What is the solution for y^'=-2xtan(y),y(0)= pi/2 ?

The solution for y^'=-2xtan(y),y(0)= pi/2 is y=arcsin(e^{-x^2})+2pin,y=pi+arcsin(-e^{-x^2})+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-2xtan(y),y(0)= pi/2

y^{^{'}} = - 2 x tan (y), y (0) = \frac{π}{2}

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

Solution steps

y^{'} (x) = - 2 x tan (y)

Solve separable ODE:

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

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

\int \frac{( x + 3 ) ^{2}}{x + 2} d x

\frac{d}{d x} (tan (x) ln (x))

x \to 3 lim (4 x^{3} - 3 x^{2} + x - 5)

\frac{\partial}{\partial y} (4 x y - 2 x^{2} - y^{4})

- 8 sin (x)

What is the solution for y^'=-2xtan(y),y(0)= pi/2 ?
The solution for y^'=-2xtan(y),y(0)= pi/2 is y=arcsin(e^{-x^2})+2pin,y=pi+arcsin(-e^{-x^2})+2pin