If y = y(x), y ∈ [0, (π)/(2)) is the solution of the differential...

Given,

$\sec y\dfrac{dy}{dx} - \sin \left(x + y\right) - \sin \left(x - y\right) = 0$

$⇒ \sec y\dfrac{dy}{dx} = 2\sin x \cos y$

$⇒ \sec^{2}ydy = 2sinxdx$

$⇒ ∫ \sec^{2}ydy = ∫ 2sinxdx$

$⇒ \tan y = - 2\cos x + c$

Given, $y\left(0\right) = 0$

So, $0 = - 2\cos 0 + c$

$⇒ c = 2$

So, the particular solution is

$\tan y = - 2\cos x + 2$

at $x = \dfrac{\pi}{2}, \tan y = - 2cos\dfrac{\pi}{2} + 2$

$⇒ \tan y = 2 ⇒ \sec^{2}y = 5$

Here, differentiate both sides of $\tan y = - 2\cos x + 2,$ we get,

$\sec^{2}y\dfrac{dy}{dx} = 2\sin x$

At $x = \dfrac{\pi}{2}$

$5\dfrac{dy}{dx} = 2\sin \left(\dfrac{\pi}{2}\right)$

$⇒ 5\dfrac{dy}{dx} = 2$