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

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

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y^'=ty^3-y

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

x \to - 2 lim ((3 x^{2}) - 6 x + 2)

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

y = x, y = \frac{1}{3} x

y = x - 8

n = 4 \sum \infty \frac{n}{n - 3}

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