What is the solution for (dy)/(dx)=-1+(2x+y)^2 ?

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

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(dy)/(dx)=-1+(2x+y)^2

\frac{d y}{d x} = - 1 + (2 x + y)^{2}

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

Solution steps

\frac{d y}{d x} = - 1 + (2 x + y)^{2}

Solve using substitution:

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

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

\frac{\partial}{\partial x} (4 x e^{x y})

e^{c o s (x)} + cos (e^{x})

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

\frac{d}{d x} (\frac{x ^{3} + 5}{x})

\frac{d}{d x} (, x e^{y} + y^{2} x - x^{3}, at (1, 0))

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