Question:
If $y=y(x), \quad y \in\left[0, \frac{\pi}{2}\right)$ is the solution of the
differential equation
$\sec y \frac{d y}{d x}-\sin (x+y)-\sin (x-y)=0$, with $y(0)=0$
then $5 \mathrm{y}^{\prime}\left(\frac{\pi}{2}\right)$ is equal to
Solution:
$\sec y \frac{d y}{d x}=2 \sin x \cos y$
$\sec ^{2} y d y=2 \sin x d x$
$\tan y=-2 \cos x+c$
$c=2$
$\tan y=-2 \cos x+2 \Rightarrow$ at $x=\frac{\pi}{2}$
$\tan y=2$
$\sec ^{2} y \frac{d y}{d x}=2 \sin x$
$5 \frac{\mathrm{dy}}{\mathrm{dx}}=2$
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