What is the solution for xy^'=ytan(ln(y)) ?

The solution for xy^'=ytan(ln(y)) is y=e^{arccsc((c_{1})/x)+2pin},y=e^{pi+arccsc(-(c_{1})/x)+2pin}

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xy^'=ytan(ln(y))

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d t} = - \frac{t}{2 y}

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

\frac{d y}{d x} = \frac{sin ( 3 x )}{cos ( 4 y )}

y^{^{'}} = \frac{x ^{3}}{y}

\frac{d y}{d x} = e^{- y} (2 x - 4)

What is the solution for xy^'=ytan(ln(y)) ?
The solution for xy^'=ytan(ln(y)) is y=e^{arccsc((c_{1})/x)+2pin},y=e^{pi+arccsc(-(c_{1})/x)+2pin}