jee-main 2020 Q68

jee-main · India · session2_02sep_shift1 Differential equations Solving Separable DEs with Initial Conditions

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) $(2, 1)$

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) $(2, 1)$

Paper Questions

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