What is the solution for y^'=sin(x+y) ?

The solution for y^'=sin(x+y) is y=2arctan((-2-x-c_{1})/(x+c_{1)})+2pin-x

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y^'=sin(x+y)

y^{^{'}} = sin (x + y)

y = 2 arctan (\frac{- 2 - x - c _{1}}{x + c _{1}}) + 2 πn - x

Solution steps

y^{'} (x) = sin (x + y)

Solve using substitution:

y = 2 arctan (\frac{- 2 - x - c _{1}}{x + c _{1}}) + 2 πn - x

y = 2 arctan (\frac{- 2 - x - c _{1}}{x + c _{1}}) + 2 πn - x

\frac{d}{d t} (e^{9 t s i n (2 t)})

y^{^{''}} + 4 y^{^{'}} = t^{2} + 3 e^{t}

\int e^{- v} cos (w v) d v

\int 4 + x^{2} d x

x \to 3 lim (\frac{1 - x}{x ^{2} - 9})

What is the solution for y^'=sin(x+y) ?
The solution for y^'=sin(x+y) is y=2arctan((-2-x-c_{1})/(x+c_{1)})+2pin-x