What is the solution for y^'+3xy^3=0 ?

The solution for y^'+3xy^3=0 is y=sqrt(-1/(-3x^2+c_{1))},y=-sqrt(-1/(-3x^2+c_{1))}

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y^'+3xy^3=0

y^{^{'}} + 3 x y^{3} = 0

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

Solution steps

y^{'} (x) + 3 x y^{3} = 0

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}}

f (x) = \frac{x + 3}{x - 3}

\int_{- 1}^{2} 3 x 4 x^{2} - 2 d x

x \to 0 - lim (\frac{1}{e ^{x} - 1})

\int 0.815 x e^{- 0.3 x} d x

\frac{d}{d x} (x^{2} (x^{2} + 5)^{2})

What is the solution for y^'+3xy^3=0 ?
The solution for y^'+3xy^3=0 is y=sqrt(-1/(-3x^2+c_{1))},y=-sqrt(-1/(-3x^2+c_{1))}