Given differential equation

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

Rearranging,

$\sec y\,\frac{dy}{dx}=\sin(x+y)+\sin(x-y)$

Use trigonometric identity

We know:

$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$

Taking $A=x+y$ and $B=x-y$,

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

So the differential equation becomes

$\sec y\,\frac{dy}{dx}=2\sin x\cos y$

Since $\sec y=\dfrac{1}{\cos y}$,

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

Hence,

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

Separate variables

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

Since $\dfrac{1}{\cos^2 y}=\sec^2 y$,

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

Integrating both sides,

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

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

Use initial condition $y(0)=0$

At $x=0$, $y=0$:

$\tan 0=-2\cos 0+C$

$0=-2(1)+C$

$C=2$

Therefore,

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

Find $y'\left(\frac{\pi}{2}\right)$

From earlier,

$y'=2\sin x\cos^2 y$

First find $y$ at $x=\frac{\pi}{2}$:

$\tan y=2-2\cos\frac{\pi}{2}=2$

So,

$\tan y=2$

Therefore,

$\cos^2 y=\frac{1}{1+\tan^2 y}=\frac{1}{1+4}=\frac{1}{5}$

Also,

$\sin\frac{\pi}{2}=1$

Hence,

$y'\left(\frac{\pi}{2}\right)=2\cdot 1\cdot \frac{1}{5}=\frac{2}{5}$

Thus,

$5y'\left(\frac{\pi}{2}\right)=5\cdot \frac{2}{5}=2$

Comparison with stored answer

Derived answer = $2$.

Stored correct answer = $2$.

They agree.