What is the solution for (dy)/(dt)=3y+y^3 ?

The solution for (dy)/(dt)=3y+y^3 is y=sqrt(3/(e^{-6t-c_{1))-1}},y=-sqrt(3/(e^{-6t-c_{1))-1}}

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(dy)/(dt)=3y+y^3

\frac{d y}{d t} = 3 y + y^{3}

y = \frac{3}{e ^{- 6 t - c_{1}} - 1}, y = - \frac{3}{e ^{- 6 t - c_{1}} - 1}

Solution steps

\frac{d y}{d t} = 3 y + y^{3}

Solve Bernoulli ODE:

y = \frac{3}{e ^{- 6 t - c_{1}} - 1}, y = - \frac{3}{e ^{- 6 t - c_{1}} - 1}

y = \frac{3}{e ^{- 6 t - c_{1}} - 1}, y = - \frac{3}{e ^{- 6 t - c_{1}} - 1}

2 y^{^{''}} + 18 y = 5 cos (3 t)

\frac{d}{d x} (\frac{1}{( - x ^{2} + x - 2 ) ^{5}})

x^{2} y^{^{''}} + 5 x y^{^{'}} + 4 y = 0

\frac{d y}{d x} = \frac{1}{x ^{2}} - \frac{2 y}{x}

\frac{x ^{3}}{( x + 1 ) ^{2}}

What is the solution for (dy)/(dt)=3y+y^3 ?
The solution for (dy)/(dt)=3y+y^3 is y=sqrt(3/(e^{-6t-c_{1))-1}},y=-sqrt(3/(e^{-6t-c_{1))-1}}