What is the solution for y^'=tan(x+y) ?

The solution for y^'=tan(x+y) is y=arcsin(e^{x+c_{1}})+2pin-x,y=pi+arcsin(-e^{x+c_{1}})+2pin-x

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y^'=tan(x+y)

y^{^{'}} = tan (x + y)

y = arcsin (e^{x + c_{1}}) + 2 πn - x, y = π + arcsin (- e^{x + c_{1}}) + 2 πn - x

Solution steps

y^{'} (x) = tan (x + y)

Solve using substitution:

y = arcsin (e^{x + c_{1}}) + 2 πn - x, y = π + arcsin (- e^{x + c_{1}}) + 2 πn - x

y = arcsin (e^{x + c_{1}}) + 2 πn - x, y = π + arcsin (- e^{x + c_{1}}) + 2 πn - x

\int θ^{4} + 4 θ^{2} d θ

\frac{d y}{d x} = 1 + y^{2}

3 x \frac{d y}{d x} - 2 y = \frac{x ^{3}}{y ^{2}}

\frac{d}{d x} (e^{- \frac{x ^{2}}{2} - \frac{y ^{2}}{2}})

e^{x}

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