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

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

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y^'=(y-1)^2

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

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

Solution steps

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

\frac{d}{d x} (- \frac{4 x}{( x ^{2} - 4 ) ^{2}})

\frac{d}{d x} ((\frac{x ^{2} + 6}{x ^{2} - 6})^{3})

\frac{A}{y ^{7}} + B e^{y}

n = 1 \sum \infty \frac{3 - 2 n}{4 n}

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