JEE PYQ: Differential Equation Question 17
Question 17 - 2020 (02 Sep Shift 1)
Let $y = y(x)$ be the solution of the differential equation, $\frac{2+\sin x}{y+1}\cdot\frac{dy}{dx} = -\cos x$, $y > 0$, $y(0) = 1$. If $y(\pi) = a$ and $\frac{dy}{dx}$ at $x = \pi$ is $b$, then the ordered pair $(a, b)$ is equal to:
(1) $\left(2, \frac{3}{2}\right)$
(2) $(1, -1)$
(3) $(1, 1)$
(4) $(2, 1)$
Show Answer
Answer: (3)
Solution
$\frac{dy}{y+1} = \frac{-\cos x}{2+\sin x},dx$. Integrating: $\ln|y+1| = -\ln|2+\sin x| + \ln C$. So $(y+1)(2+\sin x) = C$. At $(0,0)$: wait, $y(0) = 1$: $C = 2 \times 2 = 4$. $(y+1)(2+\sin x) = 4$. At $x = \pi$: $(a+1)(2) = 4$, $a = 1$. $\frac{dy}{dx}\big|_{x=\pi} = \frac{-(y+1)\cos x}{2+\sin x} = \frac{-2(-1)}{2} = 1$. So $(a,b) = (1,1)$.
BETA
