What is the solution for y^'=(x+y+3)^2 ?

The solution for y^'=(x+y+3)^2 is y=tan(x+c_{1})-x-3

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y^'=(x+y+3)^2

y^{^{'}} = (x + y + 3)^{2}

y = tan (x + c_{1}) - x - 3

Solution steps

y^{'} (x) = (x + y + 3)^{2}

Solve using substitution:

y = tan (x + c_{1}) - x - 3

y = tan (x + c_{1}) - x - 3

\int \frac{1}{x - 3} d x

y = 12 - x, y = x, y = 1

\frac{\partial}{\partial x} (ln (x + 2 y))

e^{t} cos (6 t)

\int_{- 3}^{1} 2 e^{x} d x

What is the solution for y^'=(x+y+3)^2 ?
The solution for y^'=(x+y+3)^2 is y=tan(x+c_{1})-x-3