What is the solution for y^'-(y-1)^2=0 ?

The solution for y^'-(y-1)^2=0 is y=-1/(t+c_{1)}+t/(t+c_{1)}+(c_{1})/(t+c_{1)}

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y^'-(y-1)^2=0

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

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

Solution steps

y^{'} (t) - (y - 1)^{2} = 0

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}}

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What is the solution for y^'-(y-1)^2=0 ?
The solution for y^'-(y-1)^2=0 is y=-1/(t+c_{1)}+t/(t+c_{1)}+(c_{1})/(t+c_{1)}