What is the solution for (dy)/(dt)=t^2y^3,y(0)=-1 ?

The solution for (dy)/(dt)=t^2y^3,y(0)=-1 is y=-sqrt(-3/(2t^3-3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dt)=t^2y^3,y(0)=-1

\frac{d y}{d t} = t^{2} y^{3}, y (0) = - 1

y = - - \frac{3}{2 t ^{3} - 3}

Solution steps

\frac{d y}{d t} = t^{2} y^{3}

Solve separable ODE:

y = - - \frac{3}{2 t ^{3} - 3}

y = - - \frac{3}{2 t ^{3} - 3}

\frac{x + 1}{2 x ^{2} + 1}

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

\frac{\partial}{\partial x} (arcsec (x + yz))

f (x) = \frac{x ^{2}}{2} + \frac{2}{x ^{2}}

\int (tan^{5} (x)) d x

What is the solution for (dy)/(dt)=t^2y^3,y(0)=-1 ?
The solution for (dy)/(dt)=t^2y^3,y(0)=-1 is y=-sqrt(-3/(2t^3-3))