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

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

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y^'=cos(x-y)

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

y = - 2 arccot (- x - c_{1}) - 2 πn + x

Solution steps

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

Solve using substitution:

y = - 2 arccot (- x - c_{1}) - 2 πn + x

y = - 2 arccot (- 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=-2arccot(-x-c_{1})-2pin+x