What is the solution for (dy)/(dx)=1-cos(2x-y+3) ?

The solution for (dy)/(dx)=1-cos(2x-y+3) is y=-2arctan(x+c_{1})-2pin+2x+3

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(dy)/(dx)=1-cos(2x-y+3)

\frac{d y}{d x} = 1 - cos (2 x - y + 3)

y = - 2 arctan (x + c_{1}) - 2 πn + 2 x + 3

Solution steps

\frac{d y}{d x} = 1 - cos (2 x - y + 3)

Solve using substitution:

y = - 2 arctan (x + c_{1}) - 2 πn + 2 x + 3

y = - 2 arctan (x + c_{1}) - 2 πn + 2 x + 3

f (x) = \frac{8}{x ^{3}}

\frac{5}{s ^{2} ( s + 2 )}

\int_{0}^{1} x cos (\frac{nπ x}{2}) d x

ln (\frac{1}{2 x + 3})

\frac{1}{( 5 \cdot s + 1 )}

What is the solution for (dy)/(dx)=1-cos(2x-y+3) ?
The solution for (dy)/(dx)=1-cos(2x-y+3) is y=-2arctan(x+c_{1})-2pin+2x+3