What is the solution for y^'sec(t)=-(y^3)/2 ?

The solution for y^'sec(t)=-(y^3)/2 is y=sqrt(-1/(-sin(t)+c_{1))},y=-sqrt(-1/(-sin(t)+c_{1))}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'sec(t)=-(y^3)/2

y^{^{'}} sec (t) = - \frac{y ^{3}}{2}

y = - \frac{1}{- sin ( t ) + c _{1}}, y = - - \frac{1}{- sin ( t ) + c _{1}}

Solution steps

y^{'} (t) sec (t) = - \frac{y ^{3}}{2}

Solve separable ODE:

y = - \frac{1}{- sin ( t ) + c _{1}}, y = - - \frac{1}{- sin ( t ) + c _{1}}

y = - \frac{1}{- sin ( t ) + c _{1}}, y = - - \frac{1}{- sin ( t ) + c _{1}}

What is the solution for y^'sec(t)=-(y^3)/2 ?
The solution for y^'sec(t)=-(y^3)/2 is y=sqrt(-1/(-sin(t)+c_{1))},y=-sqrt(-1/(-sin(t)+c_{1))}