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

The solution for y^'=-3xy^3 is y=sqrt(-1/(-3x^2+c_{1))},y=-sqrt(-1/(-3x^2+c_{1))}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-3xy^3

y^{^{'}} = - 3 x y^{3}

y = - \frac{1}{- 3 x ^{2} + c _{1}}, y = - - \frac{1}{- 3 x ^{2} + c _{1}}

Solution steps

y^{'} (x) = - 3 x y^{3}

Solve separable ODE:

y = - \frac{1}{- 3 x ^{2} + c _{1}}, y = - - \frac{1}{- 3 x ^{2} + c _{1}}

y = - \frac{1}{- 3 x ^{2} + c _{1}}, y = - - \frac{1}{- 3 x ^{2} + c _{1}}

\int (6 x y) d x

e^{3 x}, 2

\frac{d}{d x} (x^{p (x)})

w

\int x^{3} x d x

What is the solution for y^'=-3xy^3 ?
The solution for y^'=-3xy^3 is y=sqrt(-1/(-3x^2+c_{1))},y=-sqrt(-1/(-3x^2+c_{1))}