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

The solution for y^'=(x+y+1)^2,y(0)=0 is y=tan(x+pi/4+pin)-x-1

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y^'=(x+y+1)^2,y(0)=0

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

y = tan (x + \frac{π}{4} + πn) - x - 1

Solution steps

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

Solve using substitution:

y = tan (x + \frac{π}{4} + πn) - x - 1

y = tan (x + \frac{π}{4} + πn) - x - 1

\frac{1}{x + 2}, 1

\frac{d y}{d x} = \frac{( 1 + x )}{x y ^{10}}

\frac{\partial}{\partial y} (x e^{- 2 x y})

\frac{( 4 x ^{2} + 2 x + 2 )}{x}

\int x 2 x - 1 d x

What is the solution for y^'=(x+y+1)^2,y(0)=0 ?
The solution for y^'=(x+y+1)^2,y(0)=0 is y=tan(x+pi/4+pin)-x-1