Differential Equation: y' = 2(3x + y)^2 - 1

$y' = 2(3x + y)^2 - 1$

$\dfrac{dz}{dx} - 3 = 2z^2 - 1$

$\dfrac{dz}{dx} = 2z^2 + 2$

$\dfrac{dz}{dx} = 2(z^2 + 1)$

$\dfrac{dz}{z^2 + 1} = 2\,dx$

$\arctan z = 2x + c$

$\arctan (3x + y) = 2x + c$

You can now apply the initial condition to solve for the constant of integration c.