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

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

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

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

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

Solution steps

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

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}}

x e^{y - x^{2}} d x - d y = 0

\frac{\partial}{\partial x} (4 y + 9 x^{2})

\frac{d}{d x} (x^{(\frac{1}{2})})

ma c l a u r in \frac{1}{( 1 - x ^{2} )}

x \to 0 lim (\frac{( ln ( x ) ) ^{2}}{x})

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