What is the solution for y^'=4y^3 ?

The solution for y^'=4y^3 is y=sqrt(-1/(2(4t+c_{1)))},y=-sqrt(-1/(2(4t+c_{1)))}

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y^'=4y^3

y^{^{'}} = 4 y^{3}

y = - \frac{1}{2 ( 4 t + c _{1} )}, y = - - \frac{1}{2 ( 4 t + c _{1} )}

Solution steps

y^{'} (t) = 4 y^{3}

Solve separable ODE:

y = - \frac{1}{2 ( 4 t + c _{1} )}, y = - - \frac{1}{2 ( 4 t + c _{1} )}

y = - \frac{1}{2 ( 4 t + c _{1} )}, y = - - \frac{1}{2 ( 4 t + c _{1} )}

\frac{d y}{d x} = (x^{5} (y - 6)), y (0) = 8

y x^{2} d x - 3 y d x = d y

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}, y (0) = 4

(x^{2} + 1) (tan (y)) y^{^{'}} = x

y^{^{'}} = (1 + 4 x) y, y (0) = 1

What is the solution for y^'=4y^3 ?
The solution for y^'=4y^3 is y=sqrt(-1/(2(4t+c_{1)))},y=-sqrt(-1/(2(4t+c_{1)))}