If $$y = y(x)$$, $$y \in \left[0, \frac{\pi}{2}\right)$$ is the solution of the differential equation $$\sec y \frac{dy}{dx} - \sin(x + y) - \sin(x - y) = 0$$, with $$y(0) = 0$$, then $$5y'\left(\frac{\pi}{2}\right)$$ is equal to _________.

We begin with the given differential equation

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

First, we recall the standard trigonometric identity

$$\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta.$$

Using this identity with $$\alpha=x$$ and $$\beta=y,$$ we have

$$\sin(x+y)+\sin(x-y)=2\sin x\cos y.$$

Substituting in the differential equation gives

$$\sec y \dfrac{dy}{dx}-2\sin x\cos y=0.$$

Now we isolate the derivative. Remembering that $$\sec y=\dfrac1{\cos y},$$ we write

$$\dfrac{1}{\cos y}\dfrac{dy}{dx}=2\sin x\cos y.$$

Multiplying both sides by $$\cos y$$ we obtain

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

To separate the variables we divide by $$\cos^2 y$$ and multiply by $$dx$$:

$$\dfrac{dy}{\cos^2 y}=2\sin x\,dx.$$

Since $$\dfrac{1}{\cos^2 y}=\sec^2 y,$$ the equation takes the form

$$\sec^2 y\,dy = 2\sin x\,dx.$$

We now integrate both sides. We use the formula $$\displaystyle\int \sec^2 y\,dy=\tan y$$ and $$\displaystyle\int\sin x\,dx=-\cos x.$$ Thus

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

where $$C$$ is the constant of integration.

To determine $$C,$$ we apply the initial condition $$y(0)=0.$$ Substituting $$x=0$$ and $$y=0$$ gives

$$\tan 0 = -2\cos 0 + C \quad\Longrightarrow\quad 0 = -2(1) + C,$$

so

$$C = 2.$$

Hence the implicit solution is

$$\tan y = -2\cos x + 2 = 2(1-\cos x).$$

We are asked to find $$y'\left(\dfrac{\pi}{2}\right).$$ For this we return to the already isolated derivative

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

We need $$\sin x$$ and $$\cos^2 y$$ at $$x=\dfrac{\pi}{2}.$$

First, $$\sin\left(\dfrac{\pi}{2}\right)=1.$$

Next we find $$y$$ at $$x=\dfrac{\pi}{2}.$$ Plugging $$x=\dfrac{\pi}{2}$$ into the relation $$\tan y = 2(1-\cos x)$$ gives

$$\tan y = 2\Bigl(1-\cos\frac{\pi}{2}\Bigr)=2(1-0)=2.$$

Because $$y\in\left[0,\dfrac{\pi}{2}\right),$$ we take the principal value, so $$y=\tan^{-1}2.$$

To compute $$\cos^2 y,$$ we use the identity $$\sec^2 y = 1+\tan^2 y.$$ Hence

$$\sec^2 y = 1 + \tan^2 y = 1 + 2^2 = 5,$$

so

$$\cos^2 y = \dfrac{1}{\sec^2 y} = \dfrac{1}{5}.$$

Now we substitute back into the derivative formula:

$$y'\left(\dfrac{\pi}{2}\right) = 2\bigl(\sin\dfrac{\pi}{2}\bigr)\cos^2 y = 2(1)\left(\dfrac{1}{5}\right)=\dfrac{2}{5}.$$

Finally, we require $$5y'\left(\dfrac{\pi}{2}\right):$$

$$5y'\left(\dfrac{\pi}{2}\right)=5\left(\dfrac{2}{5}\right)=2.$$

So, the answer is $$2$$.