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

\frac{d}{d x} (\frac{3 x - 4}{12 x ^{2}})

F (x) = x + 1

y^{^{'}} = \frac{( x ^{2} - y ^{2} )}{x y}, y (2) = 2

cos^{4} (x)

f (x) = x^{3} - 4 x, at x = 2

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