Solve the differential equation $\frac{dy}{dx}=\tan\left(4x\right)$

$\left(1+x^4\right)\cdot dy+x\cdot\left(1+4y^2\right)\cdot dx=0$

$\frac{dy}{dx}=\frac{2x}{3y^2}$

$\frac{dx}{dy}=y+xy$

$dx+e^{3x}dy=0$

$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$

$\frac{dy}{dx}=e^{3x+2y}$

$y'=x+8$