What is the solution for y^'=-ty^3 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-ty^3

y^{^{'}} = - t y^{3}

y = - \frac{1}{- t ^{2} + c _{1}}, y = - - \frac{1}{- t ^{2} + c _{1}}

Solution steps

y^{'} (t) = - t y^{3}

Solve separable ODE:

y = - \frac{1}{- t ^{2} + c _{1}}, y = - - \frac{1}{- t ^{2} + c _{1}}

y = - \frac{1}{- t ^{2} + c _{1}}, y = - - \frac{1}{- t ^{2} + c _{1}}

\frac{d y}{d x} y = (3 + x)^{\frac{5}{x}}

f^{^{'}} (x) 4 f (x) = 2 x - 6

y y^{^{'}} = (1 - y^{2}) sin (x)

H (x) x d x - x (- ln (H (x))) (H (x) d x + x d H (x)) = 0

\frac{d y}{d x} y = (csc (x) + cot (x))^{- 1}

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