JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 24)
If $$y = y(x),y \in \left[ {0,{\pi \over 2}} \right)$$ is the solution of the differential equation $$\sec y{{dy} \over {dx}} - \sin (x + y) - \sin (x - y) = 0$$, with y(0) = 0, then $$5y'\left( {{\pi \over 2}} \right)$$ is equal to ______________.
Answer
2
Explanation
$$\sec y{{dy} \over {dx}} = 2\sin x\cos y$$
$${\sec ^2}ydy = 2\sin xdx$$
$$\tan y = - 2\cos x + c$$
$$c = 2$$
$$\tan y = - 2\cos x + 2 \Rightarrow $$ at $$x = {\pi \over 2}$$
$$\tan y = 2$$
$${\sec ^2}y{{dy} \over {dx}} = 2\sin x$$
$$ \therefore $$ $$5{{dy} \over {dx}} = 2$$
$${\sec ^2}ydy = 2\sin xdx$$
$$\tan y = - 2\cos x + c$$
$$c = 2$$
$$\tan y = - 2\cos x + 2 \Rightarrow $$ at $$x = {\pi \over 2}$$
$$\tan y = 2$$
$${\sec ^2}y{{dy} \over {dx}} = 2\sin x$$
$$ \therefore $$ $$5{{dy} \over {dx}} = 2$$
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