What is the solution for y^3-y^'+0.5y=0,y(0)= 1/3 ?

The solution for y^3-y^'+0.5y=0,y(0)= 1/3 is y=sqrt((0.5e^t)/(5.5-e^t))

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y^3-y^'+0.5y=0,y(0)= 1/3

y^{3} - y^{^{'}} + 0.5 y = 0, y (0) = \frac{1}{3}

y = \frac{0.5 e ^{t}}{5.5 - e ^{t}}

Solution steps

y^{3} - y^{'} (t) + 0.5 y = 0

Solve separable ODE:

y = \frac{0.5 e ^{t}}{5.5 - e ^{t}}

y = \frac{0.5 e ^{t}}{5.5 - e ^{t}}

d v = \frac{x}{4 + x ^{2}} d x

x^{^{'}} = (1 - x) (1 - cos (π) x)

y^{^{'}} = - \frac{0.0491}{c} \cdot 19.62 (y + 1)

(y)^{6} (\frac{\frac{d y}{d x} - 5}{3}) = tan (3 x) - \frac{5}{3} (y)^{6}

\frac{d p}{d t} = 3 p - p^{2}

What is the solution for y^3-y^'+0.5y=0,y(0)= 1/3 ?
The solution for y^3-y^'+0.5y=0,y(0)= 1/3 is y=sqrt((0.5e^t)/(5.5-e^t))