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

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

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(dy)/(dx)=(y-x+1)^2+1

\frac{d y}{d x} = (y - x + 1)^{2} + 1

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

Solution steps

\frac{d y}{d x} = (y - x + 1)^{2} + 1

Solve using substitution:

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

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

x \to 0 lim (\frac{0}{x ^{3}})

y^{^{'}} = 3 - \frac{y}{1000}

y = 49.99 sin (\frac{x}{5.0073}) + 49.99

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

x \to 0 lim (\frac{1}{x ^{2} + x} - \frac{1}{x})

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