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

The solution for y^'=(x+y-1)^2-1 is y=-1/(x+c_{1)}-x+1

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y^'=(x+y-1)^2-1

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

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

Solution steps

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

Solve using substitution:

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

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

\frac{\partial}{\partial x} (\frac{t}{x - 1})

\int 2 x - 2 y d x

\frac{1}{1 - z ^{2}}, 0

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0, y (0) = 3

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

What is the solution for y^'=(x+y-1)^2-1 ?
The solution for y^'=(x+y-1)^2-1 is y=-1/(x+c_{1)}-x+1