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

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

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

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

y = 2 tan (2 x + c_{1}) - 4 x

Solution steps

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

Solve using substitution:

y = 2 tan (2 x + c_{1}) - 4 x

y = 2 tan (2 x + c_{1}) - 4 x

x \to 1 + lim (\frac{x ^{2}}{x ^{2} - 1})

\frac{d y}{d x} = 3 e^{2 x} y^{2}

ln (x + 2)

\int x^{3} 5^{- x} d x

\int (x^{2} + 2 x - 1) cos (3 x) d x

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