y^'=(2cos(y))/(sin(y))

Lösung

y^{^{'}} = \frac{2 cos ( y )}{sin ( y )}

y = arccos (e^{- 2 t - c_{1}}) + 2 πn, y = - arccos (e^{- 2 t - c_{1}}) + 2 πn

Schritte zur Lösung

y^{'} (t) = \frac{2 cos ( y )}{sin ( y )}

Solve separable ODE:

y = arccos (e^{- 2 t - c_{1}}) + 2 πn, y = - arccos (e^{- 2 t - c_{1}}) + 2 πn

y = arccos (e^{- 2 t - c_{1}}) + 2 πn, y = - arccos (e^{- 2 t - c_{1}}) + 2 πn

x \to 0 - lim (\frac{x - 1}{x})

\int \frac{cos ( 2 x )}{1 + sin ^{2} ( 2 x )} d x

x \to 0 lim (\frac{cos ( x ^{2} )}{x ^{2}})

y^{^{'}} = 6 + t - y

d er i v a t i v e y = x \cdot e^{x}