What is the solution for y^'+1/2 y=y^3 ?

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

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y^'+1/2 y=y^3

y^{^{'}} + \frac{1}{2} y = y^{3}

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

Solution steps

y^{'} (t) + \frac{1}{2} y = y^{3}

Solve separable ODE:

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

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

\frac{1}{2} e^{- t} cos (t) + \frac{1}{2} e^{- t} sin (t) - \frac{1}{2} e^{- 2 t}

\frac{d}{d x} (cos^{2} (ln (x)))

y^{^{''}} + 8 y^{^{'}} + 6 y = 0

\frac{4 s + 3}{s ( s ^{2} + 6 s + 34 )}

f^{^{'}} (x) = x + 1

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