What is the solution for y^'=y-y^3,y(0)=2 ?

The solution for y^'=y-y^3,y(0)=2 is y=(2e^t)/(sqrt(-3+4e^{2t))}

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y^'=y-y^3,y(0)=2

y^{^{'}} = y - y^{3}, y (0) = 2

y = \frac{2 e ^{t}}{- 3 + 4 e ^{2 t}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 e ^{t}}{- 3 + 4 e ^{2 t}}

y = \frac{2 e ^{t}}{- 3 + 4 e ^{2 t}}

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

((\frac{1}{y}) - (\frac{x}{y})) d x + (x^{2} \cdot cos (y)) d y = 0, y (1) = π

\frac{d}{d x} (ln (y)) = ln ((x^{2} - 9)^{\frac{1}{3}})

- x y^{^{'}} + y = y^{- 1}

y^{^{'}} - 2 y = 0, y (x, 1) = x^{2}

What is the solution for y^'=y-y^3,y(0)=2 ?
The solution for y^'=y-y^3,y(0)=2 is y=(2e^t)/(sqrt(-3+4e^{2t))}