What is the solution for y^'=2y(1-3y) ?

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

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y^'=2y(1-3y)

y^{^{'}} = 2 y (1 - 3 y)

y = - \frac{e ^{2 t + c_{1}}}{1 - 3 e ^{2 t + c_{1}}}

Solution steps

y^{'} (t) = 2 y (1 - 3 y)

Solve separable ODE:

y = - \frac{e ^{2 t + c_{1}}}{1 - 3 e ^{2 t + c_{1}}}

y = - \frac{e ^{2 t + c_{1}}}{1 - 3 e ^{2 t + c_{1}}}

t y^{^{'}} - y = 0

ln (x), y = 2, 1, e^{2}

\frac{d}{d x} (3 4 - x)

\frac{d r}{d t} = - 4 r t

6 π r^{2}

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