What is the solution for y^'=(y-1)^2,y(0)=1.01 ?

The solution for y^'=(y-1)^2,y(0)=1.01 is y=-t/(-t+100)+(101)/(-t+100)

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y^'=(y-1)^2,y(0)=1.01

y^{^{'}} = (y - 1)^{2}, y (0) = 1.01

y = - \frac{t}{- t + 100} + \frac{101}{- t + 100}

Solution steps

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

Solve separable ODE:

y = - \frac{t}{- t + 100} + \frac{101}{- t + 100}

y = - \frac{t}{- t + 100} + \frac{101}{- t + 100}

(x + 4) (2 y + 6) d y = x y^{2} d x

y^{3} (x^{4} + 1) \frac{d y}{d x} - x^{3} (y^{4} + 1) = 0

y^{^{'}} = y \cdot (1 + e^{x})

y^{^{'}} + y^{2} = 3

\frac{d y}{d x} = \frac{x ^{2}}{y}, y (0) = 2

What is the solution for y^'=(y-1)^2,y(0)=1.01 ?
The solution for y^'=(y-1)^2,y(0)=1.01 is y=-t/(-t+100)+(101)/(-t+100)