What is the solution for y^'=(1-y)^2 ?

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

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y^'=(1-y)^2

y^{^{'}} = (1 - y)^{2}

y = - \frac{1}{t + c _{1}} + \frac{t}{t + c _{1}} + \frac{c _{1}}{t + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{t + c _{1}} + \frac{t}{t + c _{1}} + \frac{c _{1}}{t + c _{1}}

y = - \frac{1}{t + c _{1}} + \frac{t}{t + c _{1}} + \frac{c _{1}}{t + c _{1}}

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

\frac{d}{df} (H (f)) = 3 u (f)^{2} f

x^{^{'}} = 0.0175 - 0.3821 x

2 \frac{d x}{d t} + \frac{d x}{d t} - 2 x = 1

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

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