin x)+\ln c$

$(1+y)(2+\sin x)=c$

$4.1=c\implies c=4$

$1+y={4\over2+\sin x}\implies y={4\over2+\sin x}-1$

$y(\pi)=2-1=1=a$

${dy\over dx}={-4\over(2+\sin x)^2}.\cos x=1\text{ at }x=\pi \implies b=1$

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