If y=y(x), y in [0, frac{pi}{2}) is the solut | vedclass

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(B) Given the differential equation: $\sec y \frac{dy}{dx} - (\sin(x+y) + \sin(x-y)) = 0$.Using the identity $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$

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$2$

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(B) Given the differential equation: $\sec y \frac{dy}{dx} - (\sin(x+y) + \sin(x-y)) = 0$.Using the identity $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$,we get:$\sec y \frac{dy}{dx} - 2 \sin x \cos y = 0$.$\sec y \frac{dy}{dx} = 2 \sin x \cos y$.Dividing by $\cos y$ (since $\cos y 
eq 0$ for $y \in [0, \frac{\pi}{2})$):$\sec^2 y \frac{dy}{dx} = 2 \sin x$.Integrating both sides with respect to $x$:$\int \sec^2 y dy = \int 2 \sin x dx$.$	an y = -2 \cos x + C$.Given $y(0) = 0$,we substitute $x=0$ and $y=0$:$	an(0) = -2 \cos(0) + C \Rightarrow 0 = -2(1) + C \Rightarrow C = 2$.So,$	an y = 2 - 2 \cos x$.Differentiating with respect to $x$:$\sec^2 y \frac{dy}{dx} = 2 \sin x$.At $x = \frac{\pi}{2}$,$	an y = 2 - 2 \cos(\frac{\pi}{2}) = 2 - 0 = 2$.Since $	an y = 2$,$\sec^2 y = 1 + 	an^2 y = 1 + 2^2 = 5$.Substituting into the derivative equation:$5 \frac{dy}{dx} = 2 \sin(\frac{\pi}{2}) = 2(1) = 2$.Thus,$5y'(\frac{\pi}{2}) = 2$.

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

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