What is the solution for (dy)/(dx)tan(y)= 1/x ?

The solution for (dy)/(dx)tan(y)= 1/x is y=arccos((c_{1})/x)+2pin,y=-arccos((c_{1})/x)+2pin

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(dy)/(dx)tan(y)= 1/x

\frac{d y}{d x} tan (y) = \frac{1}{x}

y = arccos (\frac{c _{1}}{x}) + 2 πn, y = - arccos (\frac{c _{1}}{x}) + 2 πn

Solution steps

\frac{d y}{d x} tan (y) = \frac{1}{x}

Solve separable ODE:

y = arccos (\frac{c _{1}}{x}) + 2 πn, y = - arccos (\frac{c _{1}}{x}) + 2 πn

y = arccos (\frac{c _{1}}{x}) + 2 πn, y = - arccos (\frac{c _{1}}{x}) + 2 πn

x \to 1 lim (18 x^{100} - 5)

\int_{1}^{3} (3 x^{2} + 4) d x

\int \frac{x ^{3}}{x ^{2} - 3} d x

\frac{d}{d x} (80 e^{- \frac{x}{10}} - 40 x)

\frac{d y}{d x} = \frac{- 0.1}{cos ( y )}, y (0) = \frac{π}{2}

What is the solution for (dy)/(dx)tan(y)= 1/x ?
The solution for (dy)/(dx)tan(y)= 1/x is y=arccos((c_{1})/x)+2pin,y=-arccos((c_{1})/x)+2pin