What is the solution for y^'=-2-y+y^2 ?

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

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y^'=-2-y+y^2

y^{^{'}} = - 2 - y + y^{2}

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

Solution steps

y^{'} (t) = - 2 - y + y^{2}

Solve for

t, y^{'} (t) = \frac{d y}{d t} = \frac{1}{\frac{d t}{d y}} = \frac{1}{t ^{^{'}}}

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

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{- 2 - y + y ^{2}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{- 2 - y + y ^{2}} d y

\int \frac{1}{- 2 - y + y ^{2}} d y = - \frac{1}{3} (ln (\frac{2}{3} + \frac{2}{3} y) - ln (- \frac{4}{3} + \frac{2}{3} y)) + c_{1}

t = - \frac{1}{3} (ln (\frac{2}{3} + \frac{2}{3} y) - ln (- \frac{4}{3} + \frac{2}{3} y)) + c_{1}

Simplify

t = - \frac{1}{3} ln (\frac{1 + y}{- 2 + y}) + c_{1}

Isolate

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

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

\frac{d y}{d t} = \frac{t ^{2}}{y}

\frac{d x}{d t} + 20 x = 24

y^{^{'}} = cos (x) (3 y - 1)

ln (\frac{d y}{d x}) = a x + b y

2 y y^{^{'}} = 15 t^{2}, y (0) = 4

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