What is the solution for y'=sin^2(x-y) ?

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

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y'=sin^2(x-y)

y^{^{'}} = sin^{2} (x - y)

y = - arctan (x + c_{1}) - πn + x

Solution steps

y^{'} (x) = sin^{2} (x - y)

Solve using substitution:

y = - arctan (x + c_{1}) - πn + x

y = - arctan (x + c_{1}) - πn + x

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\int \frac{( x + 1 )}{( x ^{2} + 2 x + 2 ) ^{3}} d x

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