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

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

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

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

y = tan (x + c_{1}) - x - 1

Solution steps

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

Solve using substitution:

y = tan (x + c_{1}) - x - 1

y = tan (x + c_{1}) - x - 1

\int sec^{2} (θ) tan^{3} (θ) d θ

\frac{d}{d x} (x^{(\frac{1}{5} l n (x))})

sin (\frac{1}{x})

\int x + 6 d x

y = x, y = 2 x, y = - x + 4

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