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

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

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

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

y = tan (2 x + c_{1}) - 2 x

Solution steps

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

Solve using substitution:

y = tan (2 x + c_{1}) - 2 x

y = tan (2 x + c_{1}) - 2 x

\frac{d}{d x} (13 x e^{- x})

\frac{\partial}{\partial v} (\frac{1}{4} (u - v))

f (x) = (x - 1)^{2}

n = 1 \sum \infty e^{\frac{1}{n}}

x \to 0 lim (1 + x)

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