What is the solution for y^'=(x+y+4)^2 ?

The solution for y^'=(x+y+4)^2 is y=tan(x+c_{1})-x-4

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y^'=(x+y+4)^2

y^{^{'}} = (x + y + 4)^{2}

y = tan (x + c_{1}) - x - 4

Solution steps

y^{'} (x) = (x + y + 4)^{2}

Solve using substitution:

y = tan (x + c_{1}) - x - 4

y = tan (x + c_{1}) - x - 4

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What is the solution for y^'=(x+y+4)^2 ?
The solution for y^'=(x+y+4)^2 is y=tan(x+c_{1})-x-4