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

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

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y^'=sin(x-y)

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

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

Solution steps

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

Solve using substitution:

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

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

\frac{d}{d x} (c x^{- 6})

\int (x + 1) (2 x - 1) d x

\frac{1}{x + 6}

\int \frac{1}{x ( ln ( x ) ) ^{15}} d x

\frac{d}{d x} ((2 x^{3} - 7)^{8})

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