What is the solution for (dy)/(dx)=(y^3)/(3x+1),y(0)=-1 ?

The solution for (dy)/(dx)=(y^3)/(3x+1),y(0)=-1 is y=-sqrt(-3/(2ln(1+3x)-3))

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(dy)/(dx)=(y^3)/(3x+1),y(0)=-1

\frac{d y}{d x} = \frac{y ^{3}}{3 x + 1}, y (0) = - 1

y = - - \frac{3}{2 ln ( 1 + 3 x ) - 3}

Solution steps

\frac{d y}{d x} = \frac{y ^{3}}{3 x + 1}

Solve separable ODE:

y = - - \frac{3}{2 ln ( 1 + 3 x ) - 3}

y = - - \frac{3}{2 ln ( 1 + 3 x ) - 3}

\frac{2}{t ^{3}}

y^{^{'}} + 3 y = x e^{- 2 x}

\frac{\partial}{\partial y} (- sin (x) sin (y))

n = 0 \sum \infty \frac{( x ^{n} )}{n}

What is the solution for (dy)/(dx)=(y^3)/(3x+1),y(0)=-1 ?
The solution for (dy)/(dx)=(y^3)/(3x+1),y(0)=-1 is y=-sqrt(-3/(2ln(1+3x)-3))