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

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

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y^'=cos(x+y)

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

y = 2 arctan (x + c_{1}) + 2 πn - x

Solution steps

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

Solve using substitution:

y = 2 arctan (x + c_{1}) + 2 πn - x

y = 2 arctan (x + c_{1}) + 2 πn - x

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