What is the solution for y^'=(ty^2-1)y ?

The solution for y^'=(ty^2-1)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^2-1)y

y^{^{'}} = (t y^{2} - 1) 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^{2} - 1) 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}}

- 2 x^{2}, at 1

\frac{d}{d x} (e^{x^{3} + 5 x})

\int \frac{( sin ^{2} ( x ) )}{cos ( x )} d x

\int csc (x) cot^{2} (x) d x

\frac{d}{d x} (6^{2 x})

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