If sin(x+y)dy/dx=5 then

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sin (x + y) \frac{d y}{d x} = 5

y^{2} = a e^{- 2 x} + (2 / 5) (c o s x - 2 s i n x)

y \frac{d y}{d x} + y^{2} + s i n x

If y(x) is a solution of $(\frac{2 + s i n x}{1 + y}) \frac{d y}{d x} = - c o s x a n d y (0) = 1,$ then find the value of $y (\frac{π}{2}) .$

y = y (x), y \in [0, \frac{π}{2})

sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0

y (0) = 0

5 y^{'} (\frac{π}{2})

y \frac{d y}{d x} = (x + s i n x)

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